## Chapter 09 Some Applications Of Trigonometry

### 9.1 Heights and Distances

In the previous chapter, you have studied about trigonometric ratios. In this chapter, you will be studying about some ways in which trigonometry is used in the life around you.

Let us consider Fig. 8.1 of prvious chapter, which is redrawn below in Fig. 9.1.

Fig. 9.1

In this figure, the line $\mathrm{AC}$ drawn from the eye of the student to the top of the minar is called the line of sight. The student is looking at the top of the minar. The angle $\mathrm{BAC}$, so formed by the line of sight with the horizontal, is called the angle of elevation of the top of the minar from the eye of the student.

Thus, the line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer. The angle of elevation of the point viewed is the angle formed by the line of sight with the horizontal when the point being viewed is above the horizontal level, i.e., the case when we raise our head to look at the object (see Fig. 9.2).

Fig. 9.2

Now, consider the situation given in Fig. 8.2. The girl sitting on the balcony is looking down at a flower pot placed on a stair of the temple. In this case, the line of sight is below the horizontal level. The angle so formed by the line of sight with the horizontal is called the angle of depression.

Thus, the angle of depression of a point on the object being viewed is the angle formed by the line of sight with the horizontal when the point is below the horizontal level, i.e., the case when we lower our head to look at the point being viewed (see Fig. 9.3).

Fig. 9.3

Now, you may identify the lines of sight, and the angles so formed in Fig. 8.3. Are they angles of elevation or angles of depression?

Let us refer to Fig. 9.1 again. If you want to find the height $\mathrm{CD}$ of the minar without actually measuring it, what information do you need? You would need to know the following:

(i) the distance $\mathrm{DE}$ at which the student is standing from the foot of the minar

(ii) the angle of elevation, $\angle \mathrm{BAC}$, of the top of the minar

(iii) the height $\mathrm{AE}$ of the student.

Assuming that the above three conditions are known, how can we determine the height of the minar?

In the figure, $\mathrm{CD}=\mathrm{CB}+\mathrm{BD}$. Here, $\mathrm{BD}=\mathrm{AE}$, which is the height of the student.

To find $\mathrm{BC}$, we will use trigonometric ratios of $\angle \mathrm{BAC}$ or $\angle \mathrm{A}$.

In $\triangle \mathrm{ABC}$, the side $\mathrm{BC}$ is the opposite side in relation to the known $\angle \mathrm{A}$. Now, which of the trigonometric ratios can we use? Which one of them has the two values that we have and the one we need to determine? Our search narrows down to using either $\tan \mathrm{A}$ or $\cot \mathrm{A}$, as these ratios involve $\mathrm{AB}$ and $\mathrm{BC}$.

Therefore, $\tan \mathrm{A}=\dfrac{\mathrm{BC}}{\mathrm{AB}}$ or $\cot \mathrm{A}=\dfrac{\mathrm{AB}}{\mathrm{BC}}$, which on solving would give us $\mathrm{BC}$.

By adding $\mathrm{AE}$ to $\mathrm{BC}$, you will get the height of the minar.

Now let us explain the process, we have just discussed, by solving some problems.

**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.

Fig. 9.4

**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., AB. Also, ACB is a triangle, right-angled at B.

To solve the problem, we choose the trigonometric ratio $\tan 60^{\circ}$ (or $\cot 60^{\circ}$ ), as the ratio involves $\mathrm{AB}$ and $\mathrm{BC}$.

$\begin{array}{rlrl} & \text { Now, } & \tan 60^{\circ} & =\dfrac{\mathrm{AB}}{\mathrm{BC}} \\ & \text { i.e., } & \sqrt{3} & =\dfrac{\mathrm{AB}}{15} \\ \text { i.e., }& & \mathrm{AB} & =15 \sqrt{3}\end{array}$

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 $\sqrt{3}=1.73$ )

Fig. 9.5

**Solution** : In Fig. 9.5, the electrician is required to reach the point $\mathrm{B}$ on the pole $\mathrm{AD}$.

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

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

Therefore, $$ \mathrm{BC}=\dfrac{3.7 \times 2}{\sqrt{3}}=4.28 \mathrm{~m} \text { (approx.) } $$

i.e., the length of the ladder should be $4.28 \mathrm{~m}$.

Now, $$ \dfrac{\mathrm{DC}}{\mathrm{BD}}=\cot 60^{\circ}=\dfrac{1}{\sqrt{3}} $$

i.e., $$ \mathrm{DC}=\dfrac{3.7}{\sqrt{3}}=2.14 \mathrm{~m} \text { (approx.) } $$

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 $\mathrm{E}$ and we are required to find the height of the chimney.

