Concise Mathematics Class 10 ICSE Solutions for Chapter 22 - Heights and Distances

EXERCISE 22(A)

1. The height of a tree is $\sqrt{3}$ times the length of its shadow. Find the angle of elevation of the sun.

Ans: Let the length of the shadow of the tree be $‘x^{\prime }$.

Therefore the height of the tree is $\sqrt{3}x$.

Here, $\theta$ is the angle of elevation of the sun. then,

From $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan\theta ={\displaystyle \frac{\sqrt{3}x}{x}}$

$\Rightarrow tan\theta =\sqrt{3}$

$\Rightarrow \theta =ta{n}^{-1}\sqrt{3}$

$\Rightarrow \theta ={60}^o$

So, the angle of elevation is 60°.

2. The angle of elevation of the top of a tower, from a point on the ground and at a distance of 160 m from its foot, is found to be 60°. Find the height of the tower.  tower. 

Ans: Let the height of the tower be $‘h^{\prime }$.

The angle of elevation is 60°.

From $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{160}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h}{160}}$

$\Rightarrow h=160\sqrt{3}$m

$\Rightarrow \theta =277.13$m

So, the height of the tower is 277.13m.

3. A ladder is placed along a wall such that its upper end is resting against a vertical wall. The foot of the ladder is 2.4 m from the wall and the ladder is making an angle of 68° with the ground. Find the height, upto which the reaches. 

Ans: Let height upto which the ladder reaches be $‘h^{\prime }$.

The angle of elevation is 68°.

From $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{68}^o={\displaystyle \frac{h}{2.4}}$

$\Rightarrow 2.475={\displaystyle \frac{h}{2.4}}$

$\Rightarrow h=2.475\times 2.4$

$\Rightarrow \theta =5.94$m

So, the height up to which the ladder reaches is 5.94m.

4. Two persons are standing on the opposite sides of the tower. They observe the angles of elevation of the top of the tower to be 30° and 38° respectively. Find the distance between them, if the height of the tower is 50m. 

Ans: Let the distance between both the persons be $x+y$.

The angle of elevation of A is 30°.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{38}^o={\displaystyle \frac{50}{x}}$

$\Rightarrow 0.7813={\displaystyle \frac{50}{x}}$

$\Rightarrow x={\displaystyle \frac{50}{0.7813}}$

$\Rightarrow x=63.99$m

$\Rightarrow x\approx 64$m

From triangle $\mathrm{\Delta }$BCD,

$\Rightarrow tan\theta ={\displaystyle \frac{BC}{BD}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{50}{y}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{50}{y}}$

$\Rightarrow y=50\sqrt{3}$

$\Rightarrow y=50\times 1.732$

$\Rightarrow y=86.6$m

Therefore the distance between both the persons will be,

$d=x+y$

$\Rightarrow d=64+86.6$

$\Rightarrow d=150.6$m

5. A kite is attached to a string. Find the length of the string, when the height of the kite is 60m and the string makes an angle 30° with the ground. 

Ans: Let the length of the string be L.

From triangle $\mathrm{\Delta }$ABC,

$sin\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow sin{30}^o={\displaystyle \frac{60}{L}}$

$\Rightarrow {\displaystyle \frac{1}{2}}={\displaystyle \frac{60}{x}}$

$\Rightarrow x=60\times 2$

$\Rightarrow x=120$m

So, the length of the string is 120m

6. A boy, l.6m tall, is 20m away from a tower and observes the angle of elevation on the top of the tower to be (i) 45° (ii) 60°. Find the height of the tower in each case.

Ans: Let the height of the tower be ‘$h$’.

(i) If $\theta ={45}^o$

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{h-1.6}{20}}$

$\Rightarrow 1={\displaystyle \frac{h-1.6}{20}}$

$\Rightarrow h-1.6=20$

$\Rightarrow h=20+1.6$

$\Rightarrow h=21.6$m

So, the height of the tower is 21.6m.

(ii) If $\theta ={60}^o$

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h-1.6}{20}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h-1.6}{20}}$

$\Rightarrow h-1.6=20\sqrt{3}$

$\Rightarrow$ $h=(20\times 1.732)+1.6$

$\Rightarrow h=34.64+1.6$

$\Rightarrow h=36.24$m

So, the height of the tower is 36.24m.

