Exercise 9.1 (Revised) - Chapter 9 - Some Applications Of Trigonometry - Ncert Solutions class 10 - Maths

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NCERT Class 10 Maths: Chapter 9 - Some Applications of Trigonometry Solutions

Ex 9.1 Question 1.

A circus artist is climbing a $20 \mathrm{~m}$ long rope, which is tightly stretched and tied from the top of a vertical pole to the ground. Find the height of the pole, if the angle made by the rope with the ground level is $30^{\circ}$ (see figure).

Answer.

In right triangle $\mathrm{ABC}$,
$
\begin{aligned}
& \sin 30^{\circ}=\frac{A B}{A C} \\
& \Rightarrow \frac{1}{2}=\frac{A B}{20} \\
& A B=20 / 2 \\
& \Rightarrow A B=10 \mathrm{~m}
\end{aligned}
$

Hence, the height of the pole is $10 \mathrm{~m}$.
Ex 9.1 Question 2.

A tree breaks due to storm and the broken part bends so that the top of the tree touches the ground making an angle $30^{\circ}$ with it. The distance between the foot of the tree to the point where the top touches the ground is $8 \mathrm{~m}$. Find the height of the tree.

Answer.

Let AC be the broken part of tree

In right triangle $\mathrm{ABC}$,
$
\cos 30^{\circ}=\frac{\mathrm{BC}}{\mathrm{AC}}
$

$\begin{aligned}
& \Rightarrow \frac{\sqrt{3}}{2}=\frac{8}{A C} \\
& \Rightarrow A C=\frac{16}{\sqrt{3}} \mathrm{~m} \\
& \text { Again, } \tan 30^{\circ}=\frac{A B}{B C} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{8} \\
& \Rightarrow A B=\frac{8}{\sqrt{3}} \mathrm{~m} \\
& \therefore \text { Height of the tree }=A B+A D=A B+A C \\
& =\frac{8}{\sqrt{3}}+\frac{16}{\sqrt{3}}=\frac{24}{\sqrt{3}}
\end{aligned}$

$=\frac{24}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=8 \sqrt{3} \mathrm{~m}$

Ex 9.1 Question 3.

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 $1.5 \mathrm{~m}$ and is inclined at an angle of $30^{\circ}$ to the ground, whereas for elder children, she wants to have a steep slide at a height of $3 \mathrm{~m}$ and inclined at an angle of $60^{\circ}$ to the ground. What should be the length of the slide in each case?

Answer.

In right triangle $\mathrm{ABC}$,

$
\begin{aligned}
& \sin 30^{\circ}=\frac{A B}{A C} \\
& \Rightarrow \frac{1}{2}=\frac{1.5}{A C} \\
& \Rightarrow A C=3 \mathrm{~m}
\end{aligned}
$

In right triangle $\mathrm{PQR}$,
$
\begin{aligned}
& \sin 60^{\circ}=\frac{P Q}{P R} \\
& \Rightarrow \frac{\sqrt{3}}{2}=\frac{3}{P R} \\
& \Rightarrow P R=2 \sqrt{3} \mathrm{~m}
\end{aligned}
$

Hence, the lengths of the slides are $3 \mathrm{~m}$ and $2 \sqrt{3} \mathrm{~m}$ respectively.

Ex 9.1 Question 4.

The angle of elevation of the top of a tower from a point on the ground, which is $30 \mathrm{~m}$ away from the foot of the tower is $30^{\circ}$. Find the height of the tower.

Answer.

In right triangle $\mathrm{ABC}, \mathrm{AB}$ be the height of the tower.
$
\tan 30^{\circ}=\frac{A B}{B C}
$

$\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{30} \\
& \Rightarrow A B=\frac{30}{\sqrt{3}} \mathrm{~m} \\
& \Rightarrow \frac{30}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=10 \sqrt{3} \mathrm{~m}
\end{aligned}$

Ex 9.1 Question 5.

A kite is flying at a height of $60 \mathrm{~m}$ 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 $60^{\circ}$. Find the length of the string, assuming that there is no slack in the string.

Answer.

