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Maths 3 Marks Question

Some Applications of Trigonometry · Jharkhand Board (JAC) STD 10 (Jharkhand - English Medium). 49 questions.

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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?
Answer
Let 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°.
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Question 23 Marks
As observed from the top of a 75m tall light house, the angle of depression of two ships are 30° and 45°. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between the two ships.
Answer
Let, 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 $75(\sqrt{3}-1)\text{m}.$
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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°, What is the height of the tower?
Answer
Given,
angle of elevation, $\angle\text{C}=60^\circ$
distance between point and tower, BC = 20m
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 $20\sqrt{3}\text{m.}$
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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° and 45°. Find the height of the rock.
Answer
Let AB be the height of Rock which is H m. and makes an angle of elevations 45° and 30° respectively from the bottom and top of tower whose height is 100m.
Let AE = h m, BC = xm 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 $\triangle\text{ADE,}$
$\tan30^\circ=\frac{\text{AE}}{\text{DE}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{h}}{\text{x}}$
$\Rightarrow\ 100+\text{h}=\sqrt{3}\text{h}$
⇒ h = 136.65
⇒ H = 100 + 136.65
⇒ H = 236.65
Hence the height of rock is 236.65m
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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.
Answer
Let, angle of elevation $=\angle\text{ACB}=60^\circ$
heigh of kite = AB = 75m
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 87m.
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Question 63 Marks
Two poles of equal heights are standing opposite to each other on either side of the road which is 80m wide. From a point between them on the road the angles of elevation of the top of the poles are 60° and 30° respectively. Find the height of the poles and the distance of the point from the poles.
Answer
Let, 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°

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 $30(\sqrt{3}+1)\text{m}.$
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Question 73 Marks
An electric pole is 10m 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° with the horizontal through the foot of the pole, find the length of the wire.
Answer

Height of the electric pole H = 10m = 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 $\text{l}=10\sqrt{2}\text{m}$
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Question 83 Marks
A parachutist is descending vertically and makes angles of elevation of 45° and 60° at two observing points 100m 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.
Answer
Let BC be the height h of the parachutist and makes an angle of elevations 45° and 60° 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 $50(1+\sqrt{3})\text{m}=136.6\text{m}$​​​​​​​
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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° 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.
Answer
Let AB be the tree of height h. And the top of tree makes an angle 30° with ground. The distance between foot of tree to the point where the top touches the ground is 10m.
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 17.3m.
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Question 103 Marks
The tops of two towers of height x and y, standing on level ground, subtend angles of 30º and 60º respectively at the centre of the line joining their feet, then find x : y.
Answer
Let 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° and 60°.

$\because\ \tan\theta=\frac{\text{P}}{\text{B}}$
$\therefore\ \tan30^\circ=\frac{\text{AB}}{\text{BM}}=\frac{\text{x}}{\text{BM}}$
$\Rightarrow\ \frac{1}{\sqrt{3}}=\frac{\text{x}}{\text{BM}}$
$\Rightarrow\ \text{BM}=\text{x}\sqrt{3}$
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
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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°. What is the height of the tower?
Answer
Given,
Distance between point of observation and foot of tower = 20m =BC
Angle of elevation of top of tower = 60° = 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 $\text{H}=20\sqrt{3}\text{m}.$
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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° and 60°. If the height of the tower is 150m, find the distance between the objects.
Answer

Height 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
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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° and from the same point, the angle of elevation of the top of the tower is 45°. Find the height of the flag-staff.
Answer
Let BC be the tower height of 5m flag height is h m and an angle of elevation of top of tower is 45° and an angle of elevation of the top of flag is 60°.
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
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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° and 60°. Find the height of the tower.
Answer
Let 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° and 30° 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.
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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° than when it was 60°. Find the height of the tower.
Answer
Let, height of tower AB = h
length of shadow at 30°, CD = 40m
length of shadow at 60°, 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 $20\sqrt{3}\text{m.}$
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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° and 45°. Find the height of the flag-staff and that of the tower.
Answer
Let BC be the tower of height xm and AB be the flag staff of height y, 70m away from the tower, makes an angle of elevation are 60° and 45° respectively from top and bottom of the flag staff.
Let AB = ym, BC = xm and CD = 70m.
$\angle\text{ADC}=45^\circ$ 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 51.24m and height of tower is 70m.
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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°. If the height of the second tree is 80m, find the height of the first tree.
Answer

Distance 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.
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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.
Answer
Let 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 = 4m 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.
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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° with the ground. The distance between the foot of the tree to the point where the top touches the ground is 8m. Find the height of the tree.
Answer

Let 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
$=\frac{16}{\sqrt{3}}+\frac{8}{\sqrt{3}}=\frac{24}{\sqrt{3}}=\frac{24\sqrt{3}}{\sqrt{3}\cdot\sqrt{3}}=8\sqrt{3}$
$=8\sqrt{3}\text{m}$
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Question 203 Marks
From the top of a 7m high building, the angle of elevation of the top of a cable tower is 60° and the angle of depression of its foot is 45°. Determine the height of the tower.
Answer

