Question 13 Marks
If the ratio of the height of a tower and the length of its shadow is $\sqrt{3}:1,$ what is the angle of elevation of the Sun$?$
AnswerLet $C$ the angle of elevation of sun is $\theta.$
Given that: Height of tower is $\sqrt{3}$ meters and length of shadow is $1.$
Here we have to find angle of elevation of sun.
In a triangle $ABC,$
$\Rightarrow\ \tan\theta=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan\theta=\frac{\sqrt{3}}{1}$
$\Rightarrow\ \tan\theta=\sqrt{3}$ $[\because\ \tan60^\circ=\sqrt{3}]$
$\Rightarrow\ \theta=60^\circ$
Hence the angle of elevation of sun is $60^\circ .$
View full question & answer→Question 23 Marks
As observed from the top of a $75m$ tall light house, the angle of depression of two ships are $30^\circ $ and $45^\circ .$ If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
AnswerLet, distance between two ships, $CD = xm$
distance between light house and ship $BD = h$
Given,
angle of depression $\angle\text{C}=30^\circ,\ \angle\text{D}=45^\circ$
length of light house $AB = 75m$
In $\triangle\text{ABD},$
$\tan\text{D}=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\ \tan45^\circ=\frac{75}{\text{h}}$
$\Rightarrow\ 1=\frac{75}{\text{h}}$
$\Rightarrow\ \text{h}=75\text{m}$
In $\triangle\text{ACB},$
$\tan\text{C}=\frac{\text{AB}}{\text{CB}}$
$\Rightarrow\ \tan30^\circ=\frac{75}{\text{CD}+\text{DB}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{75}{\text{x}+\text{h}}$
$\Rightarrow\ \text{x}+\text{h}=75\sqrt{3}$
$\Rightarrow\ \text{x}=75\sqrt{3}-\text{h}$
$\Rightarrow\ \text{x}=75\sqrt{3}-75$
$\Rightarrow\ \text{x}=75(\sqrt{3}-1)$
Hence, distance between two ship $75(\sqrt{3}-1)\text{m}.$
View full question & answer→Question 33 Marks
From a point on the ground, $20m$ away from the foot of a vertical tower, the angle elevation of the top of the tower is $60^\circ ,$ What is the height of the tower$?$
AnswerGiven,
angle of elevation, $\angle\text{C}=60^\circ$
distance between point and tower, $BC = 20\ m$
height of tower $AB = h\ m$
In $\triangle\text{ABC,}$
we have to find height of tower,
$\tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}}{20}\Rightarrow\ \sqrt{3}=\frac{\text{h}}{20}$
$\Rightarrow\ \text{h}=20\sqrt{3}\text{m}$
Hence, height of tower is $20\sqrt{3}\text{m.}$
View full question & answer→Question 43 Marks
The angles of elevation of the top of a rock from the top and foot of a $100m$ high tower are respectively $30^\circ $ and $45^\circ .$ Find the height of the rock.
AnswerLet $AB$ be the height of Rock which is $H\ m$. and makes an angle of elevations $45^\circ $ and $30^\circ $ respectively from the bottom and top of tower whose height is 100m.
Let $AE = h\ m, BC = x\ m$ and $CD = 100.$ $\angle\text{ACB}=45^\circ,\ \angle\text{ADE}=30^\circ$
We have to find the height of the rock.
We have the corresponding figure as,
So we use trigonometric ratios.
In $\triangle\text{ABC,}$
$\tan45^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ 1=\frac{100+\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=100+\text{h}$
Again in $\triangle\text{ADE,}$
$\tan30^\circ=\frac{\text{AE}}{\text{DE}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ 100+\text{h}=\sqrt{3}\text{h}$
$\Rightarrow h = 136.65$
$\Rightarrow H = 100 + 136.65$
$\Rightarrow H = 236.65$
Hence the height of rock is $236.65\ m$
View full question & answer→Question 53 Marks
A kite is flying at a height of $75$ metres from the ground level, attached to a string inclined at $60$ to the horizontal. Find the length of the string to the nearest metre.
AnswerLet, angle of elevation $=\angle\text{ACB}=60^\circ$
heigh of kite $= AB = 75\ m$
length of string $= AC = h$
In $\triangle\text{ABC,}$ By using trigonometric ratio
$\sin60^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\frac{75}{\text{h}}$
$\Rightarrow\ \text{h}=\frac{150}{\sqrt{3}}$
$\Rightarrow\ \text{h}=86.6\text{m}$
Hence, length of string is $87\ m.$
View full question & answer→Question 63 Marks
Two poles of equal heights are standing opposite to each other on either side of the road which is $80\ m$ wide. From a point between them on the road the angles of elevation of the top of the poles are $60^\circ $ and $30^\circ $ respectively. Find the height of the poles and the distance of the point from the poles.
AnswerLet, width of river, $BD = x + y$
Given,
angle of depression, $\angle\text{B}=30^\circ,\ \angle\text{D}=45^\circ$
height of bridge $AC = 30^\circ $
In $\triangle\text{ABC},$
$\tan\text{B}=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{30}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{30}{\text{x}}$
$\Rightarrow\ \text{x}=30\sqrt{3}$
In $\triangle\text{ACD},$
$\tan\text{D}=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\ \tan45^\circ=\frac{30}{\text{y}}$
$\Rightarrow\ 1=\frac{30}{\text{y}}$
$\Rightarrow\ \text{y}=30$
width of river, $\text{BD}=\text{x}+\text{y}$
$\Rightarrow\ \text{BD}=30\sqrt{3}+30$
$\Rightarrow\ \text{BD}=30(\sqrt{3}+1)$
Hence, width of river $30(\sqrt{3}+1)\text{m}.$
View full question & answer→Question 73 Marks
An electric pole is $10\ m$ high. A steel wire tied to top of the pole is affixed at a point on the ground to keep the pole up right. If the wire makes an angle of $45^\circ $ with the horizontal through the foot of the pole, find the length of the wire.
AnswerHeight of the electric pole $H = 10\ m = AB$ angle made by steel wire with ground (horizontal) $\theta=45^\circ$
Let length of rope wire $= l = AC$
If we represent above data is form of figure thin it forms a right triangle $ABC$
Here, $\sin\theta=\frac{\text{Opposite side}}{\text{Hypotenuse}}$
$\Rightarrow\ \sin45^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\ \frac{1}{\sqrt{2}}=\frac{10\text{m}}{\text{l}}$
$\Rightarrow\ \text{l}=10\sqrt{2}\text{m}$
$\therefore$ length of wire $\text{l}=10\sqrt{2}\text{m}$
View full question & answer→Question 83 Marks
A parachutist is descending vertically and makes angles of elevation of $45^\circ $ and $60^\circ $ at two observing points $100\ m$ apart from each other on the left side of himself. Find the maximum height from which he falls and the distance of the point where he falls on the ground form the just observation point.
AnswerLet $BC$ be the height h of the parachutist and makes an angle of elevations $45^\circ $ and $60^\circ $ respectively at two observing points $100$ apart from each other.
Let $AD = 100, CD = x, BC = h$ and $\angle\text{CAB}=45^\circ,\ \angle\text{CDB}=60^\circ$
So we use trigonometric ratios.
In triangle $BCD,$
$\tan60^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\sqrt{3}}$
Now in triangle $ABC,$
$\tan45^\circ=\frac{\text{h}}{\text{x}+100}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{x}+100}$
$\Rightarrow\ \text{x}+100=\text{h}$
$\Rightarrow\ \frac{\text{h}}{\sqrt{3}}+100=\text{h}$
$\Rightarrow\ \text{h}+100\sqrt{3}=\sqrt{3}\text{h}$
$\Rightarrow\ \text{h}=\frac{100\sqrt{3}}{\sqrt{3}-1}$
$\Rightarrow\ \text{h}=50(3+\sqrt{3})$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\sqrt{3}}$
$\text{x}=\frac{50(3+\sqrt{3})}{\sqrt{3}}$
$=50(1+\sqrt{3})$
Hence the maximum height is $50(3+\sqrt{3})\text{m}=236.6\text{m}$ and distance is $50(1+\sqrt{3})\text{m}=136.6\text{m}$
View full question & answer→Question 93 Marks
A tree breaks due to the storm and the broken part bends so that the top of the tree touches the ground making an angle of $30^\circ $ with the ground. The distance from the foot of the tree to the point where the top touches the ground is $10$ metres. Find the height of the tree.
