Rbse Solutions for Class 10 Maths Chapter 9 Exercise 9.1 Solutions | Applications of Trigonometry

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Get complete, step-by-step solutions for NCERT Class 10 Maths Chapter 9 Exercise 9.1

on Some Applications of Trigonometry (Heights and Distances). Learn to solve problems involving Angle of Elevation and Angle of Depression using trigonometric ratios ($\sin, \cos, \tan$) and specific angle values ($30^\circ, 45^\circ, 60^\circ$). Solutions cover scenarios like finding the height of a pole from a rope’s angle (Q.1), finding the original height of a broken tree (Q.2), calculating distances between objects (Q.13, Q.14), and determining unknown heights from two different observation angles (Q.6, Q.7, Q.8, Q.9). Includes complex problems involving two poles and non-simultaneous observations (Q.10, Q.11, Q.15). Essential practice for real-world trigonometry

application.

1. Height of the Pole

Let $AB$ be the pole (Height $h$), $AC$ be the rope ($20 \text{ m}$), and $\angle ACB = 30^\circ$.

We need to find $AB$ (Opposite side) given $AC$ (Hypotenuse). Use $\sin \theta$:

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

$$\frac{1}{2} = \frac{h}{20}$$

$$h = \frac{20}{2} = 10 \text{ m}$$

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

2. Height of the Tree

Let $AC$ be the original tree, where it breaks at $B$. The broken part $BC$ touches the ground at $D$.

The original height of the tree is $AC = AB + BC$. Since $BC$ is the broken part, $BC = BD$.

In the right triangle $\triangle ABD$: $\angle ADB = 30^\circ$, $AD = 8 \text{ m}$.

  1. Find the height of the unbroken part ($AB$): Use $\tan 30^\circ$.$$\tan 30^\circ = \frac{AB}{AD}$$$$\frac{1}{\sqrt{3}} = \frac{AB}{8} \implies AB = \frac{8}{\sqrt{3}} \text{ m}$$
  2. Find the length of the broken part ($BD$): Use $\cos 30^\circ$.$$\cos 30^\circ = \frac{AD}{BD}$$$$\frac{\sqrt{3}}{2} = \frac{8}{BD} \implies BD = \frac{16}{\sqrt{3}} \text{ m}$$
  3. Total height of the tree ($AC = AB + BD$):$$AC = \frac{8}{\sqrt{3}} + \frac{16}{\sqrt{3}} = \frac{24}{\sqrt{3}} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m}$$The height of the tree is $\mathbf{8\sqrt{3} \text{ m}}$.

3. Length of the Slides

(i) For children below 5 years (Slide $L_1$)

  • Height ($h_1$): $1.5 \text{ m}$ (Opposite)
  • Angle: $30^\circ$
  • Length of slide ($L_1$): Hypotenuse. Use $\sin 30^\circ$.$$\sin 30^\circ = \frac{h_1}{L_1}$$$$\frac{1}{2} = \frac{1.5}{L_1} \implies L_1 = 2 \times 1.5 = 3.0 \text{ m}$$

(ii) For elder children (Slide $L_2$)

  • Height ($h_2$): $3 \text{ m}$ (Opposite)
  • Angle: $60^\circ$
  • Length of slide ($L_2$): Hypotenuse. Use $\sin 60^\circ$.$$\sin 60^\circ = \frac{h_2}{L_2}$$$$\frac{\sqrt{3}}{2} = \frac{3}{L_2} \implies L_2 = \frac{6}{\sqrt{3}} = \frac{6\sqrt{3}}{3} = 2\sqrt{3} \text{ m}$$The lengths of the slides are $\mathbf{3.0 \text{ m}}$ and $\mathbf{2\sqrt{3} \text{ m}}$.

4. Height of the Tower

Let $h$ be the height of the tower $AB$. Distance from the foot $BC = 30 \text{ m}$. Angle of elevation $\angle ACB = 30^\circ$. Use $\tan 30^\circ$.

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

$$\frac{1}{\sqrt{3}} = \frac{h}{30} \implies h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m}$$

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

5. Length of the String

Height of kite $AB = 60 \text{ m}$ (Opposite). Angle of inclination $\angle ACB = 60^\circ$. Length of string $AC$ (Hypotenuse). Use $\sin 60^\circ$.

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

$$\frac{\sqrt{3}}{2} = \frac{60}{AC} \implies AC = \frac{120}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \text{ m}$$

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

6. Distance Walked Towards the Building

Let $CD = x$ be the distance walked. $AB$ is the building, $30 \text{ m}$. The observer is $1.5 \text{ m}$ tall.

