SEBA Class 10 Maths Chapter 9 Solutions: Some Applications of Trigonometry | New Book 2026

Get free, step-by-step solutions to SEBA Class 10 Maths Chapter 9: Some Applications of Trigonometry (Exercise 9) from the new SCERT textbook 2026. In this article, all textbook questions—from Question 1 to 22—have been solved in simple way.

If you are a Class 10 student under the SEBA/Assam Board preparing for your upcoming HSLC examinations, these reliable, easy-to-understand solutions will help you master trigonometry applications and score full marks in your board exams.

See More:

Chapter 8.1 Introduction to Trigonometry

Chapter 8.2 Introduction to Trigonometry

Chapter 8.3 Introduction to Trigonometry

Chapter 8.4 Introduction to Trigonometry

1. A circus artist is climbing a 20 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).

$\mathbf{Sol^n.}$

Let $AB$ be the vertical pole and $AC$ be the tightly stretched rope tied to the ground at point $C$.

Given:

$AC = 20\text{ m}$

$\angle ACB = 30^\circ$

Then in right $\Delta ABC$:

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

$\implies \frac{1}{2} = \frac{AB}{20}$

$\implies AB = \frac{20}{2}$

$\implies AB = 10\text{ m}$

$\therefore$ The height of the pole is $10\text{ m}$.

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 m. Find the height of the tree.

$\mathbf{Sol^n.}$

Let $DB$ be the original tree.

Suppose it breaks at point $A$, and the broken part $AD$ bends so that its top touches the ground at point $C$.

Therefore, $AD = AC$.

Total height of the tree $= AB + AD = AB + AC$.

Given:

$BC = 8\text{ m}$

$\angle ACB = 30^\circ$

In right $\Delta ABC$:

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

$\implies \frac{1}{\sqrt{3}} = \frac{AB}{8}$

$\implies AB = \frac{8}{\sqrt{3}}\text{ m} \quad \text{— (1)}$

Also,

$\cos 30^\circ = \frac{BC}{AC}$

$\implies \frac{\sqrt{3}}{2} = \frac{8}{AC}$

$\implies AC = \frac{16}{\sqrt{3}}\text{ m} \quad \text{— (2)}$

Now,

Total height of the tree $= AB + AC$

$= \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}}$

$= \frac{24}{\sqrt{3}}$

$= \frac{24 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$= \frac{24\sqrt{3}}{3}$

$= 8\sqrt{3}\text{ m}$

$\therefore$ The height of the tree is $8\sqrt{3}\text{ m}$.

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 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 m, and inclined at an angle of $60^\circ$ to the ground. What should be the length of the slide in each case?

$\mathbf{Sol^n.}$

(i) For children below the age of 5 years:

Let $AB$ be the height of the slide and $AC$ be the length of the slide.

Given: $AB = 1.5\text{ m}$

$\angle ACB = 30^\circ$

In right $\Delta ABC$,

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

$\implies \frac{1}{2} = \frac{1.5}{AC}$

$\implies AC = 1.5 \times 2$

$\implies AC = 3\text{ m}$

(ii) For elder children:

Let $PQ$ be the height of the slide and $PR$ be the length of the slide.

Given: $PQ = 3\text{ m}$

$\angle PRQ = 60^\circ$

In right $\Delta PQR$:

$\sin 60^\circ = \frac{PQ}{PR}$

$\implies \frac{\sqrt{3}}{2} = \frac{3}{PR}$

$\implies PR = \frac{3 \times 2}{\sqrt{3}}$

$\implies PR = \frac{\sqrt{3} \times \sqrt{3} \times 2}{\sqrt{3}}$

$\implies PR = 2\sqrt{3}\text{ m}$

Hence, length of the slide for below 5 is $3\text{ m}$ and

Length of the slide for elders children is $2\sqrt{3}\text{ m}$.

4. 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 $30^\circ$. Find the height of the tower.

$\mathbf{Sol^n.}$

Let $AB$ be the height of the tower, and $C$ be the point on the ground.