AnswerLet $AB$ be the tree of height h. And the top of tree makes an angle $30^\circ $ with ground. The distance between foot of tree to the point where the top touches the ground is $10\ m.$
Let $BC = 10.$ And $\angle\text{ACB}=30^\circ$
Here we have to find height of tree.
Here we have the corresponding figure,
So we use trigonometric ratios.
In a triangle $ABC,$
$\Rightarrow\ \tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{10}$
$\Rightarrow\ \text{h}=\frac{10}{\sqrt{3}}$
Now in triangle $ABC$ we have
$\sin30^\circ=\frac{\text{h}}{\text{AC}}$
$\Rightarrow\ \frac{1}{2}=\frac{10}{\sqrt{3}\text{AC}}$
$\Rightarrow\ \text{AC}=\frac{20}{\sqrt{3}}$
So the length of the tree is,
$=\text{AB}+\text{AC}$
$=\text{h}+\text{AC}$
$=\frac{10}{\sqrt{3}}+\frac{20}{\sqrt{3}}$
$=10\sqrt{3}$
$=17.3$
Hence the height of tree is $17.3\ m.$
View full question & answer→Question 103 Marks
The tops of two towers of height $x$ and $y,$ standing on level ground, subtend angles of $30^\circ $ and $60^\circ $ respectively at the centre of the line joining their feet, then find $x : y.$
AnswerLet $AB$ and $CD$ be two towers which are apart from each other at a distance of $BD$ and $M$ is mid point of $BD.$
Angles of elevation are $30^\circ $ and $60^\circ .$
$\because\ \tan\theta=\frac{\text{P}}{\text{B}}$
$\therefore\ \tan30^\circ=\frac{\text{AB}}{\text{BM}}=\frac{\text{x}}{\text{BM}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{x}}{\text{BM}}$
$\Rightarrow\ \text{BM}=\text{x}\sqrt{3}$
Similarly, $\tan60^\circ=\frac{\text{CD}}{\text{MD}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{y}}{\text{MD}}\Rightarrow\ \text{MD}=\frac{\text{y}}{\sqrt{3}}$
But $BM = MD (\because M$ is mid point$)$
$\text{x}\sqrt{3}=\frac{\text{y}}{\sqrt{3}}\Rightarrow\ \frac{\text{x}}{\text{y}}=\frac{1}{\sqrt{3}\times\sqrt{3}}=\frac{1}{3}$
$\therefore x : y = 1 : 3$
View full question & answer→Question 113 Marks
A tower stand vertically on the ground. From a point on the ground $20m$ away from the foot of the tower, the angle of elevation of the top of the tower is $60^\circ .$ What is the height of the tower$?$
AnswerGiven,
Distance between point of observation and foot of tower $= 20m =BC$
Angle of elevation of top of tower $= 60^\circ = 0$
Height of tower $H = ? = AB$
Now from fig. $ABC$
$\triangle\text{ABC}$ is a right angle
$\frac{1}{\tan}=\frac{\text{Adjacent side}}{\text{Opposite side}}$
$\Rightarrow\ \tan\theta=\frac{\text{Opposite side (AB)}}{\text{Adjacent side (BC)}}$
i.e., $\tan60^\circ=\frac{\text{AB}}{20}$
$\Rightarrow\ \text{AB}=20\tan60^\circ$
$\Rightarrow\ \text{H}=20\sqrt{3}$
$=20\sqrt{3}$
$\therefore$ Height of tower $\text{H}=20\sqrt{3}\text{m}.$
View full question & answer→Question 123 Marks
On the same side of a tower, two objects are located. When observed from the top of the tower, their angles of depression are $45^\circ $ and $60^\circ $. If the height of the tower is $150\ m,$ find the distance between the objects.
AnswerHeight of tower, $H = AB = 150m$
Let $A$ and $B$ be two objects m the ground angle of depression of objects,
$\text{A}'[\angle\text{A}'\text{Ax}]=\beta=45^\circ=\angle\text{AA}'\text{B}[\text{Ax}][\text{A}'\text{B}]$
Angle of depression of objects $B’$
$\angle\text{xAB}'=\alpha=60^\circ=\angle\text{AB}'\text{B}[\text{Ax}][\text{A}'\text{B}]$
Let $A'B' = x, B'B = y$
In we figure the above data in figure, then it is as shown with $\angle\text{B}=90^\circ$
In any right angled triangle if one of the included angle is $\theta$ then,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan\alpha=\frac{\text{AB}}{\text{BB}'}$
$\Rightarrow\ \tan60^\circ=\frac{150}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{150}{\sqrt{3}}\ .....(1)$
$\tan\beta=\frac{\text{AB}}{\text{A}'\text{B}}$
$\Rightarrow\ \tan45^\circ=\frac{150}{\text{x}+\text{y}}$
$\Rightarrow\ \text{x}+\text{y}=150\ .......(2)$
$(1)$ and $(2)$
$\Rightarrow\ \text{x}+\frac{150}{\sqrt{3}}=150$
$\Rightarrow\ \text{x}+\frac{50\times3}{\sqrt{3}}=150$
$\Rightarrow\ \text{x}=150-50\sqrt{3}$
$\Rightarrow\ \text{x}=150-50(1.732)$
$\Rightarrow\ \text{x}=150-86.6=63.4\text{m}$
Distance between objects $A'B' = 63.4m$
View full question & answer→Question 133 Marks
A flag-staff stands on the top of a $5m$ high tower. From a point on the ground, the angle of elevation of the top of the flag-staff is $60^\circ $ and from the same point, the angle of elevation of the top of the tower is $45^\circ .$ Find the height of the flag-staff.
AnswerLet $BC$ be the tower height of $5m$ flag height is h m and an angle of elevation of top of tower is $45^\circ $ and an angle of elevation of the top of flag is $60^\circ .$
Let, $AC = h\ m$ and $BC = 5m$ and $\angle\text{ADB}=60^\circ,\ \angle\text{CDB}=45^\circ$
We have the corresponding angle as follows,
So we use trigonometric ratios.
In a triangle $\triangle\text{BCD},$
$\Rightarrow\ \tan45^\circ=\frac{\text{BC}}{\text{BD}}$
$\Rightarrow\ 1=\frac{5}{\text{x}}$
$\Rightarrow\ \text{x}=5$
Again in a triangle $ABD,$
$\Rightarrow\ \tan60^\circ=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow \sqrt{3}=\frac{5+\text{h}}{5}$
$\Rightarrow\ \text{h}=5(\sqrt{3}-1)$
$\Rightarrow\ \text{h}=3.66$
Hence the height of flag is $3.66m$
View full question & answer→Question 143 Marks
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff of height $5$ metres. At a point on the plane, the angles of elevation of the bottom and the top of the flag-staff are respectively $30^\circ $ and $60^\circ .$ Find the height of the tower.
AnswerLet $BC$ be the tower of height hm and $AB$ be the flag staff with distance 5m.Then angle of elevation from the top and bottom of flag staff are $60^\circ $ and $30^\circ $ respectively.
Let $CD = x$ and $\angle\text{ADC}=60^\circ,\ \angle\text{BDC}=30^\circ$
Here we have to find height h of tower.
So we use trigonometric ratios.
