Question 15 Marks
The angles of elevation of the top of a tower from two points on the ground at distances $a$ and $b$ metres from the base of the tower and in the same line are complementary. Prove that the height of the tower is sqrtab metre.
Answer
Let $AB$ be the tower of height h metres.
Let $C$ and $D$ be two point on the level ground such that $BC= b$ metres $ , BD= a$ metred $,\angle ACB =∝$ and $\angle ADB =\beta$
given $,∝ +\beta = 90^\circ$
In $\triangle ABC ,$
$\frac{A B}{B C}=\tan \propto$
$\Rightarrow \frac{h}{b}=\tan \propto........(1)$
In $\triangle ABD,$
$\frac{A B}{B D}=\tan \beta$
$\Rightarrow \frac{h}{a}=\tan \left(90^{\circ}-\propto\right)=\cot \propto$
Multiplying $(1)$ by $(2),$ we get ,
$\left(\frac{h}{2}\right)\left(\frac{h}{b}\right)=1$
$\Rightarrow h^2=a b$
$\therefore h=\sqrt{a} {b}$ metre
Hence, height of the tower is $\sqrt{a} b$ metre. View full question & answer→Question 25 Marks
With reference to the given figure, a man stands on the ground at point $A,$ which is on the same horizontal plane as $B,$ the foot of the vertical pole $BC.$ The height of the pole is $10 m.$ The man's eyes $2 m$ above the ground. He observes the angle of elevation of $C,$ the top of the pole, as $x^o,$ where $\tan x^o$ $=\frac{2}{5}$.
Calculate :
$(i)$ the distance $AB$ in metres.
$(ii)$ angle of elevation of the top of the pole when he is standing $15$ metres from the pole. Give your answer to the nearest degree.
Answer
$(i)$ Let $AD$ be the height of the man $, AD = 2m$
$\therefore CE = (10 - 2 )= 8m$
In $\triangle CED,$
$\frac{C E}{D E}=\tan x^{\circ}=\frac{2}{5}$
$\Rightarrow \frac{8}{D E}=\frac{2}{5} \Rightarrow D E=20 m $
Here $AB = DE$
$\therefore AB = 20 m$
$(ii)$ Let $AD$ be the height of the man, $AD = 2 m.$
$\therefore CE = (10 - 2) = 8 m$
Let $A "D"$ be the new position of the man and $\theta$ be angle of elevation of the top of the tower.
so, $D\ 'E = 15 m$
In $\triangle CED,$
$\tan \theta=\frac{C E}{D^{\prime} E}=\frac{8}{15}=0.533$
$\Rightarrow \theta=28^{\circ}$ View full question & answer→Question 35 Marks
A vertical tower stand on a horizontel plane and is surmounted by a vertical flagstaff of height h metre. At a point on the plane, the angle of elevation of the bottom of the flagstaff is $∝$ and at the top of the flagstaff is $\beta$ . prove that the height of the tower is
$\frac{ h \tan \alpha}{\tan \beta-\tan \alpha}$
Answer
Let $AB$ be the tower of height $x$ metre, surmounted by a vertical flagstaff $AD$.
Let $C$ be a point on the plane such that $\angle ACB=∝ ,\angle DCB=\beta$ and $AD = h ,$
In $\triangle ABC,$
$\frac{A B}{B C}=\tan \propto$
$\Rightarrow B C=\frac{X}{\tan \propto}$
In $\triangle DBC ,$
$\frac{B D}{B C}=\tan \beta$
$\Rightarrow B D=\left(\frac{x}{\tan \alpha}\right) \tan \beta\ ($ From $(1))$
$\Rightarrow(h+x) \tan \propto=\tan \beta$
$\Rightarrow x \tan \beta-x \tan \propto=h \tan \propto$
$\therefore=\frac{h \tan \propto}{\tan \beta-\tan \propto}$
Hence, height of the tower is $\frac{h \tan \alpha}{\tan \beta-\tan \alpha}$ View full question & answer→Question 45 Marks
At a point on level ground, the angle of elevation of a vertical tower is found to be such that its tangent is $5/12$. On walking $192$ metres towards the tower, the tangent of the angle is found to be $3/4$. Find the height of the tower.
Answer
Let $AB$ be the vertical tower and $C$ and $D$ be two point such that $CD= 192\ m$.
let $\angle ACB =\theta$ and $\angle ADB = \alpha$.