Effective height of the building $AP = 30 – 1.5 = 28.5 \text{ m}$.

Let $E$ be the initial point and $F$ be the final point of the boy. $EF = x$.

Let $BP = y$ be the initial distance, and $CP = z$ be the final distance. $x = y – z$.

  1. In $\triangle APD$ ($\angle ADE = 30^\circ$): (Initial position)$$\tan 30^\circ = \frac{AP}{y} \implies \frac{1}{\sqrt{3}} = \frac{28.5}{y} \implies y = 28.5\sqrt{3}$$
  2. In $\triangle APE$ ($\angle AFE = 60^\circ$): (Final position)$$\tan 60^\circ = \frac{AP}{z} \implies \sqrt{3} = \frac{28.5}{z} \implies z = \frac{28.5}{\sqrt{3}}$$
  3. Distance walked ($x = y – z$):$$x = 28.5\sqrt{3} – \frac{28.5}{\sqrt{3}}$$$$x = 28.5 \left(\sqrt{3} – \frac{1}{\sqrt{3}}\right) = 28.5 \left(\frac{3 – 1}{\sqrt{3}}\right) = 28.5 \left(\frac{2}{\sqrt{3}}\right)$$$$x = \frac{57}{\sqrt{3}} = \frac{57\sqrt{3}}{3} = 19\sqrt{3} \text{ m}$$The distance the boy walked is $\mathbf{19\sqrt{3} \text{ m}}$.

7. Height of the Tower

$BC$ is the building, $20 \text{ m}$. $AB$ is the tower (Height $h$).

From point $D$, $\angle CDB = 45^\circ$ (Bottom of tower/top of building) and $\angle CDA = 60^\circ$ (Top of tower). Let $CD = x$.

  1. In $\triangle BCD$ ($\angle CDB = 45^\circ$):$$\tan 45^\circ = \frac{BC}{CD} \implies 1 = \frac{20}{x} \implies x = 20 \text{ m}$$
  2. In $\triangle ACD$ ($\angle CDA = 60^\circ$):$AC = AB + BC = h + 20$.$$\tan 60^\circ = \frac{AC}{CD}$$$$\sqrt{3} = \frac{h + 20}{20} \implies h + 20 = 20\sqrt{3}$$$$h = 20\sqrt{3} – 20 = 20(\sqrt{3} – 1) \text{ m}$$The height of the tower is $\mathbf{20(\sqrt{3} – 1) \text{ m}}$.

8. Height of the Pedestal

$AB$ is the pedestal (Height $h$). $BC$ is the statue ($1.6 \text{ m}$).

From point $D$, $\angle ADB = 45^\circ$ (Top of pedestal) and $\angle ADC = 60^\circ$ (Top of statue). Let $BD = x$.

  1. In $\triangle ABD$ ($\angle ADB = 45^\circ$):$$\tan 45^\circ = \frac{AB}{BD} \implies 1 = \frac{h}{x} \implies x = h$$
  2. In $\triangle BCD$ ($\angle ADC = 60^\circ$):$AC = AB + BC = h + 1.6$.$$\tan 60^\circ = \frac{AC}{BD}$$$$\sqrt{3} = \frac{h + 1.6}{h} \quad (\text{Since } x=h)$$$$h\sqrt{3} = h + 1.6$$$$h\sqrt{3} – h = 1.6 \implies h(\sqrt{3} – 1) = 1.6$$$$h = \frac{1.6}{\sqrt{3} – 1}$$Rationalize the denominator:$$h = \frac{1.6}{\sqrt{3} – 1} \times \frac{\sqrt{3} + 1}{\sqrt{3} + 1} = \frac{1.6(\sqrt{3} + 1)}{3 – 1} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \text{ m}$$The height of the pedestal is $\mathbf{0.8(\sqrt{3} + 1) \text{ m}}$.

9. Height of the Building

$AB$ is the building (Height $h$), $CD$ is the tower ($50 \text{ m}$). Distance $BD = x$.