7. The upper part of a tree, broken over by the wind, makes an angle of 45° with the ground; and the distance from the root to the point where the top of the tree touches the ground is 15 m. What was the height of the tree before it was broken?

Ans: Let the height of the braking tree be ‘$h$’

The angle of elevation is 45°.

The height of the tree before its braking is AC + BC

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{BC}{15}}$

$\Rightarrow 1={\displaystyle \frac{BC}{15}}$

$\Rightarrow BC=15$m

Now, length of the tree broken by the wind,

$Cos\theta ={\displaystyle \frac{AB}{AC}}$

$\Rightarrow Cos{45}^o={\displaystyle \frac{15}{AC}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{2}}}={\displaystyle \frac{15}{AC}}$

$\Rightarrow 0.7071={\displaystyle \frac{15}{AC}}$

$\Rightarrow AC={\displaystyle \frac{15}{0.7071}}$

$\Rightarrow AC={\displaystyle \frac{150000}{7071}}$

$\Rightarrow AC=21.21$m

So, the height of the tree before its braking is,

$h=AC+BC$

$\Rightarrow h=21.21+15$

$\Rightarrow h=36.21$m

8. The angle of elevation of the top of an unfinished tower from a point at a distance of 80m from its base is 30°. How much higher must the tower be raised so that its angle of elevation at the same point may be 60°?

Ans: Let ‘$h$’ be the height of that tower to rise.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{BC}{80}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{BC}{80}}$

$\Rightarrow BC=80\times {\displaystyle \frac{1}{\sqrt{3}}}$

$\Rightarrow BC=80\times 0.57735$

$\Rightarrow BC=46.188$m

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{BD}{AB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{46.188+h}{80}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{46.188+h}{80}}$

$\Rightarrow 46.188+h=80\sqrt{3}$

$\Rightarrow 46.188+h=80\times 1.732$

$\Rightarrow 46.188+h=138.564$

$\Rightarrow h=138.564-46.188$

h = 92.376 m

Therefore the tower must be raised by 92.376m.

9. At a particular time, when the sun's altitude is 30°, the length of the shadow of a vertical tower is 45m. Calculate: 

(1) the height of the tower, 

(ii) the length of the shadow of the same tower, when the sun's altitude is : 

(a) 45° 

(b) 60°.

Ans: (i) Let ‘$h$’ be the height of the tower.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{45}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{45}}$

$\Rightarrow h={\displaystyle \frac{45}{\sqrt{3}}}$

$\Rightarrow h={\displaystyle \frac{45}{1.732}}$

$h=25.98$m

(ii) Length of the shadow if θ = ${45}^o$

$\Rightarrow tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{h}{AB}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{45}}$

$\Rightarrow h={\displaystyle \frac{45}{\sqrt{3}}}$

$\Rightarrow h={\displaystyle \frac{45}{1.732}}$

$h=25.98$m

Length of the shadow if θ = 60°,

$\Rightarrow tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{AB}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{25.98}{AB}}$

$\Rightarrow AB={\displaystyle \frac{25.98}{\sqrt{3}}}$

$\Rightarrow AB={\displaystyle \frac{25.98}{1.732}}$

$\Rightarrow AB=15$m

So the length of the shadow is 15m.

10. Two vertical poles are on either side of a road. A 30 m long ladder is placed between the two poles. When the ladder rests against one pole, it makes an angle 32°24' with the pole and when it is turned to rest against another pole. It makes an angle 32°24' with the road. Calculate the width of the road. 

Ans: 

Here, the width of the road is DB.