In a triangle $BCD,$
$\Rightarrow\ \tan\text{D}=\frac{\text{BC}}{\text{CD}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\sqrt{3}\text{h}$
Again in a triangle $ACD,$
$\Rightarrow\ \tan\text{D}=\frac{\text{AB}+\text{BC}}{\text{CD}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}+5}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}+5}{\text{x}}$
$\Rightarrow\ \sqrt{3}\text{x}=\text{h}+5$
$\Rightarrow\ \sqrt{3}\times\text{h}\sqrt{3}=\text{h}+5$
$\Rightarrow\ 3\text{h}=\text{h}+5$
$\Rightarrow\ 2\text{h}=5$
$\Rightarrow\ \text{h}=2.5$
Hence the height of tree is $2.5m.$
View full question & answer→Question 153 Marks
The shadow of a tower standing on a level ground is found to be $40m$ longer when Sun's altitude is $30^\circ $ than when it was $60^\circ .$ Find the height of the tower.
AnswerLet, height of tower $AB = h$
length of shadow at $30^\circ , CD = 40m$
length of shadow at $60^\circ , DB = xm$
angle of elevation $\angle\text{C}=30^\circ,\ \angle\text{D}=60^\circ$
In $\triangle\text{ABD},$
$\tan\text{D}=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\sqrt{3}}$
In $\triangle\text{ABC},$
$\tan\text{C}=\frac{\text{AB}}{\text{BC}}$ $[\because\ \text{BC}=\text{CD}+\text{DB}]$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{40+\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{40+\text{x}}$
$\Rightarrow\ \sqrt{3}\text{h}=40+\text{x}$
$\Rightarrow\ \sqrt{3}\text{h}=40+\frac{\text{h}}{\sqrt{3}}$
$\Rightarrow\ 3\text{h}=40\sqrt{3}+\text{h}$
$\Rightarrow\ 2\text{h}=40\sqrt{3}$
$\Rightarrow\ \text{h}=20\sqrt{3}$
Hence, height of tower $20\sqrt{3}\text{m.}$
View full question & answer→Question 163 Marks
A vertical tower stands on a horizontal plane and is surmounted by a vertical flag-staff. At a point on the plane $70$ metres away from the tower, an observer notices that the angles of elevation of the top and the bottom of the flagstaff are respectively $60^\circ $ and $45^\circ $. Find the height of the flag-staff and that of the tower.
AnswerLet $BC$ be the tower of height $x\ m$ and $AB$ be the flag staff of height $y, 70\ m$ away from the tower, makes an angle of elevation are $60^\circ$ and $45^\circ $ respectively from top and bottom of the flag staff.
Let $AB = y\ m, BC = x\ m$ and $CD = 70\ m.$
$\angle\text{ADC}=45^\circ$ and $\angle\text{ADC}=60^\circ$
So we use trigonometric ratios.
In a triangle $BCD,$
$\Rightarrow\ \tan\text{D}=\frac{\text{BC}}{\text{CD}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{x}}{70}$
$\Rightarrow\ 1=\frac{70}{\text{x}}$
$\Rightarrow\ \text{x}=70$
Again in a triangle $ADC,$
$\Rightarrow\ \tan\text{D}=\frac{\text{AB}+\text{BC}}{\text{CD}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{y}+70}{70}$
$\Rightarrow\ 70\sqrt{3}=70+\text{y}$
$\Rightarrow\ \text{y}=70(\sqrt{3}-1)$
$\Rightarrow\ \text{y}=51.24$
Hence the height of flag staff is $51.24\ m$ and height of tower is $70\ m.$
View full question & answer→Question 173 Marks
The horizontal distance between two trees of different heights is $60m.$ The angle of depression of the top of the first tree when seen from the top of the second tree is $45^\circ .$ If the height of the second tree is $80m,$ find the height of the first tree.
AnswerDistance between trees $= 60m [BC]$
Height of second tree $= 80m [CD]$
Let height of first tree $= 'h' m [AB]$
Angle of depression from second tree top from first tree top $\alpha=45^\circ$
The above information is represent in form of figure as shown.
In right triangle if one of the included angle is $0$ their,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
Draw $\text{CX}\bot\text{AB},$ $CX = BD = 60m$
$XB = CD = AB - AX$
$\tan\alpha=\frac{\text{AX}}{\text{CX}}$
$\tan45^\circ=\frac{\text{AX}}{60}\Rightarrow\ \text{AX}=60\text{m}$
$XB = CD = AB - AX$
$= 80 - 60$
$= 20m$
Height of second tree $= 80m$
Height of first tree $= 20m.$
View full question & answer→Question 183 Marks
The angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6m.
AnswerLet AB be tower of height h m and angle of elevation of the top of tower from two points are $\theta$ and $90^\circ-\theta$
Let, $AB = h\ m$ and $AC = 4\ m$ and $AD = 9$
The corresponding figure is as follows,
So we use trigonometric ratios.
In $\triangle\text{ABC},$
$\Rightarrow\ \tan\theta=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\ \tan\theta=\frac{\text{h}}{4}\ ...(1)$
Again in $\triangle\text{ABD},$
$\Rightarrow\ \tan(90-\theta)=\frac{\text{AB}}{\text{AD}}$
$\Rightarrow\ \cot\theta=\frac{\text{h}}{9}\ ...(2)$
Multiplying equation $(1)$ and $(2)$
$\Rightarrow\tan\theta\times\cot\theta$
$\Rightarrow \frac{\text{h}}{4}\times\frac{\text{h}}{9}$
$\Rightarrow\frac{\text{h}^2}{36}$
$\Rightarrow\ \text{h}^2=36\Rightarrow\sqrt{36}=6$
Hence the height of tower is $6m.$
View full question & answer→Question 193 Marks
A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle of $30^\circ $ with the ground. The distance between the foot of the tree to the point where the top touches the ground is $8\ m.$ Find the height of the tree.
AnswerLet initially tree height be $AB$
Let us assumed that the tree is broken at point $C$
Angle made by broken part $CB'$ with ground is $30^\circ=\theta$
Distance between foot of tree of point where it touches ground $B'A = 8m$
Height of tree $= h = AC +CB' = AC + CB$
The above information is represent in the form of figure as shown,
$\cos\theta=\frac{\text{Adjacent side}}{\text{Hypotenuse}}$
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\cos30^\circ=\frac{\text{AB}'}{\text{CB}'}$
$\frac{\sqrt{3}}{2}=\frac{\text{B}}{\text{CB}'}$
$\text{CB}'=\frac{16}{\sqrt{3}}$
$\tan30^\circ=\frac{\text{CA}}{\text{AB}'}$
$\frac{1}{\sqrt{3}}=\frac{\text{CA}}{8}$
$\text{CA}=\frac{8}{\sqrt{3}}$
Height of tree $= CB' + CA$
$=\frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=8\sqrt{3}$
$=8\sqrt{3}\text{m}$
View full question & answer→Question 203 Marks
From the top of a $7\ m$ high building, the angle of elevation of the top of a cable tower is $60^\circ $ and the angle of depression of its foot is $45^\circ $. Determine the height of the tower.
AnswerGiven,
Height of building $= 7m = AB$
Height of cable tower $= 'H'm = CD$
Angle of elevation of top of tower, from top of building $\alpha=60^\circ$
Angle of depression of bottom of tower, from top of building $\beta=45^\circ$
The above data is represented in form of figure as shown
Let $CX = 'x'm$
$CD = DX + XC = 7m + 'x'm$
$= x + 7m$
In $\triangle\text{ADX},$
$\tan45^\circ=\frac{\text{Opposite side (XD)}}{\text{Adjacent side (AX)}}$
$1=\frac{7}{\text{AX}}$
$\Rightarrow\ \text{AX}=7\text{m}$
In $\triangle\text{AXD},$
$\tan60^\circ=\frac{\text{XC}}{\text{AX}}$
$\sqrt{3}=\frac{\text{x}}{\text{H}}$
$\Rightarrow\ \text{x}=7\sqrt{3}$
But $\text{CD}=\text{x}+7$
$=7\sqrt{3}+7=7(\sqrt{3}+1)\text{m}$
Height of cable tower $=7(\sqrt{3}+1)\text{m.}$
View full question & answer→Question 213 Marks
From a point on the ground the angles of elevation of the bottom and top of a transmission tower fixed at the top of $20\ m$ high building are $45^\circ $ and $60^\circ $ respectively. Find the height of the transmission tower.