∴ Distance from the foot of the tower, $BC = 30\text{ m}$

Angle of elevation, $\angle ACB = 30^\circ$

In right $\Delta ABC$:

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

$\implies \frac{1}{\sqrt{3}} = \frac{AB}{30}$

$\implies AB = \frac{30}{\sqrt{3}}$

$= \frac{30 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$= \frac{30\sqrt{3}}{3}$

$= 10\sqrt{3}\text{ m}$

$\therefore$ The height of the tower is $10\sqrt{3}\text{ m}$.

5. 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 $60^\circ$. Find the length of the string, assuming that there is no slack in the string.  

$\mathbf{Sol^n.}$

Let $A$ be the kite flying at a vertical height $AB$ above the ground, and $AC$ be the length of the string tied at point $C$ on the ground.

Given:

Height of the kite, $AB = 60\text{ m}$

Angle of inclination, $\angle ACB = 60^\circ$

In right $\Delta ABC$:

$\sin 60^\circ = \frac{AB}{AC}$

$\implies \frac{\sqrt{3}}{2} = \frac{60}{AC}$

$\implies AC = \frac{60 \times 2}{\sqrt{3}}$

$\implies AC = \frac{120}{\sqrt{3}}$

$\implies AC = \frac{120 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$\implies AC = \frac{120\sqrt{3}}{3}$

$\implies AC = 40\sqrt{3}\text{ m}$

$\therefore$ The length of the string AC is $40\sqrt{3}\text{ m}$.

6. 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 $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked towards the building.

$\mathbf{Sol^n.}$

Let $AZ$ be the tall building and $DY$ be the initial position of the boy of height $1.5\text{ m}$.

Let $CX$ be his second position after walking towards the building.

Here, $DY = CX = BZ = 1.5\text{ m}$

$AB = AZ – BZ = 30 – 1.5 = 28.5\text{ m}$

$\angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$

In right $\Delta ABD$,

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

$\implies \frac{1}{\sqrt{3}} = \frac{28.5}{BD}$

$\implies BD = 28.5\sqrt{3}\text{ m}$

Again,

In right $\Delta ABC$,

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

$\implies \sqrt{3} = \frac{28.5}{BC}$

$\implies BC = \frac{28.5}{\sqrt{3}}\text{ m}$

∴ The distance walked by the boy towards the building is $CD = BD – BC$

$= 28.5\sqrt{3} – \frac{28.5}{\sqrt{3}}$

$= 28.5\left(\sqrt{3} – \frac{1}{\sqrt{3}}\right)$

$= 28.5\left(\frac{3 – 1}{\sqrt{3}}\right)$

$= \frac{28.5 \times 2}{\sqrt{3}}$

$= \frac{57}{\sqrt{3}}$

$= \frac{57 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}}$

$= \frac{57\sqrt{3}}{3}$

$= 19\sqrt{3}\text{ m}$

$\therefore$ The distance walked by the boy towards the building is $19\sqrt{3}\text{ m}$.

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 m high building are $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.     

$\mathbf{Sol^n.}$

Let $BC$ be the building of height $20\text{ m}$, and $AB$ be the transmission tower of height $h\text{ m}$ fixed on top of it.

Let $D$ be the point of observation on the ground.

Also, let $AB = h\text{ m}$

Given: $BC = 20\text{ m}$

$\angle BDC = 45^\circ$

$\angle ADC = 60^\circ$

In right $\Delta BCD$:

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

$\implies 1 = \frac{20}{CD}$

$\implies CD = 20\text{ m} \quad \text{— (1)}$

In right $\Delta ACD$:

$\tan 60^\circ = \frac{AC}{CD}$

$\implies \sqrt{3} = \frac{AB + BC}{CD}$

$\implies \sqrt{3} = \frac{h + 20}{20} \quad [\text{From equation(1)}]$

$\implies h + 20 = 20\sqrt{3}$

$\implies h = 20\sqrt{3} – 20$

$\implies h = 20(\sqrt{3} – 1)\text{ m}$

$\therefore$ The height of the transmission tower is $20(\sqrt{3} – 1)\text{ m}$.

8. 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 $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.

$\mathbf{Sol^n.}$

Let $BC$ be the pedestal of height $h\text{ m}$, and $AB$ be the statue of height $1.6\text{ m}$ placed on top of it.

