NCERT Solutions Class 10 Maths Chapter 9: Some Applications of Trigonometry
Detailed solutions for Exercise 9.1 (Heights and Distances).
#### **Q1. 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 $30^\circ$.**
#### **A1. Solution:**
Let $AB$ be the vertical pole and $AC$ be the rope.
Given: Length of rope ($AC$) = $20$ m, Angle ($\angle ACB$) = $30^\circ$.
In right $\Delta ABC$:
$\sin 30^\circ = \frac{AB}{AC}$
$\frac{1}{2} = \frac{AB}{20}$
$AB = \frac{20}{2} = 10$ m.
**The height of the pole is 10 m.**
#### **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 $30^\circ$ 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.**
#### **A2. Solution:**
Let $AB$ be the unbroken part and $AC$ be the broken part. Total height of tree = $AB + AC$.
Given: Distance $BC = 8$ m, Angle $\angle ACB = 30^\circ$.
In $\Delta ABC$:
$\tan 30^\circ = \frac{AB}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{AB}{8} \Rightarrow AB = \frac{8}{\sqrt{3}}$ m.
$\cos 30^\circ = \frac{BC}{AC} \Rightarrow \frac{\sqrt{3}}{2} = \frac{8}{AC} \Rightarrow AC = \frac{16}{\sqrt{3}}$ m.
Total height $= AB + AC = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = 8\sqrt{3}$ m.
**The height of the tree is $8\sqrt{3}$ m.**
#### **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 $30^\circ$ 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?**
#### **A3. Solution:**
**Case I (Below 5 years):** Height ($h_1$) = $1.5$ m, $\theta_1 = 30^\circ$.
$\sin 30^\circ = \frac{\text{height}}{\text{slide length } (L_1)}$
$\frac{1}{2} = \frac{1.5}{L_1} \Rightarrow L_1 = 3$ m.
**Case II (Elder children):** Height ($h_2$) = $3$ m, $\theta_2 = 60^\circ$.
$\sin 60^\circ = \frac{3}{L_2}$
$\frac{\sqrt{3}}{2} = \frac{3}{L_2} \Rightarrow L_2 = \frac{6}{\sqrt{3}} = 2\sqrt{3}$ m.
**Length of slides: 3 m and $2\sqrt{3}$ 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 $30^\circ$. Find the height of the tower.**
#### **A4. Solution:**
Let $h$ be the height of the tower and distance from foot be $30$ m.
In right $\Delta ABC$:
$\tan 30^\circ = \frac{h}{30}$
$\frac{1}{\sqrt{3}} = \frac{h}{30}$
$h = \frac{30}{\sqrt{3}} = 10\sqrt{3}$ m.
**The height of the tower is $10\sqrt{3}$ 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 $60^\circ$. Find the length of the string, assuming that there is no slack in the string.**
#### **A5. Solution:**
Let $L$ be the length of the string. Height of kite = $60$ m.
In right $\Delta ABC$:
$\sin 60^\circ = \frac{60}{L}$
$\frac{\sqrt{3}}{2} = \frac{60}{L}$
$L = \frac{120}{\sqrt{3}} = 40\sqrt{3}$ m.
**The length of the string is $40\sqrt{3}$ 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 $30^\circ$ to $60^\circ$ as he walks towards the building. Find the distance he walked towards the building.**
#### **A6. Solution:**
Height of building above boy's eye level $= 30 - 1.5 = 28.5$ m.
Let distance walked be $x$. Initial distance be $y$.
For $30^\circ$: $\tan 30^\circ = \frac{28.5}{y} \Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{y} \Rightarrow y = 28.5\sqrt{3}$.
For $60^\circ$: $\tan 60^\circ = \frac{28.5}{y-x} \Rightarrow \sqrt{3} = \frac{28.5}{y-x} \Rightarrow y-x = \frac{28.5}{\sqrt{3}}$.
$x = y - \frac{28.5}{\sqrt{3}} = 28.5\sqrt{3} - \frac{28.5}{\sqrt{3}} = \frac{28.5(3-1)}{\sqrt{3}} = \frac{57}{\sqrt{3}} = 19\sqrt{3}$ m.
**Distance walked is $19\sqrt{3}$ m.**
#### **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 $45^\circ$ and $60^\circ$ respectively. Find the height of the tower.**
#### **A7. Solution:**
Let $BC$ be the building of height 20 m and $AB$ be the transmission tower of height $h$. Let $D$ be the point on the ground.
In $\Delta BCD$: $\tan 45^\circ = \frac{BC}{CD} \Rightarrow 1 = \frac{20}{CD} \Rightarrow CD = 20$ m.
In $\Delta ACD$: $\tan 60^\circ = \frac{AC}{CD} \Rightarrow \sqrt{3} = \frac{20 + h}{20}$
$20\sqrt{3} = 20 + h \Rightarrow h = 20\sqrt{3} - 20 = 20(\sqrt{3} - 1)$ m.
**The height of the tower is $20(\sqrt{3} - 1)$ 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 $60^\circ$ and from the same point the angle of elevation of the top of the pedestal is $45^\circ$. Find the height of the pedestal.**
#### **A8. Solution:**
Let $h$ be the height of the pedestal and $x$ be the distance of the point from the pedestal.
In $\Delta$ for pedestal: $\tan 45^\circ = \frac{h}{x} \Rightarrow 1 = \frac{h}{x} \Rightarrow x = h$.
