NCERT Solutions for Exercise 9.1 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 elders children be $y$ 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 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:

A

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 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 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 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 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 the eq(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 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

$\sqrt{3}h=80-\frac{h}{\sqrt{3}}$
$\frac{h}{\sqrt{3}}=20$
$h=20\sqrt{3}$ m
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 question,

In triangle $\mathrm{\Delta }ADB$ ,
$$\begin{gathered} \Rightarrow \tan {30}^o=\frac{AB}{20+x}=\frac{h}{20+x} \\ \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{20+x} \\ \Rightarrow 20+x=h\sqrt{3} \\ \Rightarrow x=h\sqrt{3}-20 \end{gathered}$$ ............(i)

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 eq (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 $\mathrm{\angle }ACE={60}^o$ , angle of depression of its foot $\mathrm{\angle }BCE={45}^o$ .

According to 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 $h+7$ $=7(1+\sqrt{3}\text{)}m$

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 from the 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 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 equation (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 approx 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 traveled 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, distance traveled 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 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 take 3 sec 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 the equation (i) and (ii), we get

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

Hence the height of the tower is 6 m.