Given $, \tan \theta=\frac{5}{12}$
$\Rightarrow \frac{A B}{B C}=\frac{5}{12}$
$\Rightarrow A B=\frac{5}{12} B C......(1)$
Also, $\tan \propto=\frac{3}{4}$
$\Rightarrow \frac{A B}{B D}=\frac{3}{4}$
$\Rightarrow \frac{\frac{5}{12} B C}{B D}=\frac{3}{4}$
$\Rightarrow \frac{192+B D}{B D}=\frac{3}{4} \times \frac{12}{5}$
$\Rightarrow BD =240\ m $
$\therefore B C=(192+240)=432\ m$
By $(1), AB = 5/12 x\ \ \ x \ \ \ 432= 180\ m$
Hence, the height of the tower is $180\ m.$ View full question & answer→Question 55 Marks
The angles of depression of two ships $A$ and $B$ as observed from the top of a lighthouse $60\ m$ high are $60^\circ$ and $45^\circ$ respectively. If the two ships are on the opposite sides of the lighthouse, find the distance between the two ships, Give your answer correct to the nearest whole number.
Answer
Let $PQ$ be the lighthouse.
$\Rightarrow PQ = 60$
$\tan 60^{\circ}=\frac{P Q}{A Q}$
$\Rightarrow \sqrt{3}=\frac{60}{A Q}$
$\Rightarrow A Q=\frac{60}{\sqrt{3}}$
$\Rightarrow A Q=\frac{20 \times 3}{\sqrt{3}}$
$\Rightarrow A Q=\frac{20 \times \sqrt{3} \times \sqrt{3}}{\sqrt{3}}$
$\Rightarrow A Q=20 \sqrt{3} m $
In $\triangle PQB$
$\tan 45^{\circ}=\frac{P Q}{Q B}$
$\Rightarrow 1=\frac{60}{Q B}$
$\Rightarrow QB =60\ m $
Now,
$A B=A Q+Q B$
$=20 \sqrt{3}+60$
$=20 \times 1.732+60$
$=94.64$
$=95\ m $ View full question & answer→Question 65 Marks
The horizontal distance between two towers is $120\ m$. The angle of elevation of the top and angle of depression of the bottom of the first tower as observed from the top of the second $9$ is $30^\circ$ and $24^\circ$ respectively. find the height of the two towers. give your answers correct to $3$ significant figures.
Answer
In $\triangle AEC,$
$\tan 30^{\circ}=\frac{A E}{E C}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{A E}{120}$
$\Rightarrow A E=\frac{120}{\sqrt{3}}$
$=\frac{120}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$=\frac{120 \sqrt{3}}{3}$
$=40 \sqrt{3}=40 \times 1.732=69.28\ m $
$\operatorname{In} \triangle B E C,$
$\tan 24^{\circ}=\frac{E B}{E C}$
$\Rightarrow 0.4452=\frac{E B}{120}$
$\Rightarrow E B=53.4242\ m$
Thus, height of first tower,
$AB = AE + EB$
$= 69.28 + 53.424= 122.704 $
$= 123\ m \ \ ($Correct to $3$ singnificant figures$)$
And, height of second tower ,
$CD = EB $
$= 53 . 424\ m = 53.4\ m \ ($correct to $3$ significant figures$)$ View full question & answer→Question 75 Marks
An aeroplane, at an altitude of $250\ m,$ observes the angles of depression of two boats on the opposite banks of a river to be $45^\circ$ and $60^\circ$ respectively. Find the width of the river. Write the answer correct to the nearest whole number.
Answer
Let $A$ be the position of the airplane and let $BC$ be the river.
Let $D$ be the point in $BC$ just below the airplane .
$B$ and $C$ be two boats on the opposite banks of the river with angles of depression $60^\circ$ and $45^\circ$ from $A.$
In $\triangle ADC,$
$\tan 45^{\circ}=\frac{A D}{D C}$
$\Rightarrow 1=\frac{250}{y}$
$\Rightarrow y=250\ m =D C$
In $\triangle ADB,$
$\tan 60^{\circ}=\frac{A D}{B D}$
$\Rightarrow \sqrt{3}=\frac{250}{x}$
$\Rightarrow x=\frac{250}{\sqrt{3}}$
$=\frac{250 \sqrt{3}}{3}$
$=\frac{250 \times 1.732}{3}$
$=144.3 m =B D$
$\therefore B C=B D+D C$
$=144.3+250=394.3 \approx 394\ m$
Thus, the widht of the river is $394\ m$. View full question & answer→Question 85 Marks
In the given figure, from the top of a building $AB = 60\ m$ hight, the angle of depression of the top and bottom of a vertical lamp post $CD$ are observed to be $30^\circ $ and $60^\circ$ respectively. find :
$(1) $ the horizental distance between $AB $ and $CD$ ;
$(2)$ the height of the lamp post.