In right triangle $\mathrm{ABC}, \mathrm{AC}$ is the length of the string

$
\begin{aligned}
& \sin 60^{\circ}=\frac{A B}{A C} \\
& \Rightarrow \frac{\sqrt{3}}{2}=\frac{60}{A C} \\
& \Rightarrow A C=40 \sqrt{3} \mathrm{~m}
\end{aligned}
$

Hence the length of the string is $40 \sqrt{3} \mathrm{~m}$.

Ex 9.1 Question 6.

A $1.5 \mathrm{~m}$ tall boy is standing at some distance from a $30 \mathrm{~m}$ tall building. The angle of elevation from his eyes to the top of the building increases from $30^{\circ}$ to $60^{\circ}$ as he walks towards the building. Find the distance he walked towards the building.

Answer.

$\mathrm{AB}=30 \mathrm{~m}$ and $\mathrm{PR}=1.5 \mathrm{~m}$

$
\begin{aligned}
& \mathrm{AC}=\mathrm{AB}-\mathrm{BC} \\
& =\mathrm{AB}-\mathrm{PR}(\mathrm{As}, \mathrm{BC}=\mathrm{PR}) \\
& =30-1.5 \\
& =28.5 \mathrm{~m}
\end{aligned}
$

In right triangle ACQ,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{A C}{Q C} \\
& \Rightarrow \sqrt{3}=\frac{28.5}{Q C} \Rightarrow Q C=\frac{28.5}{\sqrt{3}} \mathrm{~m}
\end{aligned}
$

In right triangle ACP,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{A C}{P C} \Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{P Q+Q C} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5}{P Q+\frac{28.5}{\sqrt{3}}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{28.5 \times \sqrt{3}}{P Q \sqrt{3}+28.5} \\
& \Rightarrow P Q \sqrt{3}+28.5=85.5 \\
& \Rightarrow P Q \sqrt{3}=57 \\
& \Rightarrow P Q=\frac{57}{\sqrt{3}}
\end{aligned}
$

$
\Rightarrow P Q=\frac{57}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}=19 \sqrt{3} \mathrm{~m}
$

Hence, the distance the boy walked towards the building is $19 \sqrt{3} \mathrm{~m}$.

Ex 9.1 Question 7.

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 $20 \mathrm{~m}$ high building are $45^{\circ}$ and $60^{\circ}$ respectively. Find the height of the tower.

Answer.

Let the height of the tower be $h \mathrm{~m}$. Then, in right triangle CBP,

$
\begin{aligned}
& \tan 60^{\circ}=\frac{B C}{B P} \\
& \Rightarrow \sqrt{3}=\frac{A B+A C}{B P} \\
& \Rightarrow \sqrt{3}=\frac{20+h}{B P} \ldots \ldots \ldots .(i)
\end{aligned}
$

In right triangle $\mathrm{ABP}$,
$
\begin{aligned}
& \tan 45^{\circ}=\frac{A B}{B P} \\
& \Rightarrow 1=\frac{20}{B P} \Rightarrow B P=20 \mathrm{~m}
\end{aligned}
$

Putting this value in eq. (i), we get,

$
\begin{aligned}
& \sqrt{3}=\frac{20+h}{20} \\
& \Rightarrow 20 \sqrt{3}=20+h \\
& \Rightarrow h=20 \sqrt{3}-20 \\
& \Rightarrow h=20(\sqrt{3}-1) \mathrm{m}
\end{aligned}
$
$\therefore$ The height of the tower is $20(\sqrt{3}-1) \mathrm{m}$.
Ex 9.1 Question 8.

A statue, $1.6 \mathrm{~m}$ 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 $60^{\circ}$ and from the same point the angle of elevation of the top of the pedestal is $45^{\circ}$. Find the height of the pedestal.

Answer.