AnswerLet $AB$ be the building of height $20\ m$ and $BC$ be the building of height h meter.
Again let the angle of elevation of the bottom and top of tower at the point $O$ is $45^\circ $ and $60^\circ $ respectively.
In $\triangle\text{OAB},$
$\Rightarrow\ \tan45^\circ=\frac{\text{AB}}{\text{OA}}$
$\Rightarrow\ 1=\frac{20}{\text{x}}$
$\Rightarrow\ \text{x}=20$
Again in $\triangle\text{OAC},$
$\Rightarrow\ \tan60^\circ=\frac{\text{AC}}{\text{OA}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}+20}{\text{x}}$
$\Rightarrow\ \sqrt{3}\text{x}=\text{h}+20$
$\Rightarrow\ 20\sqrt{3}=\text{h}+20$
$\Rightarrow\ \text{h}=20\sqrt{3}-20$
$\Rightarrow\ \text{h}=20(\sqrt{3}-1)$
Hence the height of tower is $20(\sqrt{3}-1)\text{m}.$
View full question & answer→Question 223 Marks
Two points $A$ and $B$ are on the same side of a tower and in the same straight line with its base. The angles of depression of these points from the top of the tower are $60^\circ $ and $45^\circ $ respectively. If the height of the tower is $15\ m,$ then find the distance between the points.
AnswerLet $CD$ be the tower. $A$ and $B$ are the two points on the same side of the tower.
In $\triangle\text{DBC,}$
$\tan60^\circ=\frac{\text{DC}}{\text{BC}}$
$\Rightarrow\ \sqrt{3}=\frac{15}{\text{BC}}$
$\Rightarrow\ \text{BC}=\frac{15}{\sqrt{3}}$
$\Rightarrow\ \text{BC}=5\sqrt{3}\text{m}$
In $\triangle\text{DAC,}$
$\tan45^\circ=\frac{\text{DC}}{\text{AC}}$
$\Rightarrow\ 1=\frac{15}{\text{AC}}$
$\Rightarrow\ \text{AC}=15\text{m}$
Now, $ AC = AB + BC$
$\therefore AB = AC - BC$
$=15-5\sqrt{3}=5(3-\sqrt{3})\text{m}$
Hence, the distance between the two points $A$ and $B$ is $5(3-\sqrt{3})\text{m}.$
View full question & answer→Question 233 Marks
A vertically straight tree, $15\ m$ high, is broken by the wind in such a way that it top just touches the ground and makes an angle of $60^\circ $ with the ground. At what height from the ground did the tree break$?$
AnswerLet, angle $\angle\text{ACB}=60^\circ$
height of ground break tree $= BA = x\ m$
Inclined broken tree is $= (15 - x)m$
In $\triangle\text{ABC,}$
$\sin\text{C}=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\ \sin60^\circ=\frac{\text{x}}{15-\text{x}}$
$\Rightarrow\ \frac{\sqrt{3}}{2}=\frac{\text{x}}{15-\text{x}}$
$\Rightarrow\ 15\sqrt{3}=2\text{x}+\sqrt{3}\text{x}$
$\Rightarrow\ 15\sqrt{3}=\text{x}(2+\sqrt{3})$
$\Rightarrow\ \text{x}=\frac{2+\sqrt{3}}{15\sqrt{3}}=6.9\text{m}$
Hence, height of ground break tree is $6.9\ m.$
View full question & answer→Question 243 Marks
From a point $P$ on the ground the angle of elevation of a $10m$ tall building is $30^\circ .$ A flag is hoisted at the top of the building and the angle of elevation of the top of the flag-staff from $P$ is $45^\circ .$ Find the length of the flag-staff and the distance of the building from the point $P. (\text{Take }\sqrt{3}=1.732)$
AnswerLet, height of building $AB = h$
length of flag staff $BC = 10m$
angle of elevation $\angle\text{APB}=30^\circ,\ \angle\text{APC}=45^\circ$
In $\triangle\text{CPB},$
$\tan\text{P}=\frac{\text{BC}}{\text{PC}}$
$\Rightarrow\ \tan30^\circ=\frac{10}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{10}{\text{x}}$
$\Rightarrow\ \text{x}=10\sqrt{3}$
$\Rightarrow\ \text{x}=17.32\text{m}$
In $\triangle\text{APC},$
$\tan\text{P}=\frac{\text{AC}}{\text{PC}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{AB}+\text{BC}}{\text{x}}$
$\Rightarrow\ 1=\frac{\text{h}+10}{17.32}$
$\Rightarrow\ \text{h}=17.32-10$
$\Rightarrow\ \text{h}=7.32\text{m}$
Hence, length of flag staff $7.32m$ and distance between building and $P = 17.32m$
View full question & answer→Question 253 Marks
From the top of a $50m$ high tower, the angles of depression of the top and bottom of a pole are observed to be $45^\circ $ and $60^\circ $ respectively. Find the height of the pole.
AnswerLet $AB$ be the tower and $CD$ is the pole. Angles of depression from the top $A$ to the top and bottom of the pole are $45^\circ$ and $60^\circ $ respectively.
$AB = 50\ m,$ let $CD = h$ and $BD = EC = x$
$\because CE || DB$
$\therefore EB = CD = h$
and $AE = 50 - h$
Now in right $\triangle\text{ADB},$
$\tan\theta=\frac{\text{Perpendicular}}{\text{Base}}=\frac{\text{AB}}{\text{DB}}$
$\Rightarrow\ \tan60^\circ=\frac{50}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{50}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{50}{\sqrt{3}}\ ......(1)$
Similarly in right $\triangle\text{ACE,}$
$\tan45^\circ=\frac{\text{AE}}{\text{CE}}\Rightarrow\ 1=\frac{50-\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=50-\text{h}\Rightarrow\ \text{x}+\text{h}=50$
$\Rightarrow\ \text{h}=50-\text{x}=50-\frac{50}{\sqrt{3}} [$From $(1)]$
$=50-28.87=21.13$
$\therefore$ Height of the pole $= 21.13m$
View full question & answer→Question 263 Marks
If the angles of elevation of the top of a tower from two points at a distance of 4m and 9m from the base of the tower in the same straight line with it are complementary, find the height of the tower.
AnswerLet $AC$ be the height of tower is h meters.
Given that: Angle of elevation are $\angle\text{B}=90^\circ-\theta$ and $\angle\text{D}=\theta$ and also $CD = 4m$ and $BC = 9m.$
Here we have to find height of tower.
So we use trigonometric ratios.
In a triangle $ADC,$
$\tan\theta=\frac{\text{h}}{4}$
Again in a triangle $ABC,$
$\Rightarrow\ \tan(90^\circ-\theta)=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\ \cot\theta=\frac{\text{h}}{9}$
$\Rightarrow\ \frac{1}{\tan\theta}=\frac{\text{h}}{9}$
Put, $\tan\theta=\frac{\text{h}}{4}$
$\Rightarrow\ \frac{4}{\text{h}}=\frac{\text{h}}{9}$
$\Rightarrow\ \text{h}^2=36$
$\Rightarrow\ \text{h}=6$
Hence height of tower is $6$ meters.
View full question & answer→Question 273 Marks
The angle of elevation of the top of the building from the foot of the tower is $30^\circ $ and the angle of the top of the tower from the foot of the building is $60^\circ .$ If the tower is 50m high, find the height of the building.
AnswerLet $AD$ be the building of height $h\ m.$ and an angle of elevation of top of building from the foot of tower is $30^\circ $ and an angle of the top of tower from the foot of building is $60^\circ .$
Let $AD = h, AB = x$ and $BC = 50$ and $\angle\text{DBA}=30^\circ,\ \angle\text{CAB}=60^\circ$
So we use trigonometric ratios.