Fig. 9.6

We have $$\mathrm{AB}=\mathrm{AE}+\mathrm{BE}=\mathrm{AE}+1.5$$

and $$ \mathrm{DE}=\mathrm{CB}=28.5 \mathrm{~m} $$

To determine AE, we choose a trigonometric ratio, which involves both $\mathrm{AE}$ and DE. Let us choose the tangent of the angle of elevation.

Now, $$ \begin{aligned} \tan 45^{\circ} & =\dfrac{\mathrm{AE}}{\mathrm{DE}} \\ \end{aligned} $$

i.e., $$1 =\dfrac{\mathrm{AE}}{28.5}$$

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 $\mathrm{P}$ 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 $\mathrm{P}$ 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.

Fig. 9.7

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

We have $$ \begin{aligned} \tan 30^{\circ} & =\dfrac{\mathrm{AB}}{\mathrm{AP}} \ \end{aligned} $$

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

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 $\Delta \mathrm{PAD}$,

$$ \tan 45^{\circ}=\dfrac{\mathrm{AD}}{\mathrm{AP}}=\dfrac{10+x}{10 \sqrt{3}} $$

Therefore, $$ 1=\dfrac{10+x}{10 \sqrt{3}} $$

i.e., $$ x=10(\sqrt{3}-1)=7.32 $$

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** : In Fig. 9.8, AB is the tower and $\mathrm{BC}$ 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 $30^{\circ}$.

Fig. 9.8

Now, let $\mathrm{AB}$ be $h \mathrm{~m}$ and $\mathrm{BC}$ be $x \mathrm{~m}$. According to the question, $\mathrm{DB}$ is $40 \mathrm{~m}$ longer than $\mathrm{BC}$.

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

Now, we have two right triangles $\mathrm{ABC}$ and $\mathrm{ABD}$.

$\begin{array}{rlrl} \text {In } \Delta \mathrm{ABC} & \tan 60^{\circ} & =\dfrac{\mathrm{AB}}{\mathrm{BC}} \\ \text { or } & & \sqrt{3} & =\dfrac{h}{x} \\ \text {In } \Delta \mathrm{ABD} & \tan 30^{\circ} & =\dfrac{\mathrm{AB}}{\mathrm{BD}} \\ & \text { i.e., } & \dfrac{1}{\sqrt{3}} & =\dfrac{h}{x+40}\end{array}$

From (1), we have

$$ h=x \sqrt{3} $$

Putting this value in (2), we get $(x \sqrt{3}) \sqrt{3}=x+40$, i.e., $3 x=x+40$

$$ \begin{align*} \text{ अर्थात् } \qquad\qquad & x=20 \\ \text{ इसलिए }\qquad\qquad & h=20 \sqrt{3} \tag{[From (1)]} \end{align*} $$

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 multistoreyed building and the distance between the two buildings.

**Solution** : In Fig. 9.9, PC denotes the multistoryed building and $\mathrm{AB}$ 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 $\mathrm{PQ}$ and $\mathrm{BD}$. Therefore, $\angle \mathrm{QPB}$ and $\angle \mathrm{PBD}$ are alternate angles, and so are equal. So $\angle \mathrm{PBD}=30^{\circ}$. Similarly, $\quad \angle \mathrm{PAC}=45^{\circ}$. In right $\Delta \mathrm{PBD}$, we have

Fig. 9.9

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

In right $\Delta \mathrm{PAC}$, we have

$\dfrac{\mathrm{PC}}{\mathrm{AC}}=\tan 45^{\circ}=1$

i.e., $\quad P C=A C$

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

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

$ \mathrm{PD}=\dfrac{8}{\sqrt{3}-1}=\dfrac{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 $\mathrm{AB}$ is the width of the river. $\mathrm{P}$ 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 $D A P B$.

Fig. 9.10

Now, $$ A B=A D+D B $$

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

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

i.e., $\quad\quad\quad \dfrac{1}{\sqrt{3}}=\dfrac{3}{\mathrm{AD}}$ 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 $3(\sqrt{3}+1) \mathrm{m}$.

### EXERCISE 9.1

**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 Fig. 9.11).

Fig. 9.11

## Show Answer

**Solution**

It can be observed from the figure that $A B$ is the pole.

In $\triangle ABC$,

$ \begin{aligned} & \dfrac{AB}{AC}=\sin 30^{\circ} \\ & \dfrac{AB}{20}=\dfrac{1}{2} \\ & AB=\dfrac{20}{2}=10 \end{aligned} $

Therefore, the height of the pole is $10 m$.