In $\Delta$ for statue: $\tan 60^\circ = \frac{h + 1.6}{x}$
$\sqrt{3} = \frac{h + 1.6}{h} \Rightarrow h\sqrt{3} = h + 1.6$
$h(\sqrt{3} - 1) = 1.6 \Rightarrow h = \frac{1.6}{\sqrt{3} - 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1)$ m.
**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 $30^\circ$ 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.**
#### **A9. Solution:**
Let $AB$ be the tower (50 m) and $CD$ be the building ($h$).
In $\Delta ABC$ (tower): $\tan 60^\circ = \frac{50}{BC} \Rightarrow \sqrt{3} = \frac{50}{BC} \Rightarrow BC = \frac{50}{\sqrt{3}}$.
In $\Delta BCD$ (building): $\tan 30^\circ = \frac{h}{BC} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{50/\sqrt{3}}$
$h = \frac{50}{\sqrt{3} \times \sqrt{3}} = \frac{50}{3} = 16\frac{2}{3}$ m.
**The height of the building is $16\frac{2}{3}$ m.**
#### **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 $60^\circ$ and $30^\circ$, respectively. Find the height of the poles and the distances of the point from the poles.**
#### **A10. Solution:**
Let height of poles be $h$. Let the point be $x$ distance from one pole and $(80-x)$ from the other.
$\tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3}$.
$\tan 30^\circ = \frac{h}{80 - x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{80 - x}$
$80 - x = 3x \Rightarrow 4x = 80 \Rightarrow x = 20$ m.
$h = 20\sqrt{3}$ m.
**Height of poles is $20\sqrt{3}$ m; distances are 20 m and 60 m.**
#### **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 $60^\circ$. 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 $30^\circ$. Find the height of the tower and the width of the canal.**
#### **A11. Solution:**
Let $h$ be the height and $x$ be the width of the canal.
$\tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3}$.
$\tan 30^\circ = \frac{h}{x + 20} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 20}$
$x + 20 = 3x \Rightarrow 2x = 20 \Rightarrow x = 10$ m.
$h = 10\sqrt{3}$ m.
**Height of tower is $10\sqrt{3}$ m and width of 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 $60^\circ$ and the angle of depression of its foot is $45^\circ$. Determine the height of the tower.**
#### **A12. Solution:**
Let $AB$ be the building of height 7 m and $CD$ be the cable tower. Let the horizontal distance between them be $x$.
In $\Delta ABC$ (at the base): $\tan 45^\circ = \frac{AB}{BC} \Rightarrow 1 = \frac{7}{x} \Rightarrow x = 7$ m.
The horizontal distance is equal at the top, so let the part of the tower above the building height be $h'$.
In the upper triangle: $\tan 60^\circ = \frac{h'}{x} \Rightarrow \sqrt{3} = \frac{h'}{7} \Rightarrow h' = 7\sqrt{3}$ m.
Total height of the tower $= 7 + 7\sqrt{3} = 7(1 + \sqrt{3})$ m.
**The height of the tower is $7(1 + \sqrt{3})$ 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 $30^\circ$ 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.**
#### **A13. Solution:**
Let $AB$ be the lighthouse (75 m). Let $C$ and $D$ be the two ships.
In $\Delta ABC$ (closer ship): $\tan 45^\circ = \frac{75}{BC} \Rightarrow 1 = \frac{75}{BC} \Rightarrow BC = 75$ m.
In $\Delta ABD$ (farther ship): $\tan 30^\circ = \frac{75}{BD} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{BD} \Rightarrow BD = 75\sqrt{3}$ m.
Distance between ships $(CD) = BD - BC = 75\sqrt{3} - 75 = 75(\sqrt{3} - 1)$ m.
**The distance between the ships is $75(\sqrt{3} - 1)$ 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 $60^\circ$. After some time, the angle of elevation reduces to $30^\circ$. Find the distance travelled by the balloon during the interval.**
#### **A14. Solution:**
Height of balloon from girl’s eye level $= 88.2 - 1.2 = 87$ m.
Let $d_1$ be the initial horizontal distance and $d_2$ be the final horizontal distance.
In $\Delta$ for $60^\circ$: $\tan 60^\circ = \frac{87}{d_1} \Rightarrow \sqrt{3} = \frac{87}{d_1} \Rightarrow d_1 = \frac{87}{\sqrt{3}} = 29\sqrt{3}$ m.
In $\Delta$ for $30^\circ$: $\tan 30^\circ = \frac{87}{d_2} \Rightarrow \frac{1}{\sqrt{3}} = \frac{87}{d_2} \Rightarrow d_2 = 87\sqrt{3}$ m.
Distance travelled $= d_2 - d_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}$ m.
**The distance travelled by the balloon is $58\sqrt{3}$ 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 $30^\circ$, 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.**
#### **A15. Solution:**
Let $h$ be the height of the tower and $v$ be the uniform speed of the car.
Distance travelled in 6 seconds $(CD) = 6v$.
In $\Delta$ for $60^\circ$: $\tan 60^\circ = \frac{h}{x} \Rightarrow h = x\sqrt{3}$, where $x$ is the remaining distance.
In $\Delta$ for $30^\circ$: $\tan 30^\circ = \frac{h}{x + 6v} \Rightarrow \frac{1}{\sqrt{3}} = \frac{x\sqrt{3}}{x + 6v}$
$x + 6v = 3x \Rightarrow 2x = 6v \Rightarrow x = 3v$.
Time taken to cover distance $x = \frac{\text{Distance}}{\text{Speed}} = \frac{3v}{v} = 3$ seconds.
**The car will take 3 seconds to reach the foot of the tower.**