Let the height of the pedestal be $h \mathrm{~m}$.
$
\therefore \mathrm{BC}=h \mathrm{~m}
$

In right triangle $A C P$,

$\tan 60^{\circ}=\frac{A C}{P C}$

$
\begin{aligned}
& \Rightarrow \sqrt{3}=\frac{A B+B C}{P C} \\
& \Rightarrow \sqrt{3}=\frac{1.6+h}{P C} \ldots \ldots \ldots .(i)
\end{aligned}
$

In right triangle BCP,
$
\begin{aligned}
& \tan 45^{\circ}=\frac{\mathrm{BC}}{\mathrm{PC}} \\
& \Rightarrow 1=\frac{h}{\mathrm{PC}} \Rightarrow \mathrm{PC}=h \\
& \therefore \sqrt{3}=\frac{1.6+h}{h} \text { [From eq. (i)] } \\
& \Rightarrow \sqrt{3} h=1.6+h \\
& \Rightarrow h(\sqrt{3}-1)=1.6 \\
& \Rightarrow h=\frac{1.6}{\sqrt{3}-1} \\
& \Rightarrow h=\frac{1.6(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\
& \Rightarrow h=\frac{1.6 \sqrt{3}+1}{3-1} \\
& \Rightarrow h=0.8(\sqrt{3}+1) \mathrm{m}
\end{aligned}
$

$\text { Hence, the height of the pedestal is } 0.8(\sqrt{3}+1) \mathrm{m} \text {. }$

Ex 9.1 Question 9.

The angle of elevation of the top of a building from the foot of the tower is $30^{\circ}$ and the angle of elevation of the top of the tower from the foot of the building is $60^{\circ}$. If the tower is $50 \mathrm{~m}$ high, find the height of the building.

Answer.

Let the height of the building be $h \mathrm{~m}$.

In right triangle $\mathrm{PQB}$,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{P Q}{B Q} \Rightarrow \sqrt{3}=\frac{50}{B Q} \\
& \Rightarrow B Q=\frac{50}{\sqrt{3}} \text { m.........(i) }
\end{aligned}
$

In right triangle $A B Q$,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BQ}} \Rightarrow \frac{1}{\sqrt{3}}=\frac{h}{\mathrm{BQ}} \\
& \Rightarrow \mathrm{BQ}=h \sqrt{3} \mathrm{~m} . \ldots \ldots \ldots . \text { (ii) }
\end{aligned}
$

From eq. (i) and (ii),
$
h \sqrt{3}=\frac{50}{\sqrt{3}} \Rightarrow h=\frac{50}{3}=16 \frac{2}{3} \mathrm{~m}
$

Ex 9.1 Question 10.

Two poles of equal heights are standing opposite each other on either side of the road, which is $80 \mathrm{~m}$ wide. From a point between them on the road, the angles of elevation of the top of the poles are $60^{\circ}$ and $30^{\circ}$ respectively. Find the height of the poles and the distances of the point from the poles.

Answer.

Let the height of each poles be $\mathrm{H} \mathrm{m}$
$
\mathrm{AB}=\mathrm{PQ}=\mathrm{H}
$

In right triangle $\mathrm{PRQ}$,

(in figure change $\mathrm{AB}=\mathrm{H} \mathrm{m}$ rather than $\mathrm{h} \mathrm{m}$ )
$
\begin{aligned}
& \tan 60^{\circ}=\frac{\mathrm{PQ}}{\mathrm{QR}} \Rightarrow \sqrt{3}=\frac{\mathrm{H}}{h} \\
& \Rightarrow \mathrm{H}=h \sqrt{3} \mathrm{~m} \ldots \ldots \ldots . .(\mathrm{i})
\end{aligned}
$

In right triangle ABR,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{A B}{B R} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{H}{80-h} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{h \sqrt{3}}{80-h} \text { [From eq. (i)] } \\
& \Rightarrow 80-h=3 h
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow 4 h=80 \\
& \Rightarrow h=20 \mathrm{~m} \\
& \therefore \mathrm{H}=h \sqrt{3}=20 \sqrt{3} \mathrm{~m}
\end{aligned}
$

Also, $B R=80-h=80-20=60 \mathrm{~m}$
Hence the heights of the poles are $20 \sqrt{3} \mathrm{~m}$ each and the distances of the point from poles are $20 \mathrm{~m}$ and $60 \mathrm{~m}$ respectively.
Ex 9.1 Question 11.