In a triangle $ABC,$
$\Rightarrow\ \tan60^\circ=\frac{50}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{50}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{50}{\sqrt{3}}$
Again in a triangle $ABD,$
$\Rightarrow\ \tan30^\circ=\frac{\text{AD}}{\text{AB}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{h}=\frac{\text{x}}{\sqrt{3}}$
$\Rightarrow\ \text{h}=\frac{50}{\sqrt{3}\times\sqrt{3}}$
$\Rightarrow\ \text{h}=\frac{50}{3}$
Hence the height of building is $\frac{50}{3}\text{m}.$
View full question & answer→Question 283 Marks
The angle of elevation of the top of a hill at the foot of a tower is $60^\circ $ and the angle of elevation of the top of the tower from the foot of the hill is $30^\circ .$ If the tower is $50\ m$ high, what is the height of the hill$?$
AnswerLet $h$ be the height of hill $AB$ and $CD$ be the tower of height $50m.$ Angle of elevation of the top of hill from the foot of tower is $60^\circ $ and angle of elevation of top of tower from foot of hill is $30^\circ .$ Let $AB = h$ and $\angle\text{DAC}=30^\circ,\ \angle\text{ACB}=60^\circ$
Here we have to find height of hill.
The corresponding figure is as follows,
So we use trigonometric ratios.
In $\triangle\text{ACD},$
$\Rightarrow\ \tan30^\circ=\frac{\text{CD}}{\text{AC}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{50}{\text{x}}$
$\Rightarrow\ \text{x}=50\sqrt{3}$
Again in $\triangle\text{ABC},$
$\Rightarrow\ \tan60^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{h}=\text{x}\sqrt{3}$
$\Rightarrow\ \text{h}=150$
Hence the height of hill is $150\ m.$
View full question & answer→Question 293 Marks
The angle of elevation of the top of a tower at a point on the ground is $30^\circ .$ What will be the angle of elevation, if the height of the tower is tripled$?$
AnswerLet the height of the tower $AB$ be $h$ units.
Suppose $C$ is a point on the ground such that $\angle\text{ACB}=30^\circ.$
In right $\triangle\text{ACB,}$
$\tan30^\circ=\frac{\text{AB}}{\text{AC}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{AC}}$
$\Rightarrow\ \text{AC}=\sqrt{3}\text{h}\ ....(1)$
Let the angle of elevation of the tower at $C$ be $\theta,$ if the height of the tower is tripled.
New height of the tower, $AD = 3h$ units
In right $\triangle\text{ACD,}$
$\tan\theta=\frac{\text{AD}}{\text{AC}}$
$\Rightarrow\ \tan\theta=\frac{3\text{h}}{\text{AC}}$
$\Rightarrow\ \tan\theta=\frac{3\text{h}}{\sqrt{3}\text{h}}=\sqrt{3} [$From $(1)]$
$\Rightarrow\ \tan\theta=\tan60^\circ$
$\Rightarrow\ \theta=60^\circ$
Hence, the required angle of elevation is $60^\circ .$
View full question & answer→Question 303 Marks
A person observed the angle of elevation of the top of a tower as $30^\circ .$ He walked $50\ m$ towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as $60^\circ .$ Find the height of the tower.
AnswerGiven,
Angle of elevation of top of tour, from first point of elevation $(\text{A})\alpha=30^\circ$
Let the walked $50m$ from first point $(A)$ to $B$ then $50ABm =$
Angle of elevation from second point $B \Rightarrow Gb = 60^\circ $
Now let is represent the given data in form of then it forms triangle $ACD$ with triangle
$BCD$ in it $\angle\text{C}=90^\circ$
Let height of tower, be
$Hm = CD$
$BC = xm.$
If in a right angle triangle $\theta$ is the angle then,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan\alpha=\frac{\text{CD}}{\text{AC}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{H}}{\text{AB}+\text{BC}}$
$\Rightarrow \ \frac{1}{\sqrt{3}}=\frac{\text{H}}{50+\text{x}}$
$\Rightarrow\ 50+\text{x}=\text{H}\sqrt{3}\ ......(1)$
$\tan\beta=\frac{\text{CD}}{\text{BC}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{H}}{\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{H}}{\text{x}}$
$\Rightarrow\ \text{x}=\frac{\text{H}}{\sqrt{3}}\ ......(2)$
$(2)$ in $(1)$
$\Rightarrow\ 50+\frac{\text{H}}{\sqrt{3}}=\text{H}\sqrt{3}$
$\Rightarrow\ \text{H}\sqrt{3}-\frac{\text{H}}{\sqrt{3}}=50$
$\Rightarrow\ \text{H}\Big(\frac{3-1}{\sqrt{3}}\Big)=50$
$\Rightarrow\ \text{H}=\frac{50\sqrt{3}}{2}=25\sqrt{3}$
$\therefore$ Height of tower $\text{H}=25\sqrt{3}\text{m.}$
View full question & answer→Question 313 Marks
A $1.5m$ tall boy is standing at some distance from a $30m$ tall building. The angle of elevation from his eyes to the top of the building increase from $30^\circ$ to $60^\circ $ as he walks towards the building. Find the distance he walked towards the building.
AnswerLet $BG$ be the distance of tall Boy $x$ and he walks towards the building, makes an angle of elevation at top of building increase from $30^\circ $ to $60^\circ .$
Therefore $\angle\text{A}=30^\circ$ and $\angle\text{F}=60^\circ$ given $CE = 30\ m, AB = 15\ m, FG = 1.5$ and $DE = 28.5, GC = X − x$ and $FD = X − x$
We have to find $x$
The corresponding figure is as follows,
In $\triangle\text{AED},$
$\Rightarrow\ \tan\text{A}=\frac{\text{ED}}{\text{AD}}$
$\Rightarrow\ \tan30^\circ=\frac{28.5}{\text{X}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{28.5}{\text{X}}$
$\Rightarrow\ \text{X}=49.36$
Again in $\triangle\text{EFD},$
$\Rightarrow\ \tan\text{F}=\frac{\text{DE}}{\text{FD}}$
$\Rightarrow\ \tan60^\circ=\frac{28.5}{\text{X}-\text{x}}$
$\Rightarrow\ \sqrt{3}=\frac{28.5}{49.36-\text{x}}$
$\Rightarrow\ 28.5=49.36\sqrt{3}-\sqrt{3}\text{x}$
$\Rightarrow\ \text{x}=\frac{57}{\sqrt{3}}$
$\Rightarrow\ \text{x}=19\sqrt{3}$
Hence the required distance is $19\sqrt{3}\text{m}.$
View full question & answer→Question 323 Marks
The angle of elevation of a ladder leaning against a wall is $60^\circ $ and the foot of the ladder is $9.5m$ away from the wall. Find the length of the ladder.
AnswerLet angle of elevation $=\angle\text{ACB}=60^\circ$
Distance from wall $= BC = 9.5m$
length of ladder $= h$
In $\triangle\text{ABC},$
By using trigonometrical ratio,
$\cos\text{C}=\frac{\text{Base}}{\text{Hypotenuse}}$
$\Rightarrow\ \cos60^\circ=\frac{\text{BC}}{\text{AC}}$
$\Rightarrow\ \frac{1}{2}=\frac{9.5}{\text{h}}\Rightarrow\ \text{h}=19$
Hence, length of ladder is $19m.$
View full question & answer→Question 333 Marks
A vertical tower stands on a horizontal plane and is surmounted by a flag-staff of height $7m.$ From a point on the plane, the angle of elevation of the bottom of the flag-staff is $30^\circ $ and that of the top of the flag-staff is $45^\circ .$ Find the height of the tower.
AnswerGiven
Height of flagstaff $= 7m = BC$
Let height of tower $= 'h'm = AB$
Angle of elevation of bottom of flagstaff $\alpha=30^\circ$
Angle of elevation of top of flagstaff $\beta=45^\circ$
Points of desecration be $'p'$
The above data is represented in form of figure as shown.