$\mathrm{\angle }BAC$= 32°24' 

$=32+{\displaystyle \frac{24}{60}}$

$=32+0.4$

$={32.4}^o$

From triangle $\mathrm{\Delta }$ADE,

$\Rightarrow sin\theta ={\displaystyle \frac{DA}{EA}}$

$\Rightarrow sin{32.4}^o={\displaystyle \frac{DA}{30}}$

$\Rightarrow 0.535={\displaystyle \frac{DA}{30}}$

$\Rightarrow DA=0.535\times 30$

$\Rightarrow DA=16.07$m

From triangle $\mathrm{\Delta }$ABC,

$cos\theta ={\displaystyle \frac{AB}{AC}}$

$\Rightarrow cos{32.4}^o={\displaystyle \frac{AB}{30}}$

$\Rightarrow 0.844={\displaystyle \frac{AB}{30}}$

$\Rightarrow AB=0.844\times 30$

$\Rightarrow AB=25.32$m

Therefore, width of the road will be,

$\Rightarrow W=DA+AB$

$\Rightarrow W=16.07+25.32$

$\Rightarrow W=41.4$m

11. Two climbers are at points A and B on a vertical cliff face. To an observer C, 40m from the foot of the cliff, on the level ground, A is at an elevation of 48° and B of 57°. What is the distance between the climbers?

Ans: Let the distance between the climbers is ‘$d$’.

From triangle $\mathrm{\Delta }$ADC,

$\Rightarrow tan\theta ={\displaystyle \frac{AD}{CD}}$

$\Rightarrow tan{48}^o={\displaystyle \frac{AD}{40}}$

$\Rightarrow 1.1106={\displaystyle \frac{AD}{40}}$

$\Rightarrow AD=40\times 1.1106$

$\Rightarrow AD=44.42$m

From triangle $\mathrm{\Delta }$BDC,

$tan\theta ={\displaystyle \frac{BD}{CD}}$

$\Rightarrow tan{57}^o={\displaystyle \frac{BD}{40}}$

$\Rightarrow 1.539={\displaystyle \frac{BD}{40}}$

$\Rightarrow BD=1.539\times 40$

$\Rightarrow BD=61.59$m

Therefore the distance between the climber will be,

$\Rightarrow d=BD-AD$

$\Rightarrow d=61.59-44.42$

$\Rightarrow d=17.17$

12. A man stands 9m away from a flag-pole. He observes that the angle of elevation of the top of the pole is ${28}^o$ and the angle of depression of the bottom of the pole is ${28}^o$. Calculate the height of the pole. 

Ans:

The length of the flag-pole will be BD + AD

From triangle $\mathrm{\Delta }$ADC,

$\Rightarrow tan\theta ={\displaystyle \frac{AD}{CD}}$

$\Rightarrow tan{13}^o={\displaystyle \frac{AD}{9}}$

$\Rightarrow 0.231={\displaystyle \frac{AD}{9}}$

$\Rightarrow AD=9\times 0.231$

$\Rightarrow AD=2.079$m

From triangle $\mathrm{\Delta }$BDC,

$tan\theta ={\displaystyle \frac{BD}{CD}}$

$\Rightarrow tan{28}^o={\displaystyle \frac{BD}{9}}$

$\Rightarrow 0.531={\displaystyle \frac{BD}{9}}$

$\Rightarrow BD=9\times 0.531$

$\Rightarrow BD=4.779$m

Therefore the height of the flag-pole will be,

$\Rightarrow h=BD+AD$

$\Rightarrow h=4.779+2.079$

$\Rightarrow h=6.858$m

13. From the top of a cliff 92m high, the angle of depression of a buoy is 20°. Calculate, to the nearest metre, the distance of the buoy from the foot of the cliff. 

Ans: 

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{20}^o={\displaystyle \frac{92}{AB}}$

$\Rightarrow 0.363={\displaystyle \frac{92}{AB}}$

$\Rightarrow AB={\displaystyle \frac{92}{0.363}}$

$\Rightarrow AB=253.44$m

So, the distance of the buoy from the foot of the cliff is 253.44m.

EXERCISE 22(B)

1. In the figure, given below, it is given that AB is perpendicular to BD and is of length X metres. DC = 30 m, $\mathrm{\angle }ADB$ = 30° and $\mathrm{\angle }ACB$ = 45°. Without using tables, find X.

Ans: From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{AB}{CB}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{X}{CB}}$

$\Rightarrow 1={\displaystyle \frac{X}{CB}}$

$\Rightarrow CB=X$

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{AB}{DB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{X}{DC+CB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{X}{30+X}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{X}{30+X}}$

$\Rightarrow 30+X=X\sqrt{3}$

$\Rightarrow 30=X\sqrt{3}-X$

$\Rightarrow 30=X(\sqrt{3}-1)$

$\Rightarrow X={\displaystyle \frac{30}{\sqrt{3}-1}}$. 

