NCERT Solutions for Class 10 Maths Chapter 9 Some Applications of Trigonometry

Q2: A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle 30° with it. The distance between the foot of the tree to the point where the top touches the ground is 8 m. Find the height of the tree.

Answer:

Suppose DB is a tree, and the AD is the broken height of the tree, which touches the ground at C.
Given that,
$\mathrm{\angle }ACB={30}^o$ , BC = 8 m
let AB = $x$ m and AD = $y$ m
So, AD+AB = DB = $x+y$

In right angle triangle $\mathrm{\Delta }ABC$ ,
$\tan \theta =\frac{P}{B}=\frac{x}{8}$
$\tan {30}^o=\frac{x}{8}=\frac{1}{\sqrt{3}}$
So, the value of $x$ = $8/\sqrt{3}$

Similarily,
$\cos {30}^o=\frac{BC}{AC}=\frac{8}{y}$
the value of $y$ is $16/\sqrt{3}$

So, the total height of the tree is-

$x+y=\frac{24}{\sqrt{3}}=8\sqrt{3}$

= 8 (1.732) = 13.856 m (approx)

Q3: A contractor plans to install two slides for the children to play in a park. For the children below the age of 5 years, she prefers to have a slide whose top is at a height of 1.5 m, and is inclined at an angle of 30° to the ground, whereas for elder children, she wants to have a steep slide at a height of 3m, and inclined at an angle of 60° to the ground. What should be the length of the slide in each case?

Answer:

Suppose $x$ m is the length of slides for children below 5 years and the length of slides for elder children is $y$ m.

Given that,
AF = 1.5 m, BC = 3 m, $\mathrm{\angle }AEF={30}^o$ and $\mathrm{\angle }BDC={60}^o$

In triangle $\mathrm{\Delta }$ EAF,
$\sin \theta =\frac{AF}{EF}=\frac{1.5}{x}$
$\sin {30}^o=\frac{1.5}{x}$
The value of $x$ is 3 m.

Similarily in $\mathrm{\Delta }$ CDB,
$\sin {60}^o=\frac{3}{y}$
$\frac{\sqrt{3}}{2}=\frac{3}{y}$
the value of $y$ is $2\sqrt{3}$ = 2(1.732) = 3.468

Hence, the length of the slide for children below 5 yrs. is 3 m and for the elder children is 3.468 m.

Q4: The angle of elevation of the top of a tower from a point on the ground, which is 30 m away from the foot of the tower, is 30°. Find the height of the tower.

Answer:

Let the height of the tower AB is $h$ and the angle of elevation from the ground at point C is $\mathrm{\angle }ACB={30}^o$
According to the question,
In the right triangle $\mathrm{\Delta }ABC$ ,
$\tan \theta =\frac{AB}{BC}=\frac{h}{30}$
$\tan {30}^o=\frac{1}{\sqrt{3}}=\frac{h}{30}$
the value of $h$ is $10\sqrt{3}$ = 10(1.732) = 17.32 m
Thus the height of the tower is 17.32 m

Q5: A kite is flying at a height of 60 m above the ground. The string attached to the kite is temporarily tied to a point on the ground. The inclination of the string with the ground is 60°. Find the length of the string, assuming that there is no slack in the string.

Answer:

Given that,
The length of AB = 60 m and the inclination of the string with the ground at point C is $\mathrm{\angle }ACB={60}^o$.
Let the length of the string AC be $l$.
According to the question,
In right triangle $\mathrm{\Delta }$ CBA,
$\sin {60}^o=\frac{AB}{AC}=\frac{60}{l}$
$\frac{\sqrt{3}}{2}=\frac{60}{l}$
The value of length of the string ( $l$ ) is $40\sqrt{3}$ = 40(1.732) = 69.28 m
Hence, the length of the string is 69.28 m.

Q6: A 1.5 m tall boy is standing at some distance from a 30 m tall building. The angle of elevation from his eyes to the top of the building increases from 30° to 60° as he walks towards the building. Find the distance he walked towards the building.