In right angle triangle if one of the induced angle is $\theta$ then,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan\alpha=\frac{\text{AB}}{\text{AP}}$
$\tan30^\circ=\frac{\text{h}}{\text{AP}}$
$\text{AP}=\text{h}\sqrt{3}\ ......(1)$
$\tan\beta=\frac{\text{AC}}{\text{AP}}$
$\tan45^\circ=\frac{\text{h}+7}{\text{AP}}$
$\text{AP}=\text{h}+7\ ......(2)$
From $(1)$ and $(2)$
$\text{h}\sqrt{3}=\text{h}+7$
$\text{h}\sqrt{3}-\text{h}=7$
$\text{h}(\sqrt{3}-1)=7$
$\text{h}=\frac{7}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\text{h}=\frac{7\times(\sqrt{3}+1)}{2}=3.5(\sqrt{3}+1)$
Height of tower $=3.5(\sqrt{3}+1)\text{m}.$
View full question & answer→Question 343 Marks
A statue $1.6m$ tall stands on the top of pedestal. From a point on the ground, the angle of elevation of the top of the statue is 60° and from the same point the angle of elevation of the top of the pedestal is $45^\circ .$ Find the height of the pedestal.
AnswerLet, height of the pedestal $BD = h$
distance $BC = xm$
Given,
angle of elevation $\angle\text{BCD}=45^\circ,\ \angle\text{BCA}=60^\circ$
height of statue, $AD = 1.6m$
Now, In $\triangle\text{BCD},$
$\Rightarrow\ \tan\text{C}=\frac{\text{DB}}{\text{CB}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}$
In $\triangle\text{ACB},$
$\tan\text{C}=\frac{\text{AB}}{\text{CB}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{AD}+\text{DB}}{\text{CB}}$
$\Rightarrow\ \sqrt{3}=\frac{1.6+\text{h}}{\text{x}}$
$\Rightarrow\ \sqrt{3}\text{x}=1.6+\text{h}$
$\Rightarrow\ \sqrt{3}\text{h}-\text{h}=1.6$
$\Rightarrow\ \text{h}(\sqrt{3}-1)=1.6$
$\Rightarrow\ \text{h}=\frac{1.6}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow\ \text{h}=0.8(\sqrt{3}+1)$
Hence, height of pedestal $0.8(\sqrt{3}+1)\text{m}.$
View full question & answer→Question 353 Marks
Two men on either side of the cliff $80m$ high observes the angles of a elevation of the top of the cliff to be $30^\circ $ and $60^\circ $ respectively. Find the distance between the two men.
AnswerLet, distance between two men, $BD = x + y$
Given, angle of depression $\angle\text{B}=30^\circ,\ \angle\text{D}=60^\circ$
height of bridge, $AC = 80$
In $\triangle\text{ABC},$
$\tan\text{B}=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{80}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{80}{\text{x}}$
$\Rightarrow\ \text{x}=80\sqrt{3}$
In $\triangle\text{ACD},$
$\tan\text{D}=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\ \tan60^\circ=\frac{80}{\text{y}}$
$\Rightarrow\ \sqrt{3}=\frac{80}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{80}{\sqrt{3}}$
distance between two men, $BD = x + y$
$\Rightarrow\ \text{BD}=80\sqrt{3}+\frac{80}{\sqrt{3}}$
$\Rightarrow\ \text{BD}=184.8\text{m}$
Hence, distance between two men $184.8m.$
View full question & answer→Question 363 Marks
A T.V. Tower stands vertically on a bank of a river. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is $60^\circ .$ From a point $20\ m$ away this point on the same bank, the angle of elevation of the top of the tower is $30^\circ .$ Find the height of the tower and the width of the river.
AnswerLet $AB$ be the T.V tower of height hm on a bank of river and $D$ be the point on the opposite of the river. An angle of elevation at top of tower is $60^\circ $ and from a point $20m$ away then angle of elevation of tower at the same point is $30^\circ .$ Let $AB = h$ and $BC = x.$
Here we have to find height and width of river.
The corresponding figure is here,
In $\triangle\text{CAB,}$
$\Rightarrow\ \tan60^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \sqrt{3}\text{x}=\text{h}$
$\Rightarrow\ \text{x}=\frac{\text{h}}{\sqrt{3}}$
Again in $\triangle\text{DBA,}$
$\Rightarrow\ \tan30^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{20+\text{x}}$
$\Rightarrow\ \sqrt{3}\text{h}=20+\text{x}$
$\Rightarrow\ \sqrt{3}\text{h}=20+\frac{\text{h}}{\sqrt{3}}$
$\Rightarrow\ \sqrt{3}\text{h}-\frac{\text{h}}{\sqrt{3}}=20$
$\Rightarrow\ \frac{2\text{h}}{\sqrt{3}}=20$
$\Rightarrow\ \text{h}=10\sqrt{3}$
$\Rightarrow\ \text{x}=\frac{10\sqrt{3}}{\sqrt{3}}$
$\Rightarrow\ \text{x}=10$
Hence the height of the tower is $10\sqrt{3}\text{m}$ and width of river is $10m.$
View full question & answer→Question 373 Marks
The angles of depression of the top and bottom of $8\ m$ tall building from the top of a multistoried building are $30^\circ $ and $45^\circ $ respectively. Find the height of the multistoried building and the distance between the two buildings.
AnswerLet $AD$ be the multistoried building of height $h\ m.$ And angle of depression of the top and bottom are $30^\circ $ and $45^\circ .$ We assume that $BE = 8, CD = 8$ and $BC = x, ED = x$ and $AC = h - 8.$ Here we have to find height and distance of building.
We use trignometrical ratio.
In $\triangle\text{AED,}$
$\Rightarrow\ \tan\text{E}=\frac{\text{AD}}{\text{DE}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{AD}}{\text{DE}}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\text{h}$
Again in $\triangle\text{ABC,}$
$\Rightarrow\ \tan\text{B}=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}-8}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}-8}{\text{x}}$
$\Rightarrow\ \text{h}\sqrt{3}-8\sqrt{3}=\text{x}$
$\Rightarrow\ \text{h}\sqrt{3}-8\sqrt{3}=\text{h}$ [x = h]
$\Rightarrow\ \text{h}(\sqrt{3}-1)=8\sqrt{3}$
$\Rightarrow\ \text{h}=\frac{8\sqrt{3}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow\ \text{h}=\frac{24+8\sqrt{3}}{2}$
$\Rightarrow\ \text{h}=4(3+\sqrt{3})$
and,
$\Rightarrow\ \text{x}=4(3+\sqrt{3})$
Hence the required height is $4(3+\sqrt{3})\text{meter}$ and distance is $4(3+\sqrt{3})\text{meter}.$
View full question & answer→Question 383 Marks
From a point on a bridge across a river the angles of depression of the banks on opposite side of the river are $30^\circ $ and $45^\circ $ respectively. If bridge is at the height of $30m$ from the banks, find the width of the river.
Answer
Height of the bridge $[AB] = 30m$
Angle of depression of bank $1$ i.e., $[\text{B}_1]\ \alpha=30^\circ$
Angle of depression of bank $2$ i.e., $[\text{B}_2]\ \beta=45^\circ$
Given banks are on opposite sides,
Distance between banks $B_1 B_2=B_1 B+B B_2$
The above information is represented is the form of figure as shown in right angle triangle if one of the included angle is $O$ then,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
In $\triangle\text{ABB}_1$
$\tan\alpha=\frac{\text{AB}}{\text{B}_1\text{B}}$
$\tan30^\circ=\frac{30}{\text{B}_1\text{B}}$
$\text{B}_1\text{B}=30\sqrt{3}\text{m}$
In $\triangle\text{ABB}_2$
$\tan\beta=\frac{\text{AB}}{\text{BB}_2}$
$\tan45^\circ=\frac{30}{\text{BB}_2}$
$\text{BB}_2=30\text{m}$
$\text{B}_1\text{B}_2=\text{B}_1\text{B}+\text{BB}_2$
$=30\sqrt{3}+30$
$=30(\sqrt{3}+1)$
Distance between banks $=30(\sqrt{3}+1)\text{m}.$ View full question & answer→Question 393 Marks
From the top of a building $AB, 60m$ high, the angles of depression of the top and bottom of a vertical lamp post $CD$ are observed to be $30^\circ $ and $60^\circ $ respectively. Find
- The horizontal distance between $AB$ and $CD.$
- The height of the lamp post.