$\Rightarrow X={\displaystyle \frac{30}{0.732}}$

$\Rightarrow X=40.98$m

2. 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° to 30°.

Ans: Let the height of the tree be ‘$h$’.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{AB}{CA}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{CA}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h}{CA}}$

$\Rightarrow CA={\displaystyle \frac{h}{\sqrt{3}}}$

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{AB}{DA}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{DC+CA}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{20+{\displaystyle \frac{h}{\sqrt{3}}}}}$

$\Rightarrow 20+{\displaystyle \frac{h}{\sqrt{3}}}=h\sqrt{3}$

$\Rightarrow 20=h\sqrt{3}-{\displaystyle \frac{h}{\sqrt{3}}}$

$\Rightarrow h(\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}})=20$

$\Rightarrow h(1.732-{\displaystyle \frac{1}{1.732}})=20$

$\Rightarrow h(1.732-0.577)=20$

$\Rightarrow h\left(1.155\right)=20$

$\Rightarrow h={\displaystyle \frac{20}{1.155}}$

$\Rightarrow h=17.316$m

So, the height of the tree is 17.316m.

3. Find the height of a building, when it is found that on walking towards it 40m in a horizontal line through its base the angular elevation of its top changes from 30° to 45°.

Ans: Let the height of the building is ‘$h$’.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{AB}{CA}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{h}{CA}}$

$\Rightarrow 1={\displaystyle \frac{h}{CA}}$

$\Rightarrow CA=h$

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{AB}{DA}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{DC+CA}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{40+h}}$

$\Rightarrow \sqrt{3}h=40+h$

$\Rightarrow \sqrt{3}h-h=40$

$\Rightarrow h(\sqrt{3}-1)=40$

$\Rightarrow h={\displaystyle \frac{40}{\sqrt{3}-1}}$

$\Rightarrow h={\displaystyle \frac{40}{1.732-1}}$

$\Rightarrow h={\displaystyle \frac{40}{0.732}}$

$h=56.64$m

So, the height of the building is 54.64m.

4. From the top of a lighthouse 100m high, the angles of depression of two ships are observed as 48° and 36° 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, 

Ans:

Let distance between the two ships is ‘$d$’.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{36}^o={\displaystyle \frac{100}{AB}}$

$\Rightarrow 0.726={\displaystyle \frac{100}{AB}}$

$\Rightarrow AB={\displaystyle \frac{100}{0.726}}$

$\Rightarrow AB=137.64$m

From triangle $\mathrm{\Delta }$BCD,

$tan\theta ={\displaystyle \frac{BC}{BD}}$

$\Rightarrow tan{48}^o={\displaystyle \frac{100}{BD}}$

$\Rightarrow 1.1106={\displaystyle \frac{100}{BD}}$

$\Rightarrow BD={\displaystyle \frac{100}{1.1106}}$

$\Rightarrow BD=90.04$m

The distance between two ships will be,

$\Rightarrow d=AB-BD$

$\Rightarrow d=137.64-90.04$

$\Rightarrow d=47.6$m

(ii) the ships are on the opposite sides of the light house.

Ans: 

The distance between two ships will be,

$d=AB-BD$

$\Rightarrow d=137.64+90.04$

$\Rightarrow d=227.68$m

5. Two pillars of equal heights stand on either side of a roadway, which is 150m wide. At a point in the roadway between the pillars the elevations of the tops of the pillars are 60° and 30°; find the height of the pillars and the position of the point.

Ans: Let the height of the pillar be ‘$h$’,

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{AB}{AD}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{AD}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{AD}}$

$\Rightarrow AD=\sqrt{3}h$

From triangle $\mathrm{\Delta }$CDE,

$tan\theta ={\displaystyle \frac{EC}{CD}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{CD}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h}{CD}}$

$\Rightarrow CD={\displaystyle \frac{h}{\sqrt{3}}}$

The width of the roadway is 150m.