Answer
Given that $AB$ is a building that is $60 \ m,$ high .
let $BC = DE = X$ and $CD = BE = Y$
$\Rightarrow AE = AB - BE = 60 = Y$
$(1) $ In right $\triangle AED,$
$\tan 30^{\circ} \frac{A E}{D E}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{60-y}{x}$
$\Rightarrow x=60 \sqrt{3}-y \sqrt{3}.......(1)$
In right $\triangle ABC,$
$\tan 60^{\circ}=\frac{A B}{B C}$
$\Rightarrow \sqrt{3}=\frac{60}{x}$
$\Rightarrow x=\frac{60}{\sqrt{3}}$
$\Rightarrow x=\frac{60}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$\Rightarrow x=\frac{60 \sqrt{3}}{3}$
$\Rightarrow x=20 \sqrt{3}$
$\Rightarrow x=20 \times 1.732=34.64\ m$
Thus, the horizental distance between $AB$ and $CD$ is $34.64\ m.$
$(2)$ From $(1),$ we get the height of the lamp post $= CD= y$
$x=60 \sqrt{3}-y \sqrt{3}$
$\Rightarrow 20 \sqrt{3}=60 \sqrt{3}-y \sqrt{3}$
$\Rightarrow 20=6-y$
$\Rightarrow y=40 \ m$
Thus , the geight of the lamp post is $40 \ m .$ View full question & answer→Question 95 Marks
As observed from the top of a $80\ m$ tall lighthouse, the angles of depression of two ships, on the same side of a light house in a horizontal line with its base, are $30^\circ$ and $40^\circ$ respectively. Find the distance between the two ships. Give your answer corrected to the nearest metre.
Answer
Let $AB$ represent the lighthouse.
Let the two ships be at point $D$ and $C$ having angle of depression $30^\circ$ and $40^\circ$ respectively.
Let $x$ be the distance between the two ships.
clearly $, m\angle ACB = 40^\circ$ and $m \angle ADB= 30 ^\circ$
In $\triangle ACB$
$\tan 40^{\circ}=\frac{80}{C B}$
$\Rightarrow C B=\frac{80}{0.84}=95.24\ m$
$\ln \triangle ADB$
$\tan 30^{\circ}=\frac{80}{D B}$
$\Rightarrow D B=\frac{80}{0.58}=137.93\ m$
$DC = DB - CB$
$\Rightarrow X=137.93-95.24$
$\Rightarrow X=42.69 \approx 43\ m$
The distance between the two ship is $43 \ m .$ View full question & answer→Question 105 Marks
A man observes the angle of elevation of the top of a building to be $30^\circ$ . He walks towards it in a horizontal line through its base. On covering $60 \ m,$ the angle of elevation changes to $60^\circ$ . Find the height of the building correct to the nearest metre.
Answer
Let $AB$ be a building and $M$ and $N$ are the two $p\ ($ osition of the man which makes angle of elevation of top of buildings $30^\circ$ and $60^\circ$ respectively$)$
$MN= 60\ m$
Let $AB = h $ and $NB = x\ m$
Now in right $\triangle AMB ,$
$\tan 30^{\circ}=\frac{A B}{M B}$
$\Rightarrow \tan 30^{\circ}=\frac{h}{60+x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{60+x}$
$\Rightarrow 60+\times=\sqrt{3}$
$\Rightarrow x=\sqrt{3} h-60 \ldots . ....(1)$
similaraly in right $\triangle ANB,$
$\tan 60^{\circ} \frac{A B}{N B}$
$\tan 60^{\circ}=\frac{h}{60+x}$
$\Rightarrow \sqrt{3}-\frac{h}{x}$
$\Rightarrow x=\frac{h}{\sqrt{3}} \ldots .......(2)$
from $(1)$ and $(2),$ we have ,
$\sqrt{3}-60=\frac{h}{\sqrt{3}}$
$\Rightarrow 3 h-60 \sqrt{3}=h$
$\Rightarrow 2 h=60 \sqrt{3}$
$\Rightarrow h=\frac{60 \sqrt{3}}{2}$
$\Rightarrow h=30 \sqrt{3}$
$=30 \times 1.732$
$\Rightarrow h=51.96\ m $
$\therefore$ Height of the building $= 51.96=52m \ ($ approx$)$ View full question & answer→Question 115 Marks
From a point $,36 \ m$ above the surface of a lake, the angle of elevation of a bird is observed to be $30^\circ$ and the angle of depression of its image in the water of the lake is observed to be $60^\circ$. Find the actual height of the bird above the surface of the lake.