**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.

## Show Answer

**Solution**

Let $A C$ was the original tree. Due to storm, it was broken into two parts. The broken part $A^{\prime} B$ is making $30^{\circ}$ with the ground.

In $\triangle A^{\prime} BC$,

$ \begin{aligned} & \dfrac{BC}{A^{\prime} C}=\tan 30^{\circ} \\ & \dfrac{BC}{8}=\dfrac{1}{\sqrt{3}} \\ & BC=(\dfrac{8}{\sqrt{3}}) m \\ & \dfrac{A^{\prime} C}{A^{\prime} B}=\cos 30^{\circ} \\ & \dfrac{8}{A^{\prime} B}=\dfrac{\sqrt{3}}{2} \\ & A^{\prime} B=(\dfrac{16}{\sqrt{3}}) m \end{aligned} $

Height of tree $=A^{\prime} B+BC$

$ \begin{aligned} & =(\dfrac{16}{\sqrt{3}}+\dfrac{8}{\sqrt{3}}) m=\dfrac{24}{\sqrt{3}} m \\ & =8 \sqrt{3} m \end{aligned} $

Hence, the height of the tree is $8 \sqrt{3} m$.

**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?

## Show Answer

**Solution**

It can be observed that AC and PR are the slides for younger and elder children respectively.

In $\triangle ABC$,

$ \dfrac{AB}{AC}=\sin 30^{\circ} $

$\dfrac{1.5}{AC}=\dfrac{1}{2}$

$AC=3 m$

In $\triangle PQR$,

$\dfrac{PQ}{PR}=\sin 60$

$\dfrac{3}{PR}=\dfrac{\sqrt{3}}{2}$

$PR=\dfrac{6}{\sqrt{3}}=2 \sqrt{3} m$

Therefore, the lengths of these slides are $3 m$ and $2 \sqrt{3} m$.

**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.

## Show Answer

**Solution**

Let $A B$ be the tower and the angle of elevation from point $C$ (on ground) is

$30^{\circ}$.

In $\triangle ABC$,

$ \begin{aligned} & \dfrac{AB}{BC}=\tan 30^{\circ} \\ & \dfrac{AB}{30}=\dfrac{1}{\sqrt{3}} \\ & AB=\dfrac{30}{\sqrt{3}}=10 \sqrt{3} m \end{aligned} $

Therefore, the height of the tower is $10 \sqrt{3} m$.

**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.

## Show Answer

**Solution**

Let $K$ be the kite and the string is tied to point $P$ on the ground.

In $\triangle KLP$, $\dfrac{KL}{KP}=\sin 60^{\circ}$

$\dfrac{60}{K P}=\dfrac{\sqrt{3}}{2}$

$KP=\dfrac{120}{\sqrt{3}}=40 \sqrt{3} m$

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

**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.

## Show Answer

**Solution**

Let the boy was standing at point $S$ initially. He walked towards the building and reached at point $T$.

It can be observed that

$P R=P Q-R Q$

$=(30-1.5) m=28.5 m=\dfrac{57}{2} m$

In $\triangle PAR$,

$\dfrac{PR}{AR}=\tan 30^{\circ}$

$\dfrac{57}{2 AR}=\dfrac{1}{\sqrt{3}}$

$AR=(\dfrac{57}{2} \sqrt{3}) m$

In $\triangle PRB$,

$ \begin{aligned} & \dfrac{PR}{BR}=\tan 60^{\circ} \\ & \dfrac{57}{2 BR}=\sqrt{3} \\ & BR=\dfrac{57}{2 \sqrt{3}}=(\dfrac{19 \sqrt{3}}{2}) m \end{aligned} $

$S T=A B$

$=AR-BR=(\dfrac{57 \sqrt{3}}{2}-\dfrac{19 \sqrt{3}}{2}) m$

$=(\dfrac{38 \sqrt{3}}{2}) m=19 \sqrt{3} m$

Hence, he walked $19 \sqrt{3} m$ towards the building.

**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.

## Show Answer

**Solution**

Let $B C$ be the building, $A B$ be the transmission tower, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\triangle BCD$,

$\dfrac{BC}{CD}=\tan 45^{\circ}$

$\dfrac{20}{CD}=1$

$CD=20 m$

In $\triangle ACD$,

$ \begin{aligned} & \dfrac{AC}{CD}=\tan 60^{\circ} \\ & \dfrac{AB+BC}{CD}=\sqrt{3} \\ & \dfrac{AB+20}{20}=\sqrt{3} \\ & AB=(20 \sqrt{3}-20) m \\ & \quad=20(\sqrt{3}-1) m \end{aligned} $

Therefore, the height of the transmission tower is $20(\sqrt{3}-1)$ m.