Answer:

Given that,
The height of the tallboy (DC) is 1.5 m and the height of the building (AB) is 30 m.
$\mathrm{\angle }ADF={30}^o$ and $\mathrm{\angle }AEF={60}^o$

According to the question,
In right triangle AFD,
$$\begin{gathered} \Rightarrow \tan {30}^o=\frac{AF}{DF}=\frac{28.5}{DF} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{DF} \end{gathered}$$
So, DF = $(28.5)\sqrt{3}$

In right angle triangle $\mathrm{\Delta }AFE$ ,
$\tan {60}^o=\frac{AF}{FE}=\frac{28.5}{EF}$
$\sqrt{3}=\frac{28.5}{EF}$
EF = $9.5\sqrt{3}$

So, distance walked by the boy towards the building = DF - EF = $19\sqrt{3}$

Q7: From a point on the ground, the angles of elevation of the bottom and the top of a transmission tower fixed at the top of a 20 m high building are 45° and 60° respectively. Find the height of the tower.

Answer:

Suppose BC = $h$ is the height of transmission tower and the AB be the height of the building, and AD is the distance between the building and the observer point (D).
We have,
AB = 20 m, BC = $h$ m and AD = $x$ m
$\mathrm{\angle }CDA={60}^o$ and $\mathrm{\angle }BDA={45}^o$

According to the question,
In triangle $\mathrm{\Delta }$ BDA,
$\tan {45}^o=\frac{AB}{AD}=\frac{20}{x}$
So, $x$ = 20 m

Again,
In triangle $\mathrm{\Delta }$ CAD,

$$\begin{gathered} \Rightarrow \tan {60}^o=\frac{AB+BC}{AD}=\frac{20+h}{20} \\ \Rightarrow \sqrt{3}=1+\frac{h}{20} \\ \Rightarrow h=20(\sqrt{3}-1) \\ \Rightarrow 20(0.732)=14.64m \end{gathered}$$

Answer- The height of the tower is 14.64 m

Q8: A statue, 1.6 m tall, stands on the top of a 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 the height of the pedestal be $h$ m. and the height of the statue is 1.6 m.
The angle of elevation of the top of the statue and top of the pedestal is( $\mathrm{\angle }DCB=$ ${60}^o$ )and( $\mathrm{\angle }ACB=$ ${45}^o$ ) respectively.

Now,
In triangle $\mathrm{\Delta }ABC$ ,
$\tan {45}^o=1=\frac{AB}{BC}=\frac{h}{BC}$
therefore, BC = $h$ m

In triangle $\mathrm{\Delta }CBD$ ,
$$\begin{gathered} \Rightarrow \tan {60}^o=\frac{BD}{BC}=\frac{h+1.6}{h} \\ \Rightarrow \sqrt{3}=1+\frac{1.6}{h} \end{gathered}$$
the value of $h$ is $0.8(\sqrt{3}+1)$ m
Hence the height of the pedestal is $0.8(\sqrt{3}+1)$ m

Q9: The angle of elevation of the top of a building from the foot of the tower is 30° and the angle of elevation of the top of the tower from the foot of the building is 60°. If the tower is 50 m high, find the height of the building.

Answer:

It is given that, the height of the tower (AB) is 50 m. $\mathrm{\angle }AQB={30}^o$ and $\mathrm{\angle }PBQ={60}^o$
Let the height of the building be $h$ m

According to the question,
In triangle PBQ,
$\tan {60}^o=\frac{PQ}{BQ}=\frac{50}{BQ}$
$$\begin{gathered} \sqrt{3}=\frac{50}{BQ} \\ BQ=\frac{50}{\sqrt{3}} \end{gathered}$$ .......................(i)

In triangle ABQ,

$\tan {30}^o=\frac{h}{BQ}$
$BQ=h\sqrt{3}$ .........................(ii)
On equating equations (i) and (ii), we get,

$\frac{50}{\sqrt{3}}=h\sqrt{3}$
therefore, $h$ = 50/3 = 16.66 m = height of the building.

Q10: Two poles of equal heights are standing opposite each other on either side of the road, which is 80 m wide. From a point between them on the road, the angles of elevation of the top of the poles are 60° and 30°, respectively. Find the height of the poles and the distances of the point from the poles.