Answer
Let $A$ be a point $36\ m$ above the surface of the lake and $B$ be the positin of the bird .
Let $B$ be the image of brid in the water .
Here $, AC =DE = 36\ m ,\angle BAE = 30^\circ$ and $\angle B’ AE = 60^\circ ,$
Let $BE = h \ m .$
Then $, B,D = BD=36 + h$
$(\therefore B ’$ is image of $B$ aboutv $D )$
$\therefore B’ E = B’D + DE $
$= 36 + 36 + h = 72 + h ...(1)$
In $\triangle ABE ,$
$\frac{B E}{A E}=\tan 30^{\circ}$
$\Rightarrow A E=\sqrt{3} h \ldots \ldots \ldots . . . .(2)$
$\ln \triangle A B^{\prime} E$
$\frac{B ^{\prime} E}{A E}=\tan 60^{\circ}$
$\Rightarrow \frac{72+h}{A E}=\sqrt{3}(\text { from }(1))$
$\Rightarrow 72+h=(\sqrt{3} h) \sqrt{3}(\text { from }(2))$
$\Rightarrow 72+h=3 h$
$\therefore h-36\ m$
Hence, the actual height of the brid
above the surface of the lake $= 36 + 36 = 72\ m.$ View full question & answer→Question 125 Marks
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is $60^\circ$ and the angle of depression of the bottom of the tower is $30^\circ.$ Find: the height of the pole, if the height of the tower is $75 m.$
Answer
Let $AB$ be the tower and $CD$ be the pole.
Then $\angle ACE = 60^\circ$ and $\angle BCE = 30^\circ$
Let height of the pole be $x \ m$
$\therefore CD=Be=x$
In $\triangle BEC,$
$\frac{B E}{E C}=\tan 30^{\circ}$
$\Rightarrow E C=\sqrt{3} x$
$\operatorname{In} \Delta \frac{A E}{E C}=\tan 60^{\circ}$
$\Rightarrow \frac{75-x}{E C}=\sqrt{3}$
$\Rightarrow 75- x =3 x$
$\therefore x=\frac{75}{4}=18.75 m$
$\therefore$ height of the pole is $18.75m.$ View full question & answer→Question 135 Marks
A vertical pole and a vertical tower are on the same level ground in such a way that from the top of the pole, the angle of elevation of the top of the tower is $60^\circ$ and the angle of depression of the bottom of the tower is $30^\circ.$ Find: the height of the tower, if the height of the pole is $20\ m.$
Answer
Let $AB$ be the tower and $CD$ be the pole.
Then $\angle ACE=60^\circ$ and $\angle BCE=30^\circ$
In $\triangle BEC,$
$\frac{B E}{E C}=\tan 30^{\circ}$
$\Rightarrow \frac{20}{E C}=\frac{1}{\sqrt{3}}$
$\Rightarrow E C=20 \sqrt{3}\ m$
In $\triangle AEC,$
$\frac{A E}{E C}=\tan 60^{\circ}$
$\Rightarrow A E=20 \sqrt{3} \times \sqrt{3}=60\ m$
$\therefore$ Height of the tower $=A B=A E+E B$
$=(60+20)=80\ m $ View full question & answer→Question 145 Marks
A $20 \ m$ high vertical pole and a vertical tower are on the same level ground in such a way that the angle of elevation of the top of the tower, as seen from the foot of the pole is $60^\circ$ and the angle of elevation of the top of the pole, as seen from the foot of the tower is $30^\circ$ . Find:
$(i)$ the height of the tower ;
$(ii)$ the horizontal distance between the pole and the tower.
Answer
Let $AB$ be the tower and $CD$ be the pole .
Given $, CD= 20m,\angle ADB = 60^\circ$ and $\angle CBD= 30^\circ$
In $\triangle BDC,$
$\frac{C D}{B D}=\tan 30^{\circ}$
$\Rightarrow B D=20 \text { sqrt } 3\ m$
$\ln \triangle D B A,$
$\frac{A B}{B D}=\tan 60^{\circ}=\sqrt{3}$
$\Rightarrow A B=20 \text { sqrt } 3 x \times \operatorname{sqrt} 3=60\ m $
Hence ,
$(1)$ Height of the tower $= 60\ m$
$(2)$ horizontel distan ce between the pole and tower
$=20 \times 1.732=34.64\ m$ View full question & answer→Question 155 Marks
A man standing on the bank of a river observes that the angle of elevation of a tree on the opposite bank is $60^\circ$ . When he moves $50\ m$ away from the bank, he finds the angle of elevation to be $30^\circ$ . Calculate:
$(i)$ the width of the river;
$(ii)$ the height of the tree.