  1. In $\triangle CDB$ ($\angle CBD = 60^\circ$): (Elevation of tower top from building foot)$$\tan 60^\circ = \frac{CD}{BD}$$$$\sqrt{3} = \frac{50}{x} \implies x = \frac{50}{\sqrt{3}}$$
  2. In $\triangle ABD$ ($\angle ADB = 30^\circ$): (Elevation of building top from tower foot)$$\tan 30^\circ = \frac{AB}{BD}$$$$\frac{1}{\sqrt{3}} = \frac{h}{x} \implies h = \frac{x}{\sqrt{3}}$$
  3. Substitute $x$ into the expression for $h$:$$h = \frac{1}{\sqrt{3}} \left(\frac{50}{\sqrt{3}}\right) = \frac{50}{3} \text{ m}$$The height of the building is $\mathbf{50/3 \text{ m}}$ (or $16 \frac{2}{3} \text{ m}$).

10. Height of Poles and Distances from the Point

Let $AB$ and $CD$ be the poles (Height $h$). $BD$ is the road, $80 \text{ m}$. $P$ is the point between them. $\angle APB = 60^\circ, \angle CPD = 30^\circ$. Let $BP = x$, then $PD = 80 – x$.

  1. In $\triangle ABP$ ($\angle APB = 60^\circ$):$$\tan 60^\circ = \frac{AB}{BP}$$$$\sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots (1)$$
  2. In $\triangle CDP$ ($\angle CPD = 30^\circ$):$$\tan 30^\circ = \frac{CD}{PD}$$$$\frac{1}{\sqrt{3}} = \frac{h}{80 – x} \implies h\sqrt{3} = 80 – x \quad \dots (2)$$
  3. Solve for $x$ (Substitute $h$ from (1) into (2)):$$(x\sqrt{3})\sqrt{3} = 80 – x$$$$3x = 80 – x \implies 4x = 80 \implies x = 20 \text{ m}$$
    • Distances: $BP = \mathbf{20 \text{ m}}$ and $PD = 80 – 20 = \mathbf{60 \text{ m}}$.
  4. Solve for $h$ (Substitute $x$ back into (1)):$$h = x\sqrt{3} = 20\sqrt{3} \text{ m}$$
    • Height of the poles: $\mathbf{20\sqrt{3} \text{ m}}$.

11. Height of the Tower and Width of the Canal

$AB$ is the tower (Height $h$). $BC$ is the width of the canal ($x$). $D$ is the point $20 \text{ m}$ away from $C$. $CD = 20 \text{ m}$. $\angle ACB = 60^\circ, \angle ADB = 30^\circ$.

  1. In $\triangle ABC$ ($\angle ACB = 60^\circ$):$$\tan 60^\circ = \frac{AB}{BC}$$$$\sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3} \quad \dots (1)$$
  2. In $\triangle ABD$ ($\angle ADB = 30^\circ$):$BD = BC + CD = x + 20$.$$\tan 30^\circ = \frac{AB}{BD}$$$$\frac{1}{\sqrt{3}} = \frac{h}{x + 20} \implies h\sqrt{3} = x + 20 \quad \dots (2)$$
  3. Solve for $x$ (Substitute $h$ from (1) into (2)):$$(x\sqrt{3})\sqrt{3} = x + 20$$$$3x = x + 20 \implies 2x = 20 \implies x = 10 \text{ m}$$
    • Width of the canal: $\mathbf{10 \text{ m}}$.
  4. Solve for $h$ (Substitute $x$ back into (1)):$$h = x\sqrt{3} = 10\sqrt{3} \text{ m}$$
    • Height of the tower: $\mathbf{10\sqrt{3} \text{ m}}$.

12. Height of the Cable Tower

$AB$ is the building ($7 \text{ m}$). $CD$ is the cable tower (Height $H$). $AE$ is a horizontal line from $A$. $\angle CAE = 60^\circ$ (Elevation of top). $\angle EAD = 45^\circ$ (Depression of foot $D$). $AB=DE=7 \text{ m}$. $H = CE + 7$.

  1. In $\triangle ABD$ (or $\triangle ADE$ where $E$ is on $CD$):Since $\angle EAD = 45^\circ$, the angle of elevation $\angle ADB = 45^\circ$.$$\tan 45^\circ = \frac{AB}{BD} \implies 1 = \frac{7}{BD} \implies BD = 7 \text{ m}$$Since $AE=BD$, $AE = 7 \text{ m}$.
  2. In $\triangle ACE$ ($\angle CAE = 60^\circ$):$$\tan 60^\circ = \frac{CE}{AE}$$$$\sqrt{3} = \frac{CE}{7} \implies CE = 7\sqrt{3} \text{ m}$$
  3. Total height of the tower ($H = CE + DE$):$$H = 7\sqrt{3} + 7 = 7(\sqrt{3} + 1) \text{ m}$$The height of the cable tower is $\mathbf{7(\sqrt{3} + 1) \text{ m}}$.