Therefore $CD+AD=150$

$\Rightarrow {\displaystyle \frac{h}{\sqrt{3}}}+\sqrt{3}h=150$

$\Rightarrow h({\displaystyle \frac{1}{\sqrt{3}}}+\sqrt{3})=150$

$\Rightarrow h(0.577+1.732)=150$

$\Rightarrow h\left(2.309\right)=150$

$\Rightarrow h={\displaystyle \frac{150}{2.309}}$

$\Rightarrow h=64.96$m

And the position of the point CD from pillar CE is,

$\Rightarrow CD={\displaystyle \frac{h}{\sqrt{3}}}$

$\Rightarrow CD=37.5$m

6. From the figure, given below, calculate the length of the CD.

Ans: From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{AB}{CB}}$

$\Rightarrow tan{47}^o={\displaystyle \frac{AB}{15}}$

$\Rightarrow 1.072={\displaystyle \frac{AB}{15}}$

$\Rightarrow AB=1.072\times 15$

$\Rightarrow AB=16.08$m

From triangle $\mathrm{\Delta }$AED,

$tan\theta ={\displaystyle \frac{AE}{DE}}$

$\Rightarrow tan{22}^o={\displaystyle \frac{AE}{15}}$

$\Rightarrow 0.404={\displaystyle \frac{AE}{15}}$

$\Rightarrow AE=0.404\times 15$

$\Rightarrow AE=6.06$m

Length of CD is,

$CD=AB-AE$

$\Rightarrow CD=16.08-6.06$

$\Rightarrow CD=10.02$m

7. The angle of elevation of the top of a tower is observed to be 60°. At a point, 30m vertically above the first point of observation, the elevation is found to be 45°. Find :

(i) the height of the tower, 

(ii) its horizontal distance from the points of observation.

Ans: 

Let the height of the tower be ‘$h$’.

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{BD}{AB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{AB}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h}{AB}}$

$\Rightarrow AB={\displaystyle \frac{h}{\sqrt{3}}}$

From triangle $\mathrm{\Delta }$ECD, where EA = CB

$tan\theta ={\displaystyle \frac{DC}{CE}}$

$\Rightarrow tan\theta ={\displaystyle \frac{DB-CB}{CE}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{h-30}{CE}}$

$\Rightarrow 1={\displaystyle \frac{h-30}{CE}}$

$\Rightarrow CE=h-30$

Here AB = CE

Therefore,

${\displaystyle \frac{h}{\sqrt{3}}}=h-30$

$\Rightarrow {\displaystyle \frac{h}{\sqrt{3}}}-h=-30$ ------$\times (-1)$

$\Rightarrow h(1-{\displaystyle \frac{1}{\sqrt{3}}})=30$

$\Rightarrow h={\displaystyle \frac{30}{(1-{\displaystyle \frac{1}{\sqrt{3}}})}}$

$\Rightarrow h={\displaystyle \frac{30}{(1-{\displaystyle \frac{1}{1.732}})}}$

$\Rightarrow h={\displaystyle \frac{30}{(1-0.577)}}$

$\Rightarrow h={\displaystyle \frac{30}{0.423}}$

$\Rightarrow h=70.92$m

Therefore, height of the tower is 70.92m

(ii) The horizontal distance from point of observation is,

$\Rightarrow CE=h-30$

$\Rightarrow CE=70.92-30$

$\Rightarrow CE=40.92$m

8. 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° and 60°. Find the height of the tower.

Ans: 

Let the height of the tower be ‘$h$’.

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{BD}{AB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{60}{AB}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{60}{AB}}$

$\Rightarrow AB={\displaystyle \frac{60}{\sqrt{3}}}$

From triangle $\mathrm{\Delta }$ECD, AB = CE

$tan\theta ={\displaystyle \frac{DC}{CE}}$

$\Rightarrow tan\theta ={\displaystyle \frac{DB-CB}{CE}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{60-h}{CE}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{DC}{{\displaystyle \frac{60}{\sqrt{3}}}}}$

$\Rightarrow DC={\displaystyle \frac{1}{\sqrt{3}}}\times {\displaystyle \frac{60}{\sqrt{3}}}$

$\Rightarrow DC={\displaystyle \frac{60}{{(\sqrt{3})}^2}}$

$\Rightarrow DC={\displaystyle \frac{60}{3}}$

$\Rightarrow DC=20$m

Therefore AE = CB = DB - DC = 60 - 20 = 40m

Hence, the height of the tower is 40m.