Let $D$ be the observation point on the ground.

Also, let $ BC= h\text{ m}$

Given:

$AB = 1.6\text{ m}$

$\angle BDC = 45^\circ$

$\angle ADC = 60^\circ$

In right $\Delta BCD$:

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

$\implies 1 = \frac{h}{CD}$

$\implies CD = h \quad \text{— (1)}$

In right $\Delta ACD$:

$\tan 60^\circ = \frac{AC}{CD}$

$\implies \sqrt{3} = \frac{AB + BC}{CD}$

$\implies \sqrt{3} = \frac{1.6 + h}{h} \quad [\text{From  equation (1)}]$

$\implies \sqrt{3}h = 1.6 + h$

$\implies \sqrt{3}h – h = 1.6$

$\implies h(\sqrt{3} – 1) = 1.6$

$\implies h = \frac{1.6}{\sqrt{3} – 1}$

$\implies h = \frac{1.6 \times (\sqrt{3} + 1)}{(\sqrt{3} – 1)(\sqrt{3} + 1)}$

$\implies h = \frac{1.6(\sqrt{3} + 1)}{3 – 1}$

$\implies h = \frac{1.6(\sqrt{3} + 1)}{2}$

$\implies h = 0.8(\sqrt{3} + 1)\text{ m}$

$\therefore$ The height of the pedestal is $0.8(\sqrt{3} + 1)\text{ m}$.

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 m high, find the height of the building.

$\mathbf{Sol^n.}$

Let $AB$ be the building of height $h\text{ m}$, and $CD$ be the tower of height $50\text{ m}$.

Let $BC$ be the horizontal distance between their bases.

Given:

$CD = 50\text{ m}$

$\angle DBC = 60^\circ$

$\angle ACB = 30^\circ$

In right $\Delta DCB$:

$\tan 60^\circ = \frac{CD}{BC}$

$\implies \sqrt{3} = \frac{50}{BC}$

$\implies BC = \frac{50}{\sqrt{3}}\text{ m} \quad \text{— (1)}$

In right $\Delta ABC$:

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

$\implies \frac{1}{\sqrt{3}} = \frac{h}{BC}$

$\implies h = \frac{BC}{\sqrt{3}}$

$\implies h = \frac{\frac{50}{\sqrt{3}}}{\sqrt{3}}$ [from equation (1)]

$\implies h = \frac{50}{3}\text{ m}$

$\implies h = 16\frac{2}{3}\text{ m}$

$\therefore$ The height of the building is $\frac{50}{3}\text{ m}$ (or $16\frac{2}{3}\text{ m}$).

10. 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 $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.  

$\mathbf{Sol^n.}$

Let $AB$ and $CD$ be the two poles of equal height $h\text{ m}$ standing on opposite sides of a road $BD$ of width $80\text{ m}$.

Let $O$ be the observation point on the road.

Let $BO = x\text{ m}$, then $OD = (80 – x)\text{ m}$.

Given:

$AB = CD = h\text{ m}$

$BD = 80\text{ m}$

$\angle AOB = 60^\circ$ and $\angle COD = 30^\circ$

In right $\Delta ABO$:

$\tan 60^\circ = \frac{AB}{BO}$

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

$\implies h = \sqrt{3}x \quad \text{— (1)}$

In right $\Delta CDO$:

$\tan 30^\circ = \frac{CD}{OD}$

$\implies \frac{1}{\sqrt{3}} = \frac{h}{80 – x}$

$\implies 80 – x = \sqrt{3}h \quad \text{— (2)}$

$\implies 80 – x = \sqrt{3}(\sqrt{3}x)$ [ from equation (1)]

$\implies 80 – x = 3x$

$\implies 4x = 80$

$\implies x = \frac{80}{4} = 20\text{ m}$

Now, substituting $x = 20$ in equation (1):

$h = \sqrt{3}(20) = 20\sqrt{3}\text{ m}$

Distance from pole $AB = BO = x = 20\text{ m}$

Distance from pole $CD = OD = 80 – x = 80 – 20 = 60\text{ m}$

$\therefore$ The height of each pole is $20\sqrt{3}\text{ m}$, and

the distances of the point from the poles are $20\text{ m}$ and $60\text{ m}$ respectively.