Answer:

Given that,
The height of both poles are equal DC = AB. The angle of elevation of of the top of the poles are $\mathrm{\angle }DEC={30}^o$ and $\mathrm{\angle }AEB={60}^o$ resp.
Let the height of the poles be $h$ m and CE = $x$ and BE = 80 - $x$

According to the question,
In triangle DEC,

$$\begin{gathered} \Rightarrow \tan {30}^o=\frac{DC}{CE}=\frac{h}{x} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{x} \\ \Rightarrow x=h\sqrt{3} \end{gathered}$$ ..............(i)

In triangle AEB,
$$\begin{gathered} \Rightarrow \tan {60}^o=\frac{AB}{BE}=\frac{h}{80-x} \\ \Rightarrow \sqrt{3}=\frac{h}{80-x} \\ \Rightarrow x=80-\frac{h}{\sqrt{3}} \end{gathered}$$ ..................(ii)
On equating eq (i) and eq (ii), we get

$\begin{array}{rl}& \sqrt{3}h=80-\frac{h}{\sqrt{3}}\\ & \frac{h}{\sqrt{3}}=20\end{array}$
So, $x$ = 60 m

Hence, the height of both poles is ( $h=20\sqrt{3}$ )m and the position of the point is at 60 m from the pole CD and 20 m from the pole AB.

Q11: A TV tower stands vertically on a bank of a canal. From a point on the other bank directly opposite the tower, the angle of elevation of the top of the tower is 60°. From another point 20 m away from this point on the line joining this point to the foot of the tower, the angle of elevation of the top of the tower is 30° (see Fig. 9.12). Find the height of the tower and the width of the canal.

Answer:

Suppose the $h$ is the height of the tower AB and BC = $x$ m
It is given that the width of CD is 20 m,
According to the question,

In triangle $\mathrm{\Delta }ADB$ ,

$\begin{array}{rl}& \Rightarrow \tan {30}^{\circ }=\frac{AB}{20+x}=\frac{h}{20+x}\\ & \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x}\\ & \Rightarrow 20+x=h\sqrt{3}\\ & \Rightarrow x=h\sqrt{3}-20----(i)\end{array}$

In triangle ACB,
$$\begin{gathered} \Rightarrow \tan {60}^o=\frac{h}{x}=\sqrt{3} \\ \Rightarrow x=\frac{h}{\sqrt{3}} \end{gathered}$$ .............(ii)

On equating equations (i) and (ii), we get:

$h\sqrt{3}-20=\frac{h}{\sqrt{3}}$
from here we can calculate the value of $h=10\sqrt{3}=10(1.732)=17.32m$ and the width of the canal is 10 m.

Q12: From the top of a 7 m 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:

Let the height of the cable tower be (AB = $h+7$ )m
Given,
The height of the building is 7 m and angle of elevation of the top of the tower $\mathrm{\angle }ACE={60}^o$, the angle of depression of its foot $\mathrm{\angle }BCE={45}^o$ .

According to the question,

In triangle $\mathrm{\Delta }DBC$ ,
$$\begin{gathered} \tan {45}^o=\frac{CD}{BD}=\frac{7}{BD}=1 \\ BD=7m \end{gathered}$$
since DB = CE = 7 m

In triangle $\mathrm{\Delta }ACE$ ,

$$\begin{gathered} \tan {60}^o=\frac{h}{CE}=\frac{h}{7}=\sqrt{3} \\ \therefore h=7\sqrt{3}m \end{gathered}$$

Thus, the total height of the tower equal to $h+7$ $=7(1+\sqrt{3})m$

Q13: As observed from the top of a 75 m high lighthouse from the sea-level, the angles 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:

Given that,
The height of the lighthouse (AB) is 75 m above sea level. And the angle of depression of two different ships are $\mathrm{\angle }ADB={30}^0$ and $\mathrm{\angle }ACB={45}^0$ respectively

Let the distance between both the ships be $x$ m.
According to the question,

In triangle $\mathrm{\Delta }ADB$ ,

$\tan {30}^0=\frac{AB}{BD}=\frac{75}{x+y}=\frac{1}{\sqrt{3}}$
$\therefore x+y=75\sqrt{3}$ .............(i)

In triangle $\mathrm{\Delta }ACB$ ,

$\tan {45}^0=1=\frac{75}{BC}=\frac{75}{y}$
$\therefore y=75m$ .............(ii)

From equations (i) and (ii) we get;
$x=75(\sqrt{3}-1)=75(0.732)$
$x=54.9\simeq 55m$

Hence, the distance between the two ships is approximately 55 m.