Given,
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 $=7(\sqrt{3}+1)\text{m.}$
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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 20m high building are 45° and 60° respectively. Find the height of the transmission tower.
Answer
Let AB be the building of height 20m 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° and 60° 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 $20(\sqrt{3}-1)\text{m}.$
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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° and 45° respectively. If the height of the tower is 15m, then find the distance between the points.
Answer
Let 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 $5(3-\sqrt{3})\text{m}.$
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Question 233 Marks
A vertically straight tree, 15m high, is broken by the wind in such a way that it top just touches the ground and makes an angle of 60° with the ground. At what height from the ground did the tree break?
Answer
Let, angle $\angle\text{ACB}=60^\circ$
height of ground break tree = BA = xm
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 6.9m.
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Question 243 Marks
From a point P on the ground the angle of elevation of a 10m tall building is 30°. 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°. Find the length of the flag-staff and the distance of the building from the point P. $(\text{Take }\sqrt{3}=1.732)$
Answer
Let, 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
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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° and 60° respectively. Find the height of the pole.
Answer
Let 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° and 60° 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
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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.
Answer
Let 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.
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Question 273 Marks
The angle of elevation of the top of the building from the foot of the tower is 30° and the angle of the top of the tower from the foot of the building is 60°. If the tower is 50m high, find the height of the building.
Answer
Let AD be the building of height h m. and an angle of elevation of top of building from the foot of tower is 30° and an angle of the top of tower from the foot of building is 60°.
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 $\frac{50}{3}\text{m}.$
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Question 283 Marks
The angle of elevation of the top of a hill at the foot of a tower is 60° and the angle of elevation of the top of the tower from the foot of the hill is 30°. If the tower is 50m high, what is the height of the hill?
Answer
Let 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° and angle of elevation of top of tower from foot of hill is 30°. 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 150m.
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Question 293 Marks
The angle of elevation of the top of a tower at a point on the ground is 30º. What will be the angle of elevation, if the height of the tower is tripled?
Answer

Let 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°.
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Question 303 Marks
A person observed the angle of elevation of the top of a tower as 30°. He walked 50m towards the foot of the tower along level ground and found the angle of elevation of the top of the tower as 60°. Find the height of the tower.
Answer

Given,
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 ⇒ Gb = 60°
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 $\text{H}=25\sqrt{3}\text{m.}$
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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° to 60° as he walks towards the building. Find the distance he walked towards the building.
Answer
Let 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° to 60°.
Therefore $\angle\text{A}=30^\circ$ and $\angle\text{F}=60^\circ$ given CE = 30m, AB = 15m, 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 $19\sqrt{3}\text{m}.$
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Question 323 Marks
The angle of elevation of a ladder leaning against a wall is 60° and the foot of the ladder is 9.5m away from the wall. Find the length of the ladder.
Answer
Let 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.
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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° and that of the top of the flag-staff is 45°. Find the height of the tower.
Answer

Given
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 $=3.5(\sqrt{3}+1)\text{m}.$
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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°. Find the height of the pedestal.
Answer
Let, 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 $0.8(\sqrt{3}+1)\text{m}.$
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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° and 60° respectively. Find the distance between the two men.
Answer
Let, 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.
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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°. From a point 20m away this point on the same bank, the angle of elevation of the top of the tower is 30°. Find the height of the tower and the width of the river.
Answer
Let 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° and from a point 20m away then angle of elevation of tower at the same point is 30°. 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.
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Question 373 Marks
The angles of depression of the top and bottom of 8m tall building from the top of a multistoried building are 30° and 45° respectively. Find the height of the multistoried building and the distance between the two buildings.
Answer
Let AD be the multistoried building of height h m. And angle of depression of the top and bottom are 30° and 45°. 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 $4(3+\sqrt{3})\text{meter}.$
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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 $30\ m$ 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 $B_1B_2 = B_1B + BB_2$
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 $=30(\sqrt{3}+1)\text{m}.$
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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° and 60° respectively. Find
  1. The horizontal distance between AB and CD.
  2. The height of the lamp post.
  3. 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.
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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° and 30° respectively. Find the height of the tower and the flag pole mounted on it.
Answer
Let 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° and 30°.
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 $6\sqrt{3}\text{m.}$
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Question 413 Marks
The angle of elevation of the top of a tower from a point A on the ground is 30°. On moving a distance of 20 metres towards the foot of the tower to a point B the angle of elevation increases to 60°. 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 = 17.32m
Distance of tower from point A = (20 + 10) = 30m
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Question 423 Marks
A man sitting at a height of 20m 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° and 30°respectively. Find the width of the river.
Answer
Let BD be the width of river. And the angles of depression on either side of the river are 30° and 60° 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 $\frac{80}{\sqrt{3}}\text{m.}$
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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° with the level of the ground. Determine the height of the wall.
Answer
Let AB be the wall of height h m and C be the points, makes an angle 60° and foot of the ladder is 2m 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.
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Question 443 Marks
An observed from the top of a 150m tall light house, the angles of depression of two ships approaching it are 30° and 45°. If one ship is directly behind the other, find the distance between the two ships.
Answer
Let AB be the light house of 150m and angle of depression of two ship C and D are 30° and 45° 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
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Question 453 Marks
The height of a tower is 10m. What is the length of its shadow when Sun's altitude is 45°?
Answer
Let, 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.
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Question 463 Marks
A 1.6m tall girl stands at a distance of 3.2m from a lamp-post and casts a shadow of 4.8m on the ground. Find the height of the lamp-post by using (i) trignometric ratios (ii) property of similar triangles.
Answer
Let AC be the lamp post of height h.
We assume that ED = 1.6m, BE = 4.8m and EC = 3.2m
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 $\frac{8}{3}\text{m}.$
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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° than when it was 45°. Prove that the height of tower is $\text{x}(\sqrt{3}+1)\text{ metres.}$
Answer
Let, 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 $\text{x}(\sqrt{3}+1)\text{m.}$
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Question 483 Marks
From the top of a building 15m high the angle of elevation of the top of a tower is found to be 30°. From the bottom of the same building, the angle of elevation of the top of the tower is found to be 60°. Find the height of the tower and the distance between the tower and building.
Answer
Let, 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 12.9m
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Question 493 Marks
From the top of a 120m 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° and 45°. Find the distance between the cars. $(\text{Take }\sqrt{3}=1.732)$
Answer
Let 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
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