**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.

## Show Answer

**Solution**

Let $A B$ be the statue, $B C$ be the pedestal, and $D$ be the point on the ground from where the elevation angles are to be measured.

In $\triangle BCD$,

$ \begin{aligned} & \dfrac{B C}{C D}=\tan 45^{\circ} \\ & \dfrac{B C}{C D}=1 \\ & B C=C D \end{aligned} $

In $\triangle ACD$,

$ \begin{aligned} & \dfrac{AB+BC}{CD}=\tan 60^{\circ} \\ & \dfrac{AB+BC}{BC}=\sqrt{3} \\ & 1.6+BC=BC \sqrt{3} \\ & BC(\sqrt{3}-1)=1.6 \\ & BC=\dfrac{(1.6)(\sqrt{3}+1)}{(\sqrt{3}-1)(\sqrt{3}+1)} \\ & \quad=\dfrac{1.6(\sqrt{3}+1)}{(\sqrt{3})^{2}-(1)^{2}} \\ & \quad=\dfrac{1.6(\sqrt{3}+1)}{2}=0.8(\sqrt{3}+1) \end{aligned} $

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

**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.

## Show Answer

**Solution**

Let $A B$ be the building and $C D$ be the tower.

In $\triangle CDB$, $\dfrac{CD}{BD}=\tan 60^{\circ}$

$\dfrac{50}{BD}=\sqrt{3}$

$BD=\dfrac{50}{\sqrt{3}}$

In $\triangle A B D$,

$\dfrac{AB}{BD}=\tan 30^{\circ}$

$AB=\dfrac{50}{\sqrt{3}} \times \dfrac{1}{\sqrt{3}}=\dfrac{50}{3}=16 \dfrac{2}{3}$

Therefore, the height of the building is

$ 16 \dfrac{2}{3} m $

**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.

## Show Answer

**Solution**

Let $A B$ and $C D$ be the poles and $O$ is the point from where the elevation angles are measured.

In $\triangle ABO$,

$ \begin{aligned} & \dfrac{AB}{BO}=\tan 60^{\circ} \\ & \dfrac{AB}{BO}=\sqrt{3} \\ & BO=\dfrac{AB}{\sqrt{3}} \end{aligned} $

In $\triangle CDO$, $\dfrac{CD}{DO}=\tan 30^{\circ}$

$\dfrac{CD}{80-BO}=\dfrac{1}{\sqrt{3}}$

$CD \sqrt{3}=80-BO$

$CD \sqrt{3}=80-\dfrac{AB}{\sqrt{3}}$

$CD \sqrt{3}+\dfrac{AB}{\sqrt{3}}=80$

Since the poles are of equal heights,

$C D=A B$

$CD[\sqrt{3}+\dfrac{1}{\sqrt{3}}]=80$

$CD(\dfrac{3+1}{\sqrt{3}})=80$

$CD=20 \sqrt{3} m$

$BO=\dfrac{AB}{\sqrt{3}}=\dfrac{CD}{\sqrt{3}}=(\dfrac{20 \sqrt{3}}{\sqrt{3}}) m=20 m$

$DO=BD-BO=(80-20) m=60 m$

Therefore, the height of poles is $20 \sqrt{3} m$ and the point is $20 m$ and $60 m$ far from these poles.

**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 joing this point to the foot of the tower, the angle of elevation of the top of the tower is $30^{\circ}$ (see Fig. 9.12). Find the height of the tower and the width of the canal.

Fig. 9.12

## Show Answer

**Solution**

In $\triangle ABC$,

$\dfrac{AB}{BC}=\tan 60^{\circ}$

$\dfrac{AB}{BC}=\sqrt{3}$

$BC=\dfrac{AB}{\sqrt{3}}$

In $\triangle ABD$,

$\dfrac{AB}{BD}=\tan 30^{\circ}$

$\dfrac{AB}{BC+CD}=\dfrac{1}{\sqrt{3}}$

$\dfrac{AB}{\dfrac{AB}{\sqrt{3}}+20}=\dfrac{1}{\sqrt{3}}$

$\dfrac{AB \sqrt{3}}{AB+20 \sqrt{3}}=\dfrac{1}{\sqrt{3}}$

$3 AB=AB+20 \sqrt{3}$

$2 AB=20 \sqrt{3}$

$AB=10 \sqrt{3} m$

$BC=\dfrac{AB}{\sqrt{3}}=(\dfrac{10 \sqrt{3}}{\sqrt{3}}) m=10 m$

Therefore, the height of the tower is $10 \sqrt{3} m$ and the width of the canal is $10 m$.