9. A man on a cliff observes a boat, at an angle of depression 30°, which is sailing towards the shore to the point immediately beneath him. Three minutes later, the angle of depression of the boat was found to be 60°. Assuming that the boat sails at a uniform speed, determine: 

(i) how much more time will it take to reach the shore ? 

(ii) the speed of the boat in meters per second, if the height of the cliff is 500 m.

Ans: Let the height of the cliff be ‘h’ and point A be the initial position and point D be the position of the boat after 3 min.

From triangle $\mathrm{\Delta }$DBC,

$tan\theta ={\displaystyle \frac{BC}{DB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{DB}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h}{DB}}$

$\Rightarrow DB={\displaystyle \frac{h}{\sqrt{3}}}$

From triangle $\mathrm{\Delta }$ABC,

$\Rightarrow tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{AB}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{AB}}$

$\Rightarrow AB=\sqrt{3}h$

Speed of the boat will be,

$\Rightarrow v={\displaystyle \frac{AB-DB}{3}}$

$\Rightarrow v={\displaystyle \frac{\sqrt{3}h-{\displaystyle \frac{h}{\sqrt{3}}}}{3}}$

$\Rightarrow v={\displaystyle \frac{{\displaystyle \frac{\sqrt{3}\times \sqrt{3}h-h}{\sqrt{3}}}}{3}}$

$\Rightarrow v={\displaystyle \frac{3h-h}{3\sqrt{3}}}$

$\Rightarrow v={\displaystyle \frac{2h}{3\sqrt{3}}}$

Time taken by the boat to reach the shore is,

$\Rightarrow t={\displaystyle \frac{DB}{v}}$

$\Rightarrow t={\displaystyle \frac{{\displaystyle \frac{h}{\sqrt{3}}}}{{\displaystyle \frac{2h}{3\sqrt{3}}}}}$

$\Rightarrow t={\displaystyle \frac{h}{\sqrt{3}}}\times {\displaystyle \frac{3\sqrt{3}}{2h}}$

$\Rightarrow t={\displaystyle \frac{3}{2}}$

$\Rightarrow t=1.5$min

(ii) If h = 500m, then the speed of the boat will be,

$v={\displaystyle \frac{2\times 500}{3\sqrt{3}}}$

$\Rightarrow v=192.45$m/min

$\Rightarrow v={\displaystyle \frac{192.45}{60}}$

$\Rightarrow v=3.21$m/s

As a result, the boat takes 1.5minutes longer to reach the shore. And, if the cliff is 500metres high, the boat's speed is 3.21metres per second.

10. A man in a boat rowing away from a lighthouse 150m high, takes 2 minutes to change the angle of elevation of the top of the lighthouse from 60° to 45°. Find the speed of the boat.

Ans: Let BC be the lighthouse, point D be the initial position and point A be the final position of the boat after 2min.

From triangle $\mathrm{\Delta }$DBC,

$tan\theta ={\displaystyle \frac{BC}{DB}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{150}{DB}}$
$\Rightarrow \sqrt{3}={\displaystyle \frac{150}{DB}}$

$\Rightarrow DB={\displaystyle \frac{150}{\sqrt{3}}}$m

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{150}{AB}}$

$\Rightarrow 1={\displaystyle \frac{150}{AB}}$

$\Rightarrow AB=150$m

Speed of the boat will be,

$v={\displaystyle \frac{AB-DB}{2min}}$

$\Rightarrow v={\displaystyle \frac{150-{\displaystyle \frac{150}{\sqrt{3}}}}{2\left(60\right)}}$

$\Rightarrow v={\displaystyle \frac{\sqrt{3}\times 150-150}{120\times \sqrt{3}}}$

$\Rightarrow v={\displaystyle \frac{1.732\times 150-150}{120\times 1.732}}$

$\Rightarrow v={\displaystyle \frac{109.8}{207.84}}$

$\Rightarrow v=0.528$m/s

Hence, speed of the boat is 0.528m/s

11. 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°. When he moves 40m away from the bank, he finds the angle of elevation to be 30°. Find: 

(i) the height of the tree, correct to 2 decimal places, 

(ii) the width of the river.