- The difference between the heights of the building and the lamp post.
Answer
Height of building $AB = 60m$
Height of lamp post $CD = h\ m$
Angle of depression of top of lamp post from top of building $\alpha=30^\circ$
Angle of depression of bottom of lamp from top of building $\beta=60^\circ$
The above information is represented in the form of figure as shown.
Draw $\text{DX}\bot\text{AB},DX = AC, CD = AX$
In $\triangle\text{BDX}$
$\tan\alpha=\frac{\text{Opposite side}}{\text{Adjacent side}}=\frac{\text{BX}}{\text{DX}}$
$\tan30^\circ=\frac{60-\text{CD}}{\text{DX}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{60-\text{h}}{\text{AC}}$
$\Rightarrow\ \text{AC}=(60-\text{h})\sqrt{3}\text{m}\ .......(1)$
In $\triangle\text{BCA}$
$\tan\beta=\frac{\text{AB}}{\text{AC}}\Rightarrow\ \tan60^\circ=\frac{60}{\text{AC}}$
$\Rightarrow\ \text{AC}=\frac{60}{\sqrt{3}}=20\sqrt{3}\text{m}\ ......(2)$
From $(1)$ and $(2)$
$\Rightarrow\ (60-\text{h})\sqrt{3}=20\sqrt{3}$
$\Rightarrow\ 60-\text{h}=20$
$\Rightarrow\ \text{h}=40\text{m}$
Height of lamp post $= 40m$
Distance between lamp posts building $\text{AC}=20\sqrt{3}\text{m}.$
Difference between heights of building and lamp post $= BX = 60 - h = 60 - 40 = 20m.$ View full question & answer→Question 403 Marks
On a horizontal plane there is vertical tower with a flag pole on the top of the tower. At a point $9$ metres away from the foot of the tower the angle of elevation of the top and bottom of the flag pole are $60^\circ $ and $30^\circ $ respectively. Find the height of the tower and the flag pole mounted on it.
AnswerLet $AB$ be the tower of height $h$ and $AD$ be the flag pole on tower. At the point 9m away from the foot of tower, the angle of elevation of the top and bottom of flag pole are $60^\circ $ and $30^\circ .$
Let $AD = x, BC = 9$ and $\angle\text{ACB}=30^\circ,\ \angle\text{DCB}=60^\circ.$
Here we have to find height of tower and height of flag pole.
The corresponding diagram is as follows,
In a triangle $ABC,$
$\Rightarrow\ \tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{9}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{9}$
$\Rightarrow \text{h}=\frac{9}{\sqrt{3}}$
$\Rightarrow\ \text{h}=3\sqrt{3}$
Again in a triangle $DBC,$
$\Rightarrow\ \tan\text{C}=\frac{\text{AD}+\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}+\text{x}}{9}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}+\text{x}}{9}$
$\Rightarrow\ 9\sqrt{3}=\text{h}+\text{x}$
$\Rightarrow\ 9\sqrt{3}=3\sqrt{3}+\text{x}$
$\Rightarrow\ \text{x}=6\sqrt{3}$
So height of tower is $3\sqrt{3}\text{m}$ and height of flag pole is $6\sqrt{3}\text{m.}$ View full question & answer→Question 413 Marks
The angle of elevation of the top of a tower from a point A on the ground is $30^\circ .$ On moving a distance of $20$ metres towards the foot of the tower to a point $B$ the angle of elevation increases to $60^\circ .$ Find the height of the tower and the distance of the tower from the point $A.$
Answer
Angle of elevation of top of tower from points $A, \alpha=30^\circ$
Angle of elevation of top of tower from points $B, \beta=60^\circ$
Distance between $A$ and $B, AB = 20m$
Let height of tower $CD = 'h'm$
Distance between second point $B$ from foot of tower $BC = 'x'm$
If we represent the above data in the figure, then it forms figure as shown with $\angle\text{D}=90^\circ$
In right angled triangle if one of the included angle is $\theta$ then,
$\tan\theta=\frac{\text{Opposite side}}{\text{Adjacent side}}$
$\tan\alpha=\frac{\text{CD}}{\text{AD}}$
$\tan30^\circ=\frac{\text{h}}{20+\text{x}}$
$20+\text{x}=\text{h}\sqrt{3}\ ......(1)$
$\tan\beta=\frac{\text{CD}}{\text{BD}}$
$\tan60^\circ=\frac{\text{h}}{\text{x}}$
$\text{x}=\frac{\text{h}}{\sqrt{3}}\ ......(2)$
$(2)$ in $(1)$
$\Rightarrow\ 20+\frac{\text{h}}{\sqrt{3}}=\text{h}\sqrt{3}$
$\Rightarrow\ \text{h}\Big(\sqrt{3}-\frac{1}{\sqrt{3}}\Big)=20$
$\Rightarrow\ \text{h}\Big(\frac{3-1}{\sqrt{3}}\Big)=20$
$\Rightarrow\ \text{h}=\frac{20\sqrt{3}}{2}=10\times\sqrt{3}=17.32\text{m}$
$\text{x}=\frac{10\sqrt{3}}{\sqrt{3}}=10\text{m}$
Height of tower $h = 17.32\ m$
Distance of tower from point $A = (20 + 10) = 30\ m$ View full question & answer→Question 423 Marks
A man sitting at a height of $20\ m$ on a tall tree on a small island in the middle of a river observes two poles directly opposite to each other on the two banks of the river and in line with the foot of tree. If the angles of depression of the feet of the poles from a point at which the man is sitting on the tree on either side of the river are $60^\circ $ and $30^\circ $ respectively. Find the width of the river.
AnswerLet $BD$ be the width of river. And the angles of depression on either side of the river are $30^\circ $ and $60^\circ $ respectively. It is given that $AC = 20m.$ Let $BC = x$ and $CD = y.$
And $\angle\text{ABC}=30^\circ,\ \angle\text{ADC}=60^\circ$
Here we have to find the width of river.
We have the corresponding figure as follows,
So we use trigonometric ratios.
In a triangle $ABC,$
$\Rightarrow\ \tan\text{B}=\frac{\text{AC}}{\text{BC}}$
$\Rightarrow\ \tan30^\circ=\frac{20}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{20}{\text{x}}$
$\Rightarrow\ \text{x}=20\sqrt{3}$
Again in a triangle ADC,
$\Rightarrow\ \tan\text{D}=\frac{\text{AC}}{\text{CD}}$
$\Rightarrow\ \tan60^\circ=\frac{20}{\text{y}}$
$\Rightarrow\ \sqrt{3}=\frac{20}{\text{y}}$
$\Rightarrow\ \text{y}=\frac{20}{\sqrt{3}}$
$\Rightarrow\ \text{x}+\text{y}=20\sqrt{3}+\frac{20}{20\sqrt{3}}$
$\Rightarrow\ \text{x}+\text{y}=\frac{80}{\sqrt{3}}$
Hence width of river is $\frac{80}{\sqrt{3}}\text{m.}$ View full question & answer→Question 433 Marks
A ladder is placed along a wall of a house such that its upper end is touching the top of the wall. The foot of the ladder is 2m away from the wall and the ladder is making an angle of $60^\circ $ with the level of the ground. Determine the height of the wall.
AnswerLet $AB$ be the wall of height $h\ m$ and $C$ be the points, makes an angle $60^\circ $ and foot of the ladder is $2\ m$ away from the wall. We have to find height of wall
In a triangle $ABC,$ given that $BC = 2m$ and angle $C = 60°$
Now we have to find the height of wall.
So we use trigonometrically ratios.
In a triangle $ABC,$
$\Rightarrow\ \tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{h}}{2}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}}{2}$
$\Rightarrow\ \text{h}=2\sqrt{3}$
Hence height of wall is $2\sqrt{3}$ meters. View full question & answer→Question 443 Marks
An observed from the top of a $150\ m$ tall light house, the angles of depression of two ships approaching it are $30^\circ $ and $45^\circ .$ If one ship is directly behind the other, find the distance between the two ships.