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 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 Fig. 9.12). Find the height of the tower and the width of the canal.

$\mathbf{Sol^n.}$

Let $AB$ be the height of the TV tower ($h\text{ m}$) and $BC$ be the width of the canal ($x\text{ m}$).

Let $D$ be the point $20\text{ m}$ away from point $C$, so $CD = 20\text{ m}$.

Then, $BD = CD + BC = (20 + x)\text{ m}$.

Given: $\angle ACB = 60^\circ$ and $\angle ADB = 30^\circ$

In right $\Delta ABC$:

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

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

$\implies h = \sqrt{3}x \quad \text{— (1)}$

In right $\Delta ABD$:

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

$\implies \frac{1}{\sqrt{3}} = \frac{h}{20 + x}$

$\implies 20 + x = \sqrt{3}h$

$\implies 20 + x = \sqrt{3}(\sqrt{3}x)$ [ from equation (1) ]

$\implies 20 + x = 3x$

$\implies 3x – x = 20$

$\implies 2x = 20$

$\implies x = 10\text{ m}$

Substituting $x = 10\text{ m}$ in equation (1), we get

$h = \sqrt{3}(10) = 10\sqrt{3}\text{ m}$

$\therefore$ The height of the tower is $10\sqrt{3}\text{ m}$ and the width of the canal is $10\text{ m}$.

12. From the top of a 7 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.

$\mathbf{Sol^n.}$

Let $AB$ be the building of height $7\text{ m}$, and $EC$ be the cable tower.

$AD \parallel BC$ is drawn.

Then, $ABCD$ is a rectangle, so $CD = AB = 7\text{ m}$ and $AD = BC$.

Let $ED = h\text{ m}$.

Total height of the cable tower, $EC = ED + CD = (h + 7)\text{ m}$.

Given:

$\angle EAD = 60^\circ$

$\angle DAC = 45^\circ$

$\implies \angle ACB = \angle DAC = 45^\circ \quad [\text{Alternate interior angles}]$

In right $\Delta ABC$:

$\tan 45^\circ = \frac{AB}{BC}$

$\implies 1 = \frac{7}{BC}$

$\implies BC = 7\text{ m}$

$\therefore AD = BC = 7\text{ m}$

In right $\Delta ADE$:

$\tan 60^\circ = \frac{ED}{AD}$

$\implies \sqrt{3} = \frac{h}{7}$

$\implies h = 7\sqrt{3}\text{ m}$

Now, total height of the tower:$EC = ED + CD$

$EC = 7\sqrt{3} + 7$

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

$\therefore$ The height of the tower is $7(\sqrt{3} + 1)\text{ m}$.

13. As observed from the top of a 75 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.

$\mathbf{Sol^n.}$

Let $AB$ be the lighthouse of height $75\text{ m}$.

Let $C$ and $D$ be the positions of the two ships on the sea level, such that the angles of depression from top $A$ are $45^\circ$ and $30^\circ$ respectively.

$\implies \angle ACB = 45^\circ$ and $\angle ADB = 30^\circ$

Let $BC = y\text{ m}$ and the distance between the two ships $CD = x\text{ m}$.

Then, $BD = BC + CD = (y + x)\text{ m}$.

In right $\Delta ABC$:

$\tan 45^\circ = \frac{AB}{BC}$

$\implies 1 = \frac{75}{y}$

$\implies y = 75\text{ m} \quad \text{— (1)}$

In right $\Delta ABD$:

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

$\implies \frac{1}{\sqrt{3}} = \frac{75}{x + y}$

$\implies x + y = 75\sqrt{3}$

$\implies x + 75 = 75\sqrt{3}$ [ Fromequation (1)]

$\implies x = 75\sqrt{3} – 75$

$\implies x = 75(\sqrt{3} – 1)\text{ m}$

$\therefore$ The distance between the two ships is $75(\sqrt{3} – 1)\text{ m}$.

14. 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 $60^\circ$. 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.