Ans: 

Let the height of the tree be ‘h’, point A be the initial position, point D be the final position of the person and AC be the width of the river.

From triangle $\mathrm{\Delta }$DBC,

$tan\theta ={\displaystyle \frac{AB}{AC}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{h}{AC}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{h}{DB}}$

$\Rightarrow AC={\displaystyle \frac{h}{\sqrt{3}}}$

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{AB}{AD}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{AC+DC}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{AC+40}}$

$\Rightarrow {\displaystyle \frac{h}{\sqrt{3}}}+40=\sqrt{3}h$

$\Rightarrow {\displaystyle \frac{h+40\sqrt{3}}{\sqrt{3}}}=\sqrt{3}h$

$\Rightarrow h+40\sqrt{3}=\sqrt{3}\times \sqrt{3}h$

$\Rightarrow 40\sqrt{3}=3h-h$

$\Rightarrow 40\sqrt{3}=2h$

$\Rightarrow h={\displaystyle \frac{40\sqrt{3}}{2}}$

$\Rightarrow h=20\sqrt{3}$

$\Rightarrow h=20\times 1.732$

$\Rightarrow h=34.64$m

Therefore, the height of the tree is 34.64m

Width of the river is AC,

$AC={\displaystyle \frac{h}{\sqrt{3}}}$

$\Rightarrow AC={\displaystyle \frac{34.64}{1.732}}$

$\Rightarrow AC=20$m

As a result, the height of the tree is 34.64m and the width of the river is 20m.

12. The horizontal distance between two towers is 75m and the angular depression of the top of the first tower as seen from the top of the second, which is 160m high, is 45°. Find the height of the first tower.

Ans: Let the height of the first tower is ‘$h$’,

From triangle $\mathrm{\Delta }$CDE,

$tan\theta ={\displaystyle \frac{DC}{EC}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{DC}{75}}$

$\Rightarrow DC=75m$ (because $tan{45}^o=1$)

The height of the first tower is,

$h=DB-DC$

$\Rightarrow h=160-75$

$\Rightarrow h=85$m

13. The length of the shadow of a tower standing on a level plane is found to be 2y metres longer when the sun's altitude is 30° than when it was 45°. Prove that the height of the tower is $y(\sqrt{3}+1)$ metres.

Ans: Let AB is the tower of height ‘$h$’, AC and AD is the length of the shadows.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{AB}{AC}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{h}{AC}}$

$\Rightarrow AC=h$  (because $tan{45}^o=1$)

From triangle $\mathrm{\Delta }$ABD,

$tan\theta ={\displaystyle \frac{AB}{AD}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{AB}{AC+DC}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{AC+2y}}$

$\Rightarrow AC+2y=\sqrt{3}h$

Here AC = AB, thus

$\Rightarrow h+2y=\sqrt{3}h$

$\Rightarrow h(\sqrt{3}-1)=2y$

$\Rightarrow h={\displaystyle \frac{2y}{\sqrt{3}-1}}$

On rationalising the RHS by multiplying numerator and denominator with $(\sqrt{3}+1)$ in the above equation. We get

$h={\displaystyle \frac{2y}{\sqrt{3}-1}}\times {\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}+1}}$

$\Rightarrow h={\displaystyle \frac{2y(\sqrt{3}+1)}{{(\sqrt{3})}^2-1^2}}$

$\Rightarrow h={\displaystyle \frac{2y(\sqrt{3}+1)}{3-1}}$

$\Rightarrow h={\displaystyle \frac{2y(\sqrt{3}+1)}{2}}$

$\Rightarrow h=y(\sqrt{3}+1)$

Hence proved.

14. An aeroplane flying horizontally 1km above the ground and going away from the observer is observed at an elevation of 60°. After 10 seconds, its elevation is observed to be 30°; find the uniform speed of the aeroplane in km per hour.

Ans: Let the initial position of the aeroplane be D and final position be C.