**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.

## Show Answer

**Solution**

Let $A B$ be a building and $C D$ be a cable tower.

In $\triangle ABD$,

$\dfrac{AB}{BD}=\tan 45^{\circ}$

$\dfrac{7}{BD}=1$

$BD=7 m$

In $\triangle ACE$,

$AE=BD=7 m$

$\dfrac{CE}{AE}=\tan 60^{\circ}$

$\dfrac{CE}{7}=\sqrt{3}$

$CE=7 \sqrt{3} m$

$CD=CE+ED=(7 \sqrt{3}+7) m$

$ =7(\sqrt{3}+1) m $

Therefore, the height of the cable tower is $7(\sqrt{3}+1) m$.

**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 the two ships.

## Show Answer

**Solution**

Let $A B$ be the lighthouse and the two ships be at point $C$ and $D$ respectively.

In $\triangle ABC$,

$ \begin{aligned} & \dfrac{AB}{BC}=\tan 45^{\circ} \\ & \dfrac{75}{BC}=1 \\ & BC=75 m \end{aligned} $

In $\triangle ABD$,

$ \dfrac{AB}{BD}=\tan 30^{\circ} $

$\dfrac{75}{BC+CD}=\dfrac{1}{\sqrt{3}}$

$\dfrac{75}{75+CD}=\dfrac{1}{\sqrt{3}}$

$75 \sqrt{3}=75+CD$

$75(\sqrt{3}-1) m=CD$

Therefore, the distance between the two ships is $75(\sqrt{3}-1) m$.

**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 instant is $60^{\circ}$. After some time, the angle of elevation reduces to $30^{\circ}$ (see Fig. 9.13). Find the distance travelled by the balloon during the interval.

Fig. 9.13

## Show Answer

**Solution**

Let the initial position $A$ of balloon change to $B$ after some time and $C D$ be the girl.

In $\triangle ACE$,

$\dfrac{AE}{CE}=\tan 60^{\circ}$ $\dfrac{AF-EF}{CE}=\tan 60^{\circ}$

$\dfrac{88.2-1.2}{C E}=\sqrt{3}$

$\dfrac{87}{CE}=\sqrt{3}$

$CE=\dfrac{87}{\sqrt{3}}=29 \sqrt{3} m$

In $\triangle BCG$,

$\dfrac{BG}{CG}=\tan 30^{\circ}$

$\dfrac{88.2-1.2}{CG}=\dfrac{1}{\sqrt{3}}$

$87 \sqrt{3} m=CG$

Distance travelled by balloon $=EG=CG-CE$

$=(87 \sqrt{3}-29 \sqrt{3}) m$

$=58 \sqrt{3} m$

**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.

## Show Answer

**Solution**

Let $A B$ be the tower.

Initial position of the car is $C$, which changes to $D$ after six seconds.

In $\triangle ADB$,

$\dfrac{AB}{DB}=\tan 60^{\circ}$

$\dfrac{AB}{DB}=\sqrt{3}$

$DB=\dfrac{AB}{\sqrt{3}}$

In $\triangle ABC$,

$\dfrac{AB}{BC}=\tan 30^{\circ}$

$\dfrac{AB}{BD+DC}=\dfrac{1}{\sqrt{3}}$

$ \begin{aligned} & AB \sqrt{3}=BD+DC \\ & AB \sqrt{3}=\dfrac{AB}{\sqrt{3}}+DC \\ & DC=AB \sqrt{3}-\dfrac{AB}{\sqrt{3}}=AB(\sqrt{3}-\dfrac{1}{\sqrt{3}}) \\ & =\dfrac{2 AB}{\sqrt{3}} \end{aligned} $

Time taken by the car to travel distance DC $(.$ i.e., $.\dfrac{2 AB}{\sqrt{3}})=6$ seconds

Time taken by the car to travel distance $D B(.$ i.e., $.\dfrac{AB}{\sqrt{3}})=\dfrac{6}{\dfrac{2 AB}{\sqrt{3}}} \times \dfrac{AB}{\sqrt{3}}$

$=\dfrac{6}{2}=3$ seconds

### 9.2 Summary

In this chapter, you have studied the following points :

**1.** (i) The line of sight is the line drawn from the eye of an observer to the point in the object viewed by the observer.

(ii) The angle of elevation of an object viewed, is the angle formed by the line of sight with the horizontal when it is above the horizontal level, i.e., the case when we raise our head to look at the object.

(iii) The angle of depression of an object viewed, is the angle formed by the line of sight with the horizontal when it is below the horizontal level, i.e., the case when we lower our head to look at the object.

**2.** The height or length of an object or the distance between two distant objects can be determined with the help of trigonometric ratios.