$\mathbf{Sol^n.}$

Let $A$ be the eye level of the girl at a height of $1.2\text{ m}$ from the ground.
Let $D$ and $C$ be the first and second positions of the balloon respectively.
The balloon is moving horizontally at a total height of $88.2\text{ m}$ from the ground.

Height of the balloon above the eye level of the girl:

$OD = BC = 88.2\text{ m} – 1.2\text{ m} = 87\text{ m}$

Given:

$\angle DAO = 60^\circ$

$\angle CAB = 30^\circ$

In right $\Delta AOD$:

$\tan 60^\circ = \frac{OD}{OA}$
$\implies \sqrt{3} = \frac{87}{OA}$
$\implies OA = \frac{87}{\sqrt{3}}$
$\implies OA = \frac{87}{\sqrt{3}} \times \frac{\sqrt{3}}{\sqrt{3}}$
$\implies OA = \frac{87\sqrt{3}}{3}$
$\implies OA = 29\sqrt{3}\text{ m} \quad \text{— (1)}$

In right $\Delta ABC$:

$\tan 30^\circ = \frac{BC}{AB}$
$\implies \frac{1}{\sqrt{3}} = \frac{87}{AB}$
$\implies AB = 87\sqrt{3}\text{ m} \quad \text{— (2)}$

Now, the distance travelled by the balloon during the interval is $OB$:

$OB = AB – OA$
$= 87\sqrt{3} – 29\sqrt{3}$
$= \sqrt{3}(87 – 29)$
$= 58\sqrt{3}\text{ m}$

$\therefore$ The distance travelled by the balloon during the interval is $58\sqrt{3}\text{ m}$

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.  

$\mathbf{Sol^n.}$

Let $AB$ be the tower of height $h\text{ m}$.
Let $D$ be the initial position of the car and $C$ be its position after $6\text{ seconds}$.
The angles of depression from the top of the tower $A$ to points $D$ and $C$ are $30^\circ$ and $60^\circ$ respectively, such that,

$\implies \angle ADB = 30^\circ$ and $\angle ACB = 60^\circ$

Let the uniform speed of the car be $v\text{ m/s}$.

Distance covered in $6\text{ seconds}$, $CD = \text{Speed} \times \text{Time} = 6v\text{ m}$.

Let the time taken by the car to travel from $C$ to foot $B$ be $t\text{ seconds}$.

Distance $BC = vt\text{ m}$.

Total distance $BD = BC + CD = (vt + 6v) = v(t + 6)\text{ m}$.   In right $\Delta ABC$:

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

$\implies \sqrt{3} = \frac{h}{vt}$

$\implies h = \sqrt{3}vt \quad \text{— (1)}$

In right $\Delta ABD$:

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

$\implies \frac{1}{\sqrt{3}} = \frac{h}{v(t + 6)}$

$\implies v(t + 6) = \sqrt{3}h$

Substituting the value of $h$ from equation (1), we get

$v(t + 6) = \sqrt{3}(\sqrt{3}vt)$

$\implies v(t + 6) = 3vt$

Dividing both sides by $v$ ($v \ne 0$), we get

$\implies t + 6 = 3t$

$\implies 3t – t = 6$

$\implies 2t = 6$

$\implies t = 3\text{ seconds}$

$\therefore$ The time taken by the car to reach the foot of the tower from this point is $3\text{ seconds}$.

16. 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.

$\mathbf{Sol^n.}$

Let $AB$ be the tower of height $h\text{ m}$.

Let $C$ and $D$ be two points on the same straight line from the base $B$ of the tower, such that $BC = 4\text{ m}$ and $BD = 9\text{ m}$.

Let the angle of elevation at point $C$ be $\theta$.

Since the angles are complementary, the angle of elevation at point $D$ is $(90^\circ – \theta)$.

In right $\Delta ABC$:

$\tan \theta = \frac{AB}{BC}$

$\implies \tan \theta = \frac{h}{4}$

$\implies h = 4\tan \theta \quad \text{— (1)}$

In right $\Delta ABD$:

$\tan(90^\circ – \theta) = \frac{AB}{BD}$

$\implies \cot \theta = \frac{h}{9} \quad [\because \tan(90^\circ – \theta) = \cot \theta]$

$\implies h = 9\cot \theta \quad \text{— (2)}$

Multiplying equation (1) and equation (2), we get

$h \times h = (4\tan \theta) \times (9\cot \theta)$

$\implies h^2 = 36 \times \tan \theta \times \cot \theta$

$\implies h^2 = 36 \times \tan \theta \times \frac{1}{\tan \theta}$

$\implies h^2 = 36$

$\implies h = \sqrt{36}$

$\implies h = 6\text{ m} \quad [\because \text{Height cannot be negative}]$   $\therefore$ The height of the tower is $6\text{ m}$.