Answer
Let $AB$ be the tree and $AC$ be the width of the river.
let $D$ be a point such that $CD - 50\ m$ .
Given that $\angle BCA = 60^\circ$ and $\angle BDA= 30^\circ$ .
In $\triangle BAD,$
$\frac{B A}{A D}=\tan 30^{\circ}$
$\Rightarrow B A=\frac{A D}{\sqrt{3}} ........(1)$
In $\triangle BAC,$
$\frac{B A}{A C}=\tan 60^{\circ}$
$\Rightarrow B A=A C \sqrt{3} .......(2)$
From $(1)$ and $(2),$ we get
$\frac{A D}{\sqrt{3}}-A C \sqrt{3}$
$\Rightarrow (50 + AC)-3AC$
$\therefore AC = 25\ m$
Thus , width of the river is $25\ m .$
from $(2)$
$BA - 25 x \ \ 1.732 = 432.3\ m$
Hence , height of the tree is $43.3\ m$ . View full question & answer→Question 165 Marks
A man on a cliff observes a boat, at an angle of depression $30^\circ,$ which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat is found to be $60^\circ$ . Assuming that the boat sails at a uniform speed, determine:
$(i) $ how much more time it will take to reach the shore.
$(ii)$ the speed of the boat in metre per second, if the height of the cliff is $500 \ m.$
Answer
Let $AB$ be the cliff and $C$ and $D$ be the two position of the boat such that $\angle ADE = 30^\circ$ and $\angle ACB = 60^\circ$
let speed of the boat be $X$ metre per minute and let the boat reach the shore after $t$ minutes more .
In $\triangle ABC$
$\frac{A B}{B C}=\tan 60^{\circ}=\sqrt{3}$
$\Rightarrow \frac{h}{t x}=\sqrt{3}$
$\ln \triangle ADB ,$
$\frac{A B}{D B}= \tan 30^{\circ}$
$\Rightarrow \frac{h}{3 x+t x}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{\sqrt{3} t}{3+t}=\frac{1}{\sqrt{3}}$
$\Rightarrow 3 t =3+ t$
$\therefore t =\frac{3}{2}=1.5 $ minute
Also , if $h= 500m,$ then
$\frac{500}{1.5 x}=\sqrt{3}$
$\Rightarrow x=\frac{500}{1.5 \times 1.732}=192.455$ metre per minute
Hence, the boat takes an extra $1.5$ minutes to reach the shore .
and, if the height of clife is $500\ m$. the speed of the boat is $3.21 m/ \sec$ View full question & answer→Question 175 Marks
From the top of a cliff, $60$ metres high, the angles of depression of the top and bottom of a tower are observed to be $30^\circ$ and $60^\circ$ . Find the height of the tower.
Answer
Let $AB$ be the cliff and $CD$ be the tower.
Here $AB = 60\ m \ \angle ADE = 30^\circ $ and $\angle ACB = 60^\circ$
In $\triangle ABC ,$
$\frac{A B}{B C}=\tan 60^{\circ}=\sqrt{3}$
$\Rightarrow BC =\frac{60}{\sqrt{3}}$
$\operatorname{In} \triangle ADE ,$
$\frac{ AE }{ DE }=\tan 30^{\circ}$
$\Rightarrow AE = DE\ \tan 30^{\circ}$
$=\frac{60}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\ (\because DE = BC )$
$=20\ m $
$\therefore CD = EB = AB - AE $
$= (60 - 20) = 40\ m$
Hence, height of the tower is $40 \ m$. View full question & answer→Question 185 Marks
Two pillars of equal heights stand on either side of a roadway, which is $150\ m$ wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are $60^\circ$ and $30^\circ$ ; find the height of the pillars and the position of the point.
Answer
Let $AB$ and $CD$ be the two towers of height $h \ m$.
Let $P$ be a point in the roadway $BD$ such that $BD = 150\ m, $
$\angle APB = 60^\circ$ and $\angle CPD = 30^\circ$
In $\triangle ABP$
$\frac{ AB }{ BP }=\tan 60^{\circ}$
$\Rightarrow BP =\frac{ h }{\tan 60^{\circ}}=\frac{ h }{\sqrt{3}}$
$\ln \triangle CDP$
$\frac{ CD }{ DP }=\tan 30^{\circ}$
$\Rightarrow PD =\frac{1}{\sqrt{3}}$
Now $, 150 = BP + PD$
$\Rightarrow 150=\frac{ h }{\sqrt{3}}+\frac{1}{\sqrt{3}}$
$\therefore h =\frac{150}{\sqrt{3}+\frac{1}{\sqrt{3}}}$
$=\frac{150}{2.309}=64.95 m$
Hence, height of the piller is $64.95\ m$
The point is $\frac{ BP }{\sqrt{3}}$ from the first pillar.