AnswerLet $AB$ be the light house of $150m$ and angle of depression of two ship $C$ and $D$ are $30^\circ $ and $45^\circ $ respectively.
Let $BC = x, CD = y$ and $\angle\text{ADB}=30^\circ,\ \angle\text{ACB}=45^\circ.$
We use trigonometric ratios.
In a triangle $ABC,$
$\Rightarrow\ \tan45^\circ=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ 1=\frac{150}{\text{x}}$
$\Rightarrow\ \text{x}=150$
Again in a triangle ABD,
$\Rightarrow\ \tan30^\circ=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{150}{\text{x}+\text{y}}$
$\Rightarrow\ \text{x}+\text{y}=150\sqrt{3}$
$\Rightarrow\ 150+\text{y}=150\sqrt{3}$
$\Rightarrow\ \text{y}=150\sqrt{3}-150$
$\Rightarrow\ \text{y}=150(\sqrt{3}-1)$
$\Rightarrow\ \text{y}=150\times0.732$
$\Rightarrow\ \text{y}=109.8$
Hence distance between the ships is $109.8m$ View full question & answer→Question 453 Marks
The height of a tower is $10m.$ What is the length of its shadow when Sun's altitude is $45^\circ ?$
AnswerLet, height of tower, $AB = 10m$
length of shadow, $BC = xm$
angle of elevation, $\angle\text{C}=45^\circ$
In $\triangle\text{ABC,}$
$\tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan45^\circ=\frac{10}{\text{x}}$
$\Rightarrow\ 1=\frac{10}{\text{x}}$
$\Rightarrow\ \text{x}=10\text{m}$
Hence, length of shadow is $10m.$ View full question & answer→Question 463 Marks
A $1.6\ m$ tall girl stands at a distance of $3.2\ m$ from a lamp-post and casts a shadow of $4.8\ m$ on the ground. Find the height of the lamp-post by using $(i)$ trignometric ratios $(ii)$ property of similar triangles.
AnswerLet $AC$ be the lamp post of height h.
We assume that $ED = 1.6\ m, BE = 4.8\ m$ and $EC = 3.2\ m$
We have to find the height of the lamp post, Now we have to find height of lamp post using similar triangles.
Since triangle $BDE$ and triangle $ABC$ are similar.
$\frac{\text{AC}}{\text{BC}}=\frac{\text{ED}}{\text{BE}}$
$\frac{\text{h}}{4.8+3.2}=\frac{1.6}{4.8}$
$\text{h}=\frac{8}{3}$
Again, we have to find height of lamp post using trigonometric ratios.
In $\triangle\text{ADE,}$
$\Rightarrow\ \tan\theta=\frac{1.6}{4.8}$
$\Rightarrow\ \tan\theta=\frac{1}{3}$
Again in $\triangle\text{ABC,}$
$\Rightarrow\ \tan\theta=\frac{\text{h}}{4.8+3.2}$
$\Rightarrow\ \frac{1}{3}=\frac{\text{h}}{8}$
$\Rightarrow\ \text{h}=\frac{8}{3}$
Hence the height of lamp post is $\frac{8}{3}\text{m}.$ View full question & answer→Question 473 Marks
The length of the shadow of a tower standing on level plane is found to be $2x$ metres longer when the sun's altitude is $30^\circ $ than when it was $45^\circ .$ Prove that the height of tower is $\text{x}(\sqrt{3}+1)\text{ metres.}$
AnswerLet, height of tower $AB = h$
length of shadow $DC + CB = 2x + y$
Given,
angle of elevation $\angle\text{D}=30^\circ,\ \angle\text{C}=45^\circ$
In $\triangle\text{ABC},$
$\tan\text{C}=\frac{\text{AB}}{\text{BC}}$
$\Rightarrow\ \tan45^\circ=\frac{\text{h}}{\text{y}}$
$\Rightarrow\ 1=\frac{\text{h}}{\text{y}}$
$\Rightarrow \text{y}=\text{h}$
In $\triangle\text{ABD},$
$\tan\text{D}=\frac{\text{AB}}{\text{BD}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{2\text{x}+\text{y}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{2\text{x}+\text{y}}$
$\Rightarrow\ \sqrt{3}\text{h}=2\text{x}+\text{y}$
$\Rightarrow\ \sqrt{3}\text{h}=2\text{x}+\text{h}$
$\Rightarrow\ \sqrt{3}\text{h}-\text{h}=2\text{x}$
$\Rightarrow\ \text{h}(\sqrt{3}-1)=2\text{x}$
$\Rightarrow\ \text{h}=\frac{2\text{x}}{\sqrt{3}-1}\times\frac{\sqrt{3}+1}{\sqrt{3}+1}$
$\Rightarrow\ \text{h}=\frac{2\text{x}(\sqrt{3}+1)}{2}$
$\Rightarrow\ \text{h}=\text{x}(\sqrt{3}+1)$
Hence proved, height of tower $\text{x}(\sqrt{3}+1)\text{m.}$ View full question & answer→Question 483 Marks
From the top of a building $15\ m$ high the angle of elevation of the top of a tower is found to be $30^\circ .$ From the bottom of the same building, the angle of elevation of the top of the tower is found to be $60^\circ .$ Find the height of the tower and the distance between the tower and building.
AnswerLet, height of tower $OB = 15 + h$
angle of elevation $\angle\text{D}=30^\circ,\ \angle\text{A}=60^\circ$
distance between tower and building $AB = x$
height of building $ = 15m$
In $\triangle\text{ODC,}$
$\tan\text{D}=\frac{\text{OC}}{\text{DC}}$
$\Rightarrow\ \tan30^\circ=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ \text{x}=\sqrt{3}\text{h}$
In $\triangle\text{OAB,}$
$\tan\text{A}=\frac{\text{OB}}{\text{AB}}$
$\Rightarrow\ \tan60^\circ=\frac{\text{OC}+\text{CB}}{\text{AB}}$
$\Rightarrow\ \sqrt{3}=\frac{\text{h}+15}{\text{x}}$
$\Rightarrow\ \text{h}=7.5$
and $\text{x}=\text{h}\sqrt{3}$
$\Rightarrow\ \text{x}=7.5\times\sqrt{3}$
$\Rightarrow\ \text{x}=12.9$
Now, height of tower $OB = 15 + h ⇒ 15 + 7.5 ⇒ 22.5m$
Hence, height of tower $22.5$ and distance between tower and building $12.9\ m$ View full question & answer→Question 493 Marks
From the top of a $120\ m$ high tower, a man observes two cars on the opposite sides of the tower and in straight line with the base of tower with angles of depression as $60^\circ$ and $45^\circ .$ Find the distance between the cars. $(\text{Take }\sqrt{3}=1.732)$
AnswerLet $BD$ be the tower and $A$ and $C$ be the two points on ground.
Then, $BD$ the height of the tower $= 120m$
$\angle\text{BAD}=45^\circ,\ \angle\text{BCD}=60^\circ$
$\tan45^\circ=\frac{\text{BD}}{\text{BA}}$
$\Rightarrow\ 1=\frac{120}{\text{BA}}$
$\Rightarrow\ \text{BA}=120\text{m}\ ......(\text{i})$
$\tan60^\circ=\frac{\text{BD}}{\text{BC}}$
$\Rightarrow\ \sqrt{3}=\frac{120}{\text{BC}}$
$\text{BC}=\frac{120}{\sqrt{3}}=\frac{120\times\sqrt{3}}{\sqrt{3}\times\sqrt{3}}$
$=\frac{120\sqrt{3}}{3}=40\sqrt{3}$
$=40\times1.732=69.28\text{m}\ ......(\text{ii})$
Distance between the two points $A$ and $C = AC = BA + BC$
$= 120 + 69.28$
$[\because$ Substituted value of $BA$ and $BC$ from $(i)$ and $(ii) ]$
$= 189.28m$ View full question & answer→