Hence proved.

17. A lighthouse stands tall on a cliff by the sea, watching over ships that pass by. One day a ship is seen approaching the shore and from the top of the lighthouse, the angles of depression of the ship are observed to be $30^\circ$ and $45^\circ$ as it moves from point $P$ to point $Q$. The height of the lighthouse is 50 metres (see Fig. 9.14).

Based on the information given above, answer the following questions:

(i) Find the distance of the ship from the base of the lighthouse when it is at point $Q$, where the angle of depression is $45^\circ$.

(ii) Find the measures of $\angle PBA$ and $\angle QBA$.

(iii) Find the distance travelled by the ship between points $P$ and $Q$.

(iv) If the ship continues moving towards the shore and takes 10 minutes to travel from $Q$ to $A$, calculate the speed of the ship in $\text{km/h}$, from $Q$ to $A$.

$\mathbf{Sol^n.}$

Let $AB$ be the lighthouse of height $50\text{ m}$, where $B$ is the top and $A$ is the base at ground level.
Let $P$ and $Q$ be the two positions of the ship.
Here, $BX \parallel PA$.

Angles of depression are $\angle XBP = 30^\circ$ and $\angle XBQ = 45^\circ$.

$\implies \angle BPA = \angle XBP = 30^\circ$ and
$\angle BQA = \angle XBQ = 45^\circ \quad [\text{Alternate interior angles}]$

(i) Distance of the ship from the base when it is at point $Q$ ($QA$):

In right $\Delta BAQ$:
$\tan 45^\circ = \frac{AB}{QA}$
$\implies 1 = \frac{50}{QA}$
$\implies QA = 50\text{ m}$
$\therefore$ The distance of the ship from the base of the lighthouse when it is at point $Q$ is $50\text{ m}$.

(ii) Measures of $\angle PBA$ and $\angle QBA$:

In right-angled triangle $\Delta BAQ$ ($\angle A = 90^\circ$):
$\angle QBA = 90^\circ – \angle BQA = 90^\circ – 45^\circ = 45^\circ$

In right-angled triangle $\Delta BAP$ ($\angle A = 90^\circ$):
$\angle PBA = 90^\circ – \angle BPA = 90^\circ – 30^\circ = 60^\circ$

$\therefore \mathbf{\angle PBA = 60^\circ}$ and $\mathbf{\angle QBA = 45^\circ}$.

(iii) Distance travelled by the ship between points $P$ and $Q$ ($PQ$):

In right $\Delta BAP$:
$\tan 30^\circ = \frac{AB}{PA}$
$\implies \frac{1}{\sqrt{3}} = \frac{50}{PA}$
$\implies PA = 50\sqrt{3}\text{ m}$

Now, distance $PQ = PA – QA$:
$PQ = 50\sqrt{3} – 50$
$= 50(\sqrt{3} – 1)\text{ m}$
$= 50(1.732 – 1) = 50 \times 0.732 = 36.6\text{ m}$
$\therefore$ The distance travelled by the ship between points $P$ and $Q$ is $50(\sqrt{3} – 1)\text{ m}$.

(iv) Speed of the ship from $Q$ to $A$:

Distance from $Q$ to $A = 50\text{ m} = \frac{50}{1000}\text{ km} = \frac{1}{20}\text{ km}$
Time taken $= 10\text{ minutes} = \frac{10}{60}\text{ h} = \frac{1}{6}\text{ h}$

We know, $\text{Speed} = \frac{\text{Distance}}{\text{Time}}$

$= \frac{\frac{1}{20}}{\frac{1}{6}} = \frac{6}{20} = \frac{3}{10} = 0.3\text{ km/h}$

$\therefore$ The speed of the ship from $Q$ to $A$ is $0.3\text{ km/h}$.