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 $60^{\circ}$. From another point $20 \mathrm{~m}$ 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 $30^{\circ}$ (see figure). Find the height of the tower and the width of the canal.

Answer.

Let AB be the TV tower .

In right triangle $\mathrm{ABC}$,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{A B}{B C} \Rightarrow \sqrt{3}=\frac{A B}{B C} \\
& \Rightarrow A B=B C \sqrt{3} \text { m........(i) }
\end{aligned}
$

In right triangle $A B D$,

$
\begin{aligned}
& \tan 30^{\circ}=\frac{A B}{B D} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{B C+C D} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{B C+20} \\
& \Rightarrow A B=\frac{B C+20}{\sqrt{3}} \mathrm{~m} .
\end{aligned}
$

From eq. (i) and (ii),
$
\begin{aligned}
& \mathrm{BC} \sqrt{3}=\frac{\mathrm{BC}+20}{\sqrt{3}} \\
& \Rightarrow 3 \mathrm{BC}=\mathrm{BC}+20 \\
& \Rightarrow \mathrm{BC}=10 \mathrm{~m}
\end{aligned}
$

From eq. (i), $A B=10 \sqrt{3} \mathrm{~m}$
Hence height of the tower is $10 \sqrt{3} \mathrm{~m}$ and the width of the canal is $10 \mathrm{~m}$.
Ex 9.1 Question 12.

From the top of a $7 \mathrm{~m}$ high building, the angle of elevation of the top of a cable tower is $60^{\circ}$ and the angle of depression of its foot is $45^{\circ}$. Determine the height of the tower.

Answer.

In right triangle ABD,
$
\tan 45^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}}
$

$
\begin{aligned}
& \Rightarrow 1=\frac{7}{B D} \\
& \Rightarrow \mathrm{BD}=7 \mathrm{~m} \\
& \Rightarrow \mathrm{AE}=7 \mathrm{~m}
\end{aligned}
$

In right triangle AEC,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{C E}{A E} \\
& \Rightarrow \sqrt{3}=\frac{C E}{7} \\
& \Rightarrow C E=7 \sqrt{3} \mathrm{~m} \\
& \therefore C D=C E+E D
\end{aligned}
$

$
\begin{aligned}
& =\mathrm{CE}+\mathrm{AB}(\mathrm{As} \mathrm{AB}=\mathrm{ED}) \\
& =7 \sqrt{3}+7=7(\sqrt{3}+1) \mathrm{m}
\end{aligned}
$

Hence height of the tower is $7(\sqrt{3}+1) \mathrm{m}$.
Ex 9.1 Question 13.

As observed from the top of a $75 \mathrm{~m}$ high lighthouse from the sea-level, the angles of depression of two ships are $30^{\circ}$ and $45^{\circ}$. If one ship is exactly behind the other on the same side of the lighthouse, find the distance between two ships.

Answer.

In right triangle $A B Q$,

$
\begin{aligned}
& \tan 45^{\circ}=\frac{A B}{B Q} \\
& \Rightarrow 1=\frac{75}{B Q} \\
& \Rightarrow B Q=75 \mathrm{~m} . \ldots
\end{aligned}
$

In right triangle $\mathrm{ABP}$,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{A B}{B P} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{B Q+Q P}
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{\mathrm{AB}}{75+\mathrm{QP}} \text { [From eq. (i)] } \\
& \Rightarrow 75+\mathrm{QP}=75 \sqrt{3} \\
& \Rightarrow \mathrm{QP}=75(\sqrt{3}-1) \mathrm{m}
\end{aligned}
$

Hence the distance between the two ships is $75(\sqrt{3}-1) \mathrm{m}$.
Ex 9.1 Question 14.

A $1.2 \mathrm{~m}$ tall girl spots a balloon moving with the wind in a horizontal line at a height of $88.2 \mathrm{~m}$ from the ground. The angle of elevation of the balloon from the eyes of the girl at any distant is $60^{\circ}$ After some time, the angle of elevation reduces to $30^{\circ}$ (see figure). Find the distance travelled by the balloon during the interval.

Answer.