Q14: A 1.2 m tall girl spots a balloon moving with the wind in a horizontal line at a height of 88.2 m from the ground. The angle of elevation of the balloon from the eyes of the girl at any instant is 60°. After some time, the angle of elevation reduces to 30° (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Answer:

Given that,
The height of the girl is 1.2 m. The height of the balloon from the ground is 88.2 m, and the angle of elevation of the balloon from the eye of the girl at any instant is ( $\mathrm{\angle }ACB={60}^0$ ), and after some time, $\mathrm{\angle }DCE={30}^0$.

Let the $x$ distance travelled by the balloon from position A to position D during the interval.
AB = ED = 88.2 - 1.2 =87 m

Now, In triangle $\mathrm{\Delta }BCA$ ,

$$\begin{gathered} \tan {60}^0=\sqrt{3}=\frac{AB}{BC}=\frac{87}{BC} \\ \therefore BC=29\sqrt{3} \end{gathered}$$

In triangle $\mathrm{\Delta }DCE$ ,

$$\begin{gathered} \tan {30}^0=\frac{1}{\sqrt{3}}=\frac{DE}{CE}=\frac{87}{CE} \\ \therefore CE=87\sqrt{3} \end{gathered}$$

Thus, the distance travelled by the balloon from position A to D

$=CE-BC=87\sqrt{3}-29\sqrt{3}=58\sqrt{3}$ m

Q15: A straight highway leads to the foot of a tower. A man standing at the top of the tower observes a car at an angle of depression of 30°, which is approaching the foot of the tower with a uniform speed. Six seconds later, the angle of depression of the car is found to be 60°. Find the time taken by the car to reach the foot of the tower from this point.

Answer:

Let $h$ be the height of the tower (DC) and the speed of the car be $xm{s}^{-1}$ . Therefore, the distance (AB)covered by the car in 6 seconds is 6 $x$ m. Let $t$ time required to reach the foot of the tower. So, BC = $x$ $t$

According to the question,
In triangle $\mathrm{\Delta }DAC$ ,
$$\begin{gathered} \tan {30}^0=\frac{1}{\sqrt{3}}=\frac{h}{6x+xt} \\ x(6+t)=h\sqrt{3} \end{gathered}$$ ..........................(i)

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

$$\begin{gathered} \tan {60}^0=\sqrt{3}=\frac{h}{xt} \\ \therefore h=3.xt \end{gathered}$$ ...................(ii)

Put the value of $h$ in equation (i) we get,
$$\begin{gathered} x(6+t)=(\sqrt{3}.\sqrt{3})xt \\ 6x+xt=3xt \\ 6x=2xt \end{gathered}$$
$t=3$

Hence, from point B car takes 3 seconds to reach the foot of the tower.

Q16: The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer:

Let the height of the tower be $h$ m.
we have PB = 4m and QB = 9 m
Suppose $\mathrm{\angle }BQA=\theta$ , so $\mathrm{\angle }APB=90-\theta$

According to question,

In triangle $\mathrm{\Delta }ABQ$ ,

$$\begin{gathered} \tan \theta =\frac{h}{9} \\ \therefore h=9\tan \theta \end{gathered}$$ ..............(i)

In triangle $\mathrm{\Delta }ABP$ ,

$$\begin{gathered} \tan (90-\theta )=\cot \theta =\frac{h}{4} \\ \therefore h=4\cot \theta \end{gathered}$$ .....................(ii)

Multiply equations (i) and (ii), we get

$$\begin{gathered} {h}^2=36 \\ \Rightarrow h=6m \end{gathered}$$

Hence, the height of the tower is 6 m.