18. A 30 m long rope is tightly stretched and tied from the top of pole to the ground. If the rope makes an angle of $60^\circ$ with the ground, the height of the pole is:

(a) $10\sqrt{3}\text{ m}$

(b) $30\sqrt{3}\text{ m}$

(c) $15\text{ m}$

(d) $15\sqrt{3}\text{ m}$

$\mathbf{Sol^n.}$

Let $AB$ represent the vertical pole of height $h\text{ m}$, and $AC$ represent the tightly stretched rope.

Given:

• $AC = 30\text{ m}$

• $\angle ACB = 60^\circ$

• $\angle B = 90^\circ$

In right-angled triangle $\Delta ABC$:
$\sin 60^\circ = \frac{AB}{AC}$
$\implies \frac{\sqrt{3}}{2} = \frac{AB}{30}$
$\implies AB = \frac{30 \times \sqrt{3}}{2}$
$\implies AB = 15\sqrt{3}\text{ m}$

$\therefore$ The height of the pole is $15\sqrt{3}\text{ m}$.

Correct Option: (d) $15\sqrt{3}\text{ m}$

19. If a pole 6 m high casts a shadow $2\sqrt{3}\text{ m}$ long on the ground then sun’s elevation is:

(a) $60^\circ$

(b) $45^\circ$

(c) $30^\circ$

(d) $90^\circ$

$\mathbf{Sol^n.}$

Let $AB$ represent the vertical pole of height $6\text{ m}$, and $BC$ represent the shadow cast on the ground of length $2\sqrt{3}\text{ m}$.
Let $\theta$ be the angle of elevation of the Sun ($\angle ACB$).

Given:

• $AB = 6\text{ m}$

• $BC = 2\sqrt{3}\text{ m}$

• $\angle B = 90^\circ$

In right-angled triangle $\Delta ABC$ :
$\tan \theta = \frac{AB}{BC}$
$\implies \tan \theta = \frac{6}{2\sqrt{3}}$
$\implies \tan \theta = \frac{3}{\sqrt{3}}$
$\implies \tan \theta = \frac{\sqrt{3} \times \sqrt{3}}{\sqrt{3}}$
$\implies \tan \theta = \sqrt{3}$
$\implies \tan \theta = \tan 60^\circ $
$\implies \theta = 60^\circ$

$\therefore$ The Sun’s elevation is $60^\circ$.

Correct Option: (a) $60^\circ$

20. The angles of elevation of a tower from two points at a distance of $a$ and $b$ ($a > b$) from its foot and in the same straight line from it are $30^\circ$ and $60^\circ$, then the height of the tower is:

(a) $\sqrt{a + b}$

(b) $\sqrt{a – b}$

(c) $\sqrt{ab}$

(d) $\sqrt{\frac{a}{b}}$

$\mathbf{Sol^n.}$

Let $AB$ represent the tower of height $h$. $C$ and $D$ are two points on the level ground in the same straight line from the foot of the tower $B$.

Given:

• $AB = h$

• $BC = b$

• $BD = a$

• $\angle ACB = 60^\circ$

• $\angle ADB = 30^\circ$

In right-angled triangle $\Delta ABC$:
$\tan 60^\circ = \frac{AB}{BC}$
$\implies \sqrt{3} = \frac{h}{b}$
$\implies h = b\sqrt{3} \quad \text{— (1)}$

In right-angled triangle $\Delta ABD$:
$\tan 30^\circ = \frac{AB}{BD}$
$\implies \frac{1}{\sqrt{3}} = \frac{h}{a}$
$\implies h = \frac{a}{\sqrt{3}} \quad \text{— (2)}$

Multiplying equation (1) and equation (2):
$h \times h = (b\sqrt{3}) \times \left(\frac{a}{\sqrt{3}}\right)$
$\implies h^2 = b \times a$
$\implies h^2 = ab$
$\implies h = \sqrt{ab}$

$\therefore$ The height of the tower is $\sqrt{ab}$.