That is the position of the point is $\frac{64.95}{\sqrt{3}} \ m$ from the first pillar
The position of the point is $37.5\ m$ from the first pillar. View full question & answer→Question 195 Marks
From the top of a light house $100\ m$ high, the angles of depression of two ships are observed as $48^\circ$ and $36^\circ$ respectively. Find the distance between the two ships $($in the nearest metre$)$ if: $(i)$ the ships are on the same side of the light house. $(ii) $ the ships are on the opposite sides of the light house.
Answer
Let $AB$ the lighthouse.
let the two ship be $C$ and $D$ such that $\angle ADB= 36^\circ$ And $\angle ACB= 48^\circ$
In $\triangle ABC$
$\frac{A B}{B C}=\tan 48^{\circ}$
$\Rightarrow B C=\frac{100}{1.1106}=90.04\ m $
$\text { In } \triangle ABD$
$\frac{A B}{B D}=\tan 36^{\circ}$
$\Rightarrow B D=\frac{100}{0.7265}=137.64\ m $
$(1)$ if the ships are on the same side of the light house.
then distance between the two ships $= BD -BC= 48\ m$
$(2)$ if the ship are on the opposite side of the light house,
then distance between the tow ship $= BD + BC = 228\ m$ View full question & answer→Question 205 Marks
Find the height of a building, when it is found that on walking towards it $40\ m$ in a horizontal line through its base the angular elevation of its top changes from $30^\circ$ to $45^\circ$
Answer
Let $AB$ be the building of height $h\ m.$
Let the two points be $C$ and $D$ such that $CD = 40 \ m, $
$\angle ADB =30^\circ , \angle ACB = 45^\circ$
In $\triangle ABC$
$\frac{A B}{B C}=\tan 45^{\circ}=1$
$\Rightarrow B C= h$
$\ln \triangle ABD$
$\frac{A B}{B D}=\tan 30^{\circ}$
$\Rightarrow \frac{h}{40+h}=\frac{1}{\sqrt{3}}$
$\Rightarrow \sqrt{3} h=40+h$
$\therefore h=\frac{40}{\sqrt{3}-1}$
$=\frac{40}{0.732}=54.64\ m$
Hence, height of the building is $54. 64\ m$ View full question & answer→Question 215 Marks
Find the height of a tree when it is found that on walking away from it $20\ m,$ in a horizontal line through its base, the elevation of its top changes from $60^\circ$ to $30^\circ $.
Answer
let $AB$ be the tree of height $h\ m.$
Let the two points be $C$ and $D$ such that $CD\ m$ .
$\angle ADB = 30^\circ$ and $\angle ACB= 60^\circ$
In $\triangle ABC ,$
$\frac{A B}{B C}=\tan 60^{\circ}=\sqrt{3}$
$\ln \triangle A B D$
$\frac{A B}{B D}=\tan 30^{\circ}$
$\Rightarrow \frac{h}{20+B C}=\frac{1}{\sqrt{3}}$
$\Rightarrow \sqrt{3} h=20+\frac{h}{\sqrt{3}}$
$\therefore h=\frac{20}{\sqrt{3}-\frac{1}{\sqrt{3}}}$
$=\frac{20}{1.154}=17.32 m$
Hence, height of the tree is $17.32\ m$ View full question & answer→Question 225 Marks
From the top of a hill, the angles of depression of two consecutive kilometer stones, due east, are found to be $30^\circ$ and $45^\circ$ respectively. Find the distances of the two stones from the foot of the hill.
Answer
Let $AB$ be the hill of height $'h\ ' \ km$ and $C$ and $D$ be two consecutive stones such that $CD= 1 \ km ,$
$ \angle ACB = 30^\circ$ and $\angle ADB = 45^\circ ,$
In $\triangle ABD,$
$\frac{A B}{B D}=\tan 45^{\circ}=1$
$\Rightarrow BD = h $
In $ \triangle ABC,$
$\frac{A B}{B C}=\tan 30^{\circ}$
$\Rightarrow \frac{h}{B C}=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{h}{h+1}=\frac{1}{\sqrt{3}}$
$\Rightarrow h=\frac{1}{\sqrt{3}-1}=\frac{\sqrt{3}+1}{2}$
$=\frac{2.732}{2}=1.366 \ km .$
$\therefore BD = 1.366 \ km$
$BC = BD + DC = 1.366 + 1$
$= 2.366 \ km.$
Hence, the two stone are at a distance of $1.366 \ km$ and $2.366 \ km$ from the foot of the hill. View full question & answer→Question 235 Marks
An aeroplane flying horizontally $1 \ km$ above the ground and going away from the observer is observed at an elevation of $60^\circ$ . After $10$ seconds, its elevation is observed to be $30^\circ$ ; find the uniform speed of the aeroplane in $\ km$ per hour.
