Examples - Chapter 9 Some Applications Of Trigonometry class 10 ncert solutions Maths - SaraNextGen [2024-2025]

Updated By SaraNextGen

On April 24, 2024, 11:35 AM

Examples (Revised) : Chapter 9 - Some Applications Of Trigonometry - Ncert Solutions class 10 - Maths

Example 1:

A tower stands vertically on the ground. From a point on the ground, which is $15 \mathrm{~m}$ away from the foot of the tower, the angle of elevation of the top of the tower is found to be $60^{\circ}$. Find the height of the tower.

Solution : First let us draw a simple diagram to represent the problem (see Fig. 9.4). Here AB represents the tower, $\mathrm{CB}$ is the distance of the point from the tower and $\angle \mathrm{ACB}$ is the angle of elevation. We need to determine the height of the tower, i.e., $\mathrm{AB}$. Also, $\mathrm{ACB}$ is a triangle, right-angled at $\mathrm{B}$.
To solve the problem, we choose the trigonometric ratio $\tan 60^{\circ}$ (or $\cot 60^{\circ}$ ), as the ratio involves $\mathrm{AB}$ and $\mathrm{BC}$.

Now,
$
\tan 60^{\circ}=\frac{A B}{B C}
$
i.e.,
$
\begin{aligned}
& \sqrt{3}=\frac{\mathrm{AB}}{15} \\
& \mathrm{AB}=15 \sqrt{3}
\end{aligned}
$
i.e.,

Hence, the height of the tower is $15 \sqrt{3} \mathrm{~m}$.

Example 2 :

An electrician has to repair an electric fault on a pole of height $5 \mathrm{~m}$. She needs to reach a point $1.3 \mathrm{~m}$ below the top of the pole to undertake the repair work (see Fig. 9.5). What should be the length of the ladder that she should use which, when inclined at an angle of $60^{\circ}$ to the horizontal, would enable her to reach the required position? Also, how far from the foot of the pole should she place the foot of the ladder? (You may take $\sqrt{3}=1.73)$

Solution :

In Fig. 9.5, the electrician is required to reach the point $B$ on the pole $A D$.

So,
$
\mathrm{BD}=\mathrm{AD}-\mathrm{AB}=(5-1.3) \mathrm{m}=3.7 \mathrm{~m} .
$

Here, $\mathrm{BC}$ represents the ladder. We need to find its length, i.e., the hypotenuse of the right triangle BDC.

Now, can you think which trigonometic ratio should we consider?
It should be $\sin 60^{\circ}$.

So,
$
\frac{\mathrm{BD}}{\mathrm{BC}}=\sin 60^{\circ} \text { or } \frac{3.7}{\mathrm{BC}}=\frac{\sqrt{3}}{2}
$

Therefore,
$
\mathrm{BC}=\frac{3.7 \times 2}{\sqrt{3}}=4.28 \mathrm{~m} \text { (approx.) }
$
i.e., the length of the ladder should be $4.28 \mathrm{~m}$.

Now,
$
\frac{\mathrm{DC}}{\mathrm{BD}}=\cot 60^{\circ}=\frac{1}{\sqrt{3}}
$
i.e.,
$
\mathrm{DC}=\frac{3.7}{\sqrt{3}}=2.14 \mathrm{~m} \text { (approx.) }
$

Therefore, she should place the foot of the ladder at a distance of $2.14 \mathrm{~m}$ from the pole.

Example 3 :

An observer $1.5 \mathrm{~m}$ tall is $28.5 \mathrm{~m}$ away from a chimney. The angle of elevation of the top of the chimney from her eyes is $45^{\circ}$. What is the height of the chimney?

Solution :

Here, $A B$ is the chimney, $C D$ the observer and $\angle \mathrm{ADE}$ the angle of elevation (see Fig. 9.6). In this case, $A D E$ is a triangle, right-angled at $E$ and we are required to find the height of the chimney.