Correct Option: (c) $\sqrt{ab}$

21. In the following question a statement of assertion (A) is followed by a statement of reason (R).

Assertion (A): A ladder leaning against a wall, stands at a horizontal distance of 6 m from the wall. If the height of the wall up to which the ladder reaches is 8 m, then the length of the ladder is 10 m.

Reason (R): The ladder makes an angle of $60^\circ$ with the ground.

Choose the correct option from the given alternatives.

(a) Both Assertion (A) and Reason (R) are true and Reason (R) is the correct explanation of the Assertion (A).

(b) Both Assertion (A) and Reason (R) are true, but Reason (R) is not the correct explanation of the Assertion (A).

(c) Assertion (A) is true, but Reason (R) is false.

(d) Assertion (A) is false, but Reason (R) is true.

$\mathbf{Sol^n.}$

Let $AB$ represent the vertical wall of height $8\text{ m}$, $BC$ represent the horizontal ground distance from the wall of length $6\text{ m}$, and $AC$ represent the ladder of length $10\text{ m}$. 

• Checking Assertion (A):
  In $\Delta ABC$, by using Pythagoras Theorem:
  $AC^2 = AB^2 + BC^2$
  $\implies AC = \sqrt{AB^2 + BC^2}$
  $\implies AC = \sqrt{8^2 + 6^2}$
  $\implies AC = \sqrt{64 + 36}$
  $\implies AC = \sqrt{100} = 10\text{ m}$
  Therefore, the length of the ladder is $10\text{ m}$.
  Hence, Assertion (A) is true.

• Checking Reason (R):
  Let $\angle ACB = \theta$.
  In $\Delta ABC$:
  $\tan \theta = \frac{AB}{BC}$
  $\implies \tan \theta = \frac{8}{6} = \frac{4}{3} $
  For the angle to be $60^\circ$, we must have $\tan 60^\circ = \sqrt{3} $.
  Since $\frac{4}{3} \ne \sqrt{3}$, the ladder does not make an angle of $60^\circ$ with the ground.
  Hence, Reason (R) is false.

$\therefore$ Assertion (A) is true, but Reason (R) is false.

Correct Option: (c) Assertion (A) is true, but Reason (R) is false.

22. Statement (i): If the angle of elevation of the top of a tower from a point on the ground 100 m away from the foot of the tower is $30^\circ$, then height of the tower is $100\sqrt{3}\text{ m}$.

Statement (ii): If the length of the shadow of a vertical pole is equal to its height then the angle of elevation of Sun’s altitude is $45^\circ$.

Choose the correct alternatives:

(a) Both the statements (i) and (ii) are correct.

(b) Both the statements (i) and (ii) are incorrect.

(c) Statement (i) is correct but statement (ii) is incorrect.

(d) Statement (i) is incorrect but statement (ii) is correct.

$\mathbf{Sol^n.}$

• Checking Statement (i):

  Let $AB$ represent the height of the tower ($h$), and $BC$ represent the distance of the point on the ground from the foot of the tower ($100\text{ m}$).

  The angle of elevation $\angle ACB = 30^\circ$.

  In $\Delta ABC$:

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

  $\implies \frac{1}{\sqrt{3}} = \frac{h}{100}$

  $\implies h = \frac{100}{\sqrt{3}}\text{ m}$

  $= \frac{100 \times \sqrt{3}}{\sqrt{3} \times \sqrt{3}} = \frac{100\sqrt{3}}{3}\text{ m} \neq 100\sqrt{3}\text{ m}$

  Therefore, Statement (i) is incorrect.

• Checking Statement (ii):

  Let $AB$ represent the height of the vertical pole ($H$), and $BC$ represent the length of its shadow on the ground ($L$).

  Let $\theta$ be the angle of elevation of the Sun’s altitude ($\angle ACB$).

  Given:

  $H = L$

  In right-angled triangle $\Delta ABC$:

  $\tan \theta = \frac{AB}{BC} = \frac{H}{L}$

  $\implies \tan \theta = \frac{H}{H} = 1$

  $\implies \tan \theta = \tan 45^\circ $

  $\implies \theta = 45^\circ$

  Therefore, Statement (ii) is correct.

Correct Option: (d) Statement (i) is incorrect but statement (ii) is correct.