Answer
Let $A$ be the aeroplane and $B$ be the observer on the ground.
The vertical height will be $AC = 1\ km=1000\ m$ .
After seconds. let the aeroplane be at point $D$.
Let the speed of the aeroplane be $x m / \sec.$
$\therefore CE = 10x$
In $\triangle ABC ,$
$\frac{A C}{B C}=\tan 60^{\circ}$
$\Rightarrow \frac{1000}{B C}=\sqrt{3}$
$\Rightarrow B C=\frac{1000}{\sqrt{3}} m$
In $\triangle BDE ,$
$\frac{D E}{B E}=\tan 30^{\circ}$
$\Rightarrow BE =1000 \sqrt {3}$
$\therefore CE = BE - BC $
$\Rightarrow 10 x=1000 \sqrt{3}-\frac{1000}{\sqrt{3}}$
$x=100\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right)=100 \times 1.154$
$= 115.4 m⁄ \sec$
$= 115.4 x\ \ 18\ km⁄ hr $
$= 415.67 \ km ⁄ hr$
Hence , speed of the aeroplane is $415.67 \ km /hr.$ View full question & answer→Question 245 Marks
The length of the shadow of a tower standing on level plane is found to be $2y$ metres longer when the sun’s altitude is $30^\circ$ than when it was $45^\circ$ . Prove that the height of the tower is $y(\sqrt3 + 1)$ metres.
Answer
Let $AB$ be the tower $C$ and $D$ are two points that $CD = 2y\ m ,$
$\angle ADB = 45^\circ$ and $\angle ACB= 30 ^\circ$
In $\triangle ABD,$
$\frac{A B}{B D}=\tan 45^{\circ}=1$
$\Rightarrow h = BD$
In $\triangle ABC,$
$\frac{ AB }{ BC }=\tan 30^{\circ}$
$\Rightarrow \frac{h}{B C}=\frac{1}{\sqrt{3}}$
$\Rightarrow B C=\sqrt{3} h$
$\Rightarrow C D+h=\sqrt{3} h$
$\Rightarrow 2 y=(\sqrt{3}-1) h$
$\therefore h=\frac{2 y}{\sqrt{3}-1}$
$=\frac{2 y(\sqrt{3}+1)}{(\sqrt{3})^2-1}$
$=y(\sqrt{3}+1) m$
Hence, height of the tower is $y(\sqrt{3}-1) m$. View full question & answer→Question 255 Marks
A person standing on the bank of a river observes that the angle of elevation of the top of a tree standing on the opposite bank is $60^\circ$. When he moves $40\ m$ away from the bank, he finds the angle of elevation to be $30^\circ$ . Find:
- the height of the tree, correct to $2$ decimal places,
- the width of the river
Answer
Let $AB$ be the tree of height $'h\ ' m$ and $BC$ be the width of the river.
let $D$ be the point on the opposite bank of tree such that $CD = 40\ m$.
Here $\angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$
let the speed of the boat be $'x\ '$ metre per minute
In $\triangle ABC,$
$\frac{ AB }{ BC }=\tan 60^{\circ}=\sqrt{3}$
$\Rightarrow \frac{ h }{ BC }=\sqrt{3}$
$\Rightarrow h=B C \sqrt{3}$
ln $ \triangle A D B$
$\frac{ AB }{ BD }=\tan 30^{\circ}$
$\Rightarrow \frac{ h }{40+ BC }=\frac{1}{\sqrt{3}}$
$\Rightarrow \frac{ BC \sqrt{3}}{40+ BC }=\frac{1}{\sqrt{3}}$
$\Rightarrow BC \sqrt{3} \cdot \sqrt{3}=40+ BC$
$\Rightarrow 3 BC =40+ BC$
$\Rightarrow 3 BC - BC =40$
$\Rightarrow 2 BC =40\ m$
$\Rightarrow BC =\frac{40}{2}\ m$
$\Rightarrow BC =20\ m $
$\therefore h = 20 \times 1.732 $
$= 34.64 m$
Hence, height of the tree is $34.64\ m$ and width of the river is $20\ m.$ View full question & answer→Question 265 Marks
A man in a boat rowing away from a lighthouse $150\ m$ high, takes $2$ minutes to change the angle of elevation of the top of the lighthouse from $60^\circ$ to $45^\circ$. Find the speed of the boat.