We have $\mathrm{AB}=\mathrm{AE}+\mathrm{BE}=\mathrm{AE}+1.5$ and $\mathrm{DE}=\mathrm{CB}=28.5 \mathrm{~m}$

Now,
$
\tan 45^{\circ}=\frac{\mathrm{AE}}{\mathrm{DE}}
$
i.e.,
$
1=\frac{\mathrm{AE}}{28.5}
$

Therefore,
$
\mathrm{AE}=28.5
$

So the height of the chimney $(A B)=(28.5+1.5) \mathrm{m}=30 \mathrm{~m}$

Example 4 :

From a point $P$ on the ground the angle of elevation of the top of a $10 \mathrm{~m}$ tall building is $30^{\circ}$. A flag is hoisted at the top of the building and the angle of elevation of the top of the flagstaff from $P$ is $45^{\circ}$. Find the length of the flagstaff and the distance of the building from the point P. (You may take $\sqrt{3}=1.732$ )
Solution :

In Fig. 9.7, AB denotes the height of the building, BD the flagstaff and $P$ the given point. Note that there are two right triangles $\mathrm{PAB}$ and $\mathrm{PAD}$. We are required to find the length of the flagstaff, i.e., DB and the distance of the building from the point $P$, i.e., PA.

Since, we know the height of the building $A B$, we will first consider the right $\triangle \mathrm{PAB}$.

We have
$
\tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{AP}}
$

i.e.,
$
\frac{1}{\sqrt{3}}=\frac{10}{\mathrm{AP}}
$

Therefore,
$
\mathrm{AP}=10 \sqrt{3}
$

i.e., the distance of the building from $P$ is $10 \sqrt{3} \mathrm{~m}=17.32 \mathrm{~m}$.

Next, let us suppose $\mathrm{DB}=x \mathrm{~m}$. Then $\mathrm{AD}=(10+x) \mathrm{m}$.
Now, in right $\triangle \mathrm{PAD}, \quad \tan 45^{\circ}=\frac{\mathrm{AD}}{\mathrm{AP}}=\frac{10+x}{10 \sqrt{3}}$
Therefore, $\quad 1=\frac{10+x}{10 \sqrt{3}}$
i.e., $\quad x=10(\sqrt{3}-1)=7.32$

So, the length of the flagstaff is $7.32 \mathrm{~m}$.

Example 5 :

The shadow of a tower standing on a level ground is found to be $40 \mathrm{~m}$ longer when the Sun's altitude is $30^{\circ}$ than when it is $60^{\circ}$. Find the height of the tower.

Solution : I

n Fig. 9.8, AB is the tower and $B C$ is the length of the shadow when the Sun's altitude is $60^{\circ}$, i.e., the angle of elevation of the top of the tower from the tip of the shadow is $60^{\circ}$ and $\mathrm{DB}$ is the length of the shadow, when the angle of elevation is $30^{\circ}$.

So,
$
\mathrm{DB}=(40+x) \mathrm{m}
$

Now, we have two right triangles $A B C$ and $A B D$.
In $\triangle \mathrm{ABC}, \quad \tan 60^{\circ}=\frac{\mathrm{AB}}{\mathrm{BC}}$
or,
$
\sqrt{3}=\frac{h}{x}
$

In $\triangle \mathrm{ABD}, \quad \tan 30^{\circ}=\frac{\mathrm{AB}}{\mathrm{BD}}$
i.e.,
$
\frac{1}{\sqrt{3}}=\frac{h}{x+40}
$

From (1), we have $\quad h=x \sqrt{3}$
Putting this value in (2), we get $(x \sqrt{3}) \sqrt{3}=x+40$, i.e., $3 x=x+40$
i.e.,
$
x=20
$

So,
$
h=20 \sqrt{3}
$
[From (1)]
Therefore, the height of the tower is $20 \sqrt{3} \mathrm{~m}$.