As, per question;
$
\mathrm{AB}=\mathrm{PQ}=88.2-1.2=87 \mathrm{~m}
$

In right triangle $\mathrm{ABC}$,
$
\tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}
$

$
\begin{aligned}
& \Rightarrow \sqrt{3}=\frac{87}{B C} \\
& \Rightarrow B C=\frac{87}{\sqrt{3}}=29 \sqrt{3} \mathrm{~m}
\end{aligned}
$

In right triangle $\mathrm{PQC}$,
$
\begin{aligned}
& \tan 30^{\circ}=\frac{\mathrm{PQ}}{\mathrm{CQ}} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{87}{29 \sqrt{3}+B Q} \\
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{88.2}{\frac{88.2}{\sqrt{3}}+\mathrm{BQ}} \\
& \Rightarrow 29 \sqrt{3}+B Q=87 \sqrt{3} \\
& \Rightarrow B Q=58 \sqrt{3} \mathrm{~m}
\end{aligned}
$

Hence the distance travelled by the balloon during the interval is $58 \sqrt{3} \mathrm{~m}$.
Ex 9.1 Question 15.

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 $30^{\circ}$, 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 $60^{\circ}$. Find the time taken by the car to reach the foot of the tower from this point.

Answer.

In right triangle ABP,
$
\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BP}}
$

$
\begin{aligned}
& \Rightarrow \frac{1}{\sqrt{3}}=\frac{A B}{B P} \\
& \Rightarrow \mathrm{BP}=\mathrm{AB} \sqrt{3}
\end{aligned}
$

In right triangle $A B Q$,
$
\begin{aligned}
& \tan 60^{\circ}=\frac{A B}{B Q} \\
& \Rightarrow \sqrt{3}=\frac{A B}{B Q} \\
& \Rightarrow B Q=\frac{A B}{\sqrt{3}} \ldots \ldots \ldots . .(\text { (ii) } \\
& \because P Q=B P-B Q \\
& \therefore P Q=A B \sqrt{3}-\frac{A B}{\sqrt{3}}
\end{aligned}
$

$
\begin{aligned}
& =\frac{3 A B-A B}{\sqrt{3}}=\frac{2 A B}{\sqrt{3}}=2 B Q \text { [From eq. (ii)] } \\
& \Rightarrow B Q=\frac{1}{2} P Q
\end{aligned}
$
$\because$ Time taken by the car to travel a distance $P Q=6$ seconds.
$\therefore$ Time taken by the car to travel a distance $B Q$, i.e. $\frac{1}{2} P Q=\frac{1}{2} \times 6=3$ seconds. Hence, the further time taken by the car to reach the foot of the tower is 3 seconds.
Ex 9.1 Question 16.

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

Answer.

Let $\angle \mathrm{APB}=\theta$
Then, $\angle \mathrm{AQB}=\left(90^{\circ}-\theta\right)$

[ $\angle \mathrm{APB}$ and $\angle \mathrm{AQB}$ are complementary]
In right triangle $\mathrm{ABP}$,

$
\begin{aligned}
& \tan \theta=\frac{\mathrm{AB}}{\mathrm{PB}} \\
& \Rightarrow \tan \theta=\frac{\mathrm{AB}}{9}
\end{aligned}
$

In right triangle $\mathrm{ABQ}$,
$
\begin{aligned}
& \tan \left(90^{\circ}-\theta\right)=\frac{\mathrm{AB}}{\mathrm{QB}} \\
& \Rightarrow \cot \theta=\frac{\mathrm{AB}}{4} \ldots \ldots \ldots
\end{aligned}
$

Multiplying eq. (i) and eq. (ii),
$
\begin{aligned}
& \frac{\mathrm{AB}}{9} \cdot \frac{\mathrm{AB}}{4}=\tan \theta \cdot \cot \theta \\
& \Rightarrow \frac{\mathrm{AB}^2}{36}=1 \Rightarrow \mathrm{AB}^2=36 \\
& \Rightarrow \mathrm{AB}=6 \mathrm{~m}
\end{aligned}
$

Hence, the height of the tower is $6 \mathrm{~m}$.
Proved.