Answer
let $AB$ the lighthouse and $C $ and $D$ be the two position of the boat such that $AB = 150\ m,$
$\angle ADB = 45^\circ$ and $\angle ACB= 60^\circ$
Let speed of the boat be $x$ metre per minute
Therfore $, CD = 2 x\ m :$
In $\triangle ADB,$
$\frac{A B}{D B}-\tan 45^{\circ}$
$\Rightarrow BD = 150 \ m$
In $\triangle ABD,$
$\frac{A B}{B C}=\tan 60^{\circ}=\sqrt{3}$
$\Rightarrow \frac{150}{B C}=\sqrt{3}$
$\Rightarrow B C=\frac{150}{\sqrt{3}}$
$=\frac{150}{1.732}=86.605\ m $
$\therefore CD = BD - BC$
$= 150 - 86.605 - 63.395\ m$
$\Rightarrow 2\ x=63.395$
$\Rightarrow x=\frac{63.395}{2}=31.6975\ m / \min$
$=\frac{31.6975}{60}\ m / \sec =0.53 m / \sec $
Hence , the speed of the boat is $0.53\ m⁄ \sec$ View full question & answer→Question 275 Marks
At a particular time, when the sun’s altitude is $30^\circ ,$ the length of the shadow of a vertical tower is $45\ m$. Calculate
$(i)$ the length of the tower.
$(ii)$ the length of the shadow of the same tower, when the sun’s altitude is $(a) 45^\circ\ (b) 60^\circ$
AnswerLet the lenght of the tower be $h \ m .$
$(1)$ here $\theta = 30^\circ$
$\tan 30^{\circ}=\frac{h}{45}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{45}$
$\Rightarrow h=25.98\ m$
$(a)$ here $,\theta = 45^\circ$
$\tan 45^{\circ}=\frac{25.98}{x}$
$\Rightarrow 1=\frac{25.98}{x}$
$\Rightarrow x= 25.98\ m$
Hence the length of the shadow is $25.98\ m$
$(b)$ Here $\theta = 60^\circ$
$ \tan 60^{\circ}=\frac{25.98}{x}$
$\Rightarrow \sqrt{3}=\frac{25.98}{x}$
$\Rightarrow x=\frac{25.98}{\sqrt{3}}=15 m $
Hence the length of the shadow is $15\ m.$
View full question & answer→Question 285 Marks
A boy, $1.6 \ m$ tall, is $20\ m$ away from a tower and observes theangle of elevation of the top of the tower to be
$(i) 45^\circ , (ii) 60^\circ$ . Find the height of the tower in each case.
Answer
Let the height of the tower be $h \ m$.
$(1) $ here $\theta = 45^\circ$
$\therefore \tan 45^{\circ}=\frac{h-1.6}{20}$
$\Rightarrow 1=\frac{h-1,6}{20}$
$h=21.6\ m$
so height of the tower is $21.6\ m$
$(2)$ Here $\theta = 60^\circ$
$\therefore \tan 60^{\circ}=\frac{h-1.6}{20}$
$\Rightarrow \sqrt{3}=\frac{h-1.6}{20}$
$\therefore h=20 \times 1.732+1.6$
$=36.24 \ m$
so , height of the tower is $36 .24$ View full question & answer→Question 295 Marks
Two persons are standing on the opposite sides of a tower. They observe the angles of elevation of the top of the tower to be $30^\circ$ and $38^\circ $ respectively. Find the distance between them, if the height of the tower is $50 \ m$.
Answer
let one person $A$ be at distance $X$ and second person $B$ be at a distence of $Y$ from the foot of the tower .
Given that angle of elevation of $A$ is $30^\circ$
$\tan 30^{\circ}=\frac{50}{x}$
$\Rightarrow \frac{1}{\sqrt{3}}=\frac{50}{x}$
$\therefore x=50 \sqrt{3}=86.60\ m $
the angle of elevation of $B$ is $38^\circ$
$\tan 38^{\circ}=\frac{50}{y}$
$\Rightarrow 0.7813=\frac{50}{y}$
$y \approx 64\ m$
so, distance between $A$ and $B$ is $x + y = 150.6\ m$ View full question & answer→