Example 6:

The angles of depression of the top and the bottom of an $8 \mathrm{~m}$ tall building from the top of a multi-storeyed building are $30^{\circ}$ and $45^{\circ}$, respectively. Find the height of the multi-storeyed building and the distance between the two buildings.
Solution :

In Fig. 9.9, PC denotes the multistoryed building and $A B$ denotes the $8 \mathrm{~m}$ tall building. We are interested to determine the height of the multi-storeyed building, i.e., PC and the distance between the two buildings, i.e., AC. Look at the figure carefully. Observe that $\mathrm{PB}$ is a transversal to the parallel lines $P Q$ and $B D$. Therefore, $\angle \mathrm{QPB}$ and $\angle \mathrm{PBD}$ are alternate angles, and so are equal. So $\angle \mathrm{PBD}=30^{\circ}$. Similarly, $\angle \mathrm{PAC}=45^{\circ}$. In right $\triangle \mathrm{PBD}$, we have

$
\frac{\mathrm{PD}}{\mathrm{BD}}=\tan 30^{\circ}=\frac{1}{\sqrt{3}} \text { or } \mathrm{BD}=\mathrm{PD} \sqrt{3}
$

In right $\triangle \mathrm{PAC}$, we have
$
\frac{\mathrm{PC}}{\mathrm{AC}}=\tan 45^{\circ}=1
$
i.e.,
$
\mathrm{PC}=\mathrm{AC}
$

Also,
$
\mathrm{PC}=\mathrm{PD}+\mathrm{DC} \text {, therefore, } \mathrm{PD}+\mathrm{DC}=\mathrm{AC} \text {. }
$

Since, $\mathrm{AC}=\mathrm{BD}$ and $\mathrm{DC}=\mathrm{AB}=8 \mathrm{~m}$, we get $\mathrm{PD}+8=\mathrm{BD}=\mathrm{PD} \sqrt{3}$ (Why?)
This gives $\quad P D=\frac{8}{\sqrt{3}-1}=\frac{8(\sqrt{3}+1)}{(\sqrt{3}+1)(\sqrt{3}-1)}=4(\sqrt{3}+1) \mathrm{m}$.
So, the height of the multi-storeyed building is $\{4(\sqrt{3}+1)+8\} \mathrm{m}=4(3+\sqrt{3}) \mathrm{m}$ and the distance between the two buildings is also $4(3+\sqrt{3}) \mathrm{m}$.

Example 7 :

From a point on a bridge across a river, the angles of depression of the banks on opposite sides of the river are $30^{\circ}$ and $45^{\circ}$, respectively. If the bridge is at a height of $3 \mathrm{~m}$ from the banks, find the width of the river.

Solution :

In Fig 9.10, A and B represent points on the bank on opposite sides of the river, so that $A B$ is the width of the river. $P$ is a point on the bridge at a height of 3 $\mathrm{m}$, i.e., $\mathrm{DP}=3 \mathrm{~m}$. We are interested to determine the width of the river, which is the length of the side $A B$ of the $D A P B$.
Now,
$
\mathrm{AB}=\mathrm{AD}+\mathrm{DB}
$

In right $\triangle \mathrm{APD}, \angle \mathrm{A}=30^{\circ}$.

So,
$
\tan 30^{\circ}=\frac{\mathrm{PD}}{\mathrm{AD}}
$

i.e., $\quad \frac{1}{\sqrt{3}}=\frac{3}{\mathrm{AD}}$ or $\mathrm{AD}=3 \sqrt{3} \mathrm{~m}$

Also, in right $\triangle \mathrm{PBD}, \angle \mathrm{B}=45^{\circ}$. So, $\mathrm{BD}=\mathrm{PD}=3 \mathrm{~m}$.
Now,
$
\mathrm{AB}=\mathrm{BD}+\mathrm{AD}=3+3 \sqrt{3}=3(1+\sqrt{3}) \mathrm{m} .
$

Therefore, the width of the river is $3(\sqrt{3}+1) \mathrm{m}$.