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{1}{AB}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{1}{AB}}$

$\Rightarrow AB=\sqrt{3}$

From triangle $\mathrm{\Delta }$AED,

$tan\theta ={\displaystyle \frac{BC}{AE}}$

$\Rightarrow tan{60}^o={\displaystyle \frac{1}{AE}}$

$\Rightarrow \sqrt{3}={\displaystyle \frac{1}{AE}}$

$\Rightarrow AE={\displaystyle \frac{1}{\sqrt{3}}}$

The speed of the aeroplane will be,

$\Rightarrow v={\displaystyle \frac{AB-AE}{10s}}$

$\Rightarrow v={\displaystyle \frac{\sqrt{3}-{\displaystyle \frac{1}{\sqrt{3}}}}{{\displaystyle \frac{10}{3600}}}}$

$\Rightarrow v={\displaystyle \frac{{\displaystyle \frac{{(\sqrt{3})}^2-1}{\sqrt{3}}}}{{\displaystyle \frac{10}{3600}}}}$

$\Rightarrow v={\displaystyle \frac{{(\sqrt{3})}^2-1}{\sqrt{3}}}\times {\displaystyle \frac{3600}{10}}$

$\Rightarrow v={\displaystyle \frac{3-1}{1.732}}\times 360$

$\Rightarrow v=1.1547\times 360$

$\Rightarrow v=415.69$km/h

15. From the top of a hill, the angles of depression of two consecutive kilometre stones, due east, are found to be 30° and 45° respectively. Find the distances of the two stones from the foot of the hill.

Ans: Let's assume the hill's height is ‘h’, and the stones are shown at points D and A.

From triangle $\mathrm{\Delta }$DBC,

$tan\theta ={\displaystyle \frac{BC}{DB}}$

$\Rightarrow tan{45}^o={\displaystyle \frac{h}{DB}}$

$\Rightarrow DB=h$  (because $tan{45}^o=1$)

From triangle $\mathrm{\Delta }$ABC,

$tan\theta ={\displaystyle \frac{BC}{AB}}$

$\Rightarrow tan{30}^o={\displaystyle \frac{h}{AB}}$

$\Rightarrow {\displaystyle \frac{1}{\sqrt{3}}}={\displaystyle \frac{h}{AB}}$

$\Rightarrow AB=\sqrt{3}h$

Here, 

$\Rightarrow AB=DB+1$

$\Rightarrow \sqrt{3}h=h+1$

$\Rightarrow \sqrt{3}h-h=1$

$\Rightarrow h(\sqrt{3}-1)=1$

$\Rightarrow h={\displaystyle \frac{1}{\sqrt{3}-1}}$

$\Rightarrow h={\displaystyle \frac{1}{1.732-1}}$

$\Rightarrow h={\displaystyle \frac{1}{0.732}}$

$\Rightarrow h=1.366$km

The distance between stone at point D and foot of hill is,

$DB=h$

$DB=1.366$km

The distance between stone at point A and foot of hill is,

$AB=h+1$

$\Rightarrow AB=1.366+1$

$\Rightarrow AB=2.366$km

EXERCISE 22(C)

1. Find AD,

(i) 

(ii) 

Ans: (i) From triangle $\mathrm{\Delta }$ABE,

$tan\theta ={\displaystyle \frac{AE}{CD}}$

$\Rightarrow tan{32}^o={\displaystyle \frac{AE}{20}}$

$\Rightarrow 0.624={\displaystyle \frac{AE}{20}}$

$\Rightarrow AE=20\times 0.624$

$\Rightarrow AE=12.48$m

And, 

$\Rightarrow AD=AE+BC$

$\Rightarrow AD=12.48+5$

$\Rightarrow AD=17.48$m

(ii) From triangle $\mathrm{\Delta }$ABE,

$tan\theta ={\displaystyle \frac{AD}{CD}}$

$\Rightarrow CD={\displaystyle \frac{AD}{tan{48}^o}}$

And,

$\Rightarrow sin\theta ={\displaystyle \frac{AD}{AC}}$

$\Rightarrow AC={\displaystyle \frac{AD}{sin{48}^o}}$

$\Rightarrow BD=BC+CD$

$BD={\displaystyle \frac{AD}{sin{48}^o}}+{\displaystyle \frac{AD}{tan{48}^o}}$

By pythagoras theorem we get,

$\Rightarrow A{B}^2=A{D}^2+B{D}^2$

$\Rightarrow {30}^2=A{D}^2+{({\displaystyle \frac{AD}{sin{48}^o}}+{\displaystyle \frac{AD}{tan{48}^o}})}^2$