13. Distance Between Two Ships

$AB$ is the lighthouse ($75 \text{ m}$). $C$ and $D$ are the two ships. Angles of depression are $30^\circ$ and $45^\circ$. Due to alternate interior angles, $\angle ACB = 45^\circ$ and $\angle ADB = 30^\circ$. Let $CD = x$ be the distance between ships, and $BC = y$.

  1. In $\triangle ABC$ ($\angle ACB = 45^\circ$):$$\tan 45^\circ = \frac{AB}{BC} \implies 1 = \frac{75}{y} \implies y = 75 \text{ m}$$
  2. In $\triangle ABD$ ($\angle ADB = 30^\circ$):$BD = BC + CD = y + x = 75 + x$.$$\tan 30^\circ = \frac{AB}{BD}$$$$\frac{1}{\sqrt{3}} = \frac{75}{75 + x} \implies 75 + x = 75\sqrt{3}$$
  3. Distance between ships ($x$):$$x = 75\sqrt{3} – 75 = 75(\sqrt{3} – 1) \text{ m}$$The distance between the two ships is $\mathbf{75(\sqrt{3} – 1) \text{ m}}$.

14. Distance Travelled by the Balloon

Total height of balloon from ground is $88.2 \text{ m}$. The girl’s height is $1.2 \text{ m}$.

Effective height of balloon from the girl’s eyes $H = 88.2 – 1.2 = 87 \text{ m}$.

Let $A$ be the initial position and $B$ be the final position of the balloon. $P$ is the girl’s eye level.

$AC = H = 87 \text{ m}$ (Initial). $BD = H = 87 \text{ m}$ (Final). $CD$ is the distance travelled. $\angle APC = 60^\circ, \angle BPD = 30^\circ$. Let $PC = x, CD = y$.

  1. In $\triangle APC$ ($\angle APC = 60^\circ$):$$\tan 60^\circ = \frac{AC}{PC}$$$$\sqrt{3} = \frac{87}{x} \implies x = \frac{87}{\sqrt{3}} = 29\sqrt{3} \text{ m}$$
  2. In $\triangle BPD$ ($\angle BPD = 30^\circ$):$PD = PC + CD = x + y$.$$\tan 30^\circ = \frac{BD}{PD}$$$$\frac{1}{\sqrt{3}} = \frac{87}{x + y} \implies x + y = 87\sqrt{3}$$
  3. Distance travelled ($y$):$$y = 87\sqrt{3} – x$$$$y = 87\sqrt{3} – 29\sqrt{3} = (87 – 29)\sqrt{3} = 58\sqrt{3} \text{ m}$$The distance travelled by the balloon is $\mathbf{58\sqrt{3} \text{ m}}$.

15. Time Taken to Reach the Foot of the Tower

$AB$ is the tower (Height $h$). $C$ is the initial car position, $D$ is the position after $6 \text{ s}$. $\angle ACB = 30^\circ, \angle ADB = 60^\circ$. Let $DB = x$, $CD = y$. Time taken to cover $CD$ is $6 \text{ s}$. Time taken to cover $DB$ is $t$.

  1. In $\triangle ABD$ ($\angle ADB = 60^\circ$):$$\tan 60^\circ = \frac{AB}{DB}$$$$\sqrt{3} = \frac{h}{x} \implies h = x\sqrt{3}$$
  2. In $\triangle ABC$ ($\angle ACB = 30^\circ$):$CB = CD + DB = y + x$.$$\tan 30^\circ = \frac{AB}{CB}$$$$\frac{1}{\sqrt{3}} = \frac{h}{y + x} \implies h\sqrt{3} = y + x$$
  3. Solve for $y$ in terms of $x$ (Substitute $h = x\sqrt{3}$):$$(x\sqrt{3})\sqrt{3} = y + x$$$$3x = y + x \implies y = 2x$$
  4. Find the time $t$:Since the speed is uniform, Time is proportional to Distance.The car covers distance $y = 2x$ in $6 \text{ s}$.Let $S$ be the uniform speed.$$S = \frac{\text{Distance}}{\text{Time}}$$$$\frac{y}{6} = \frac{x}{t}$$Substitute $y = 2x$:$$\frac{2x}{6} = \frac{x}{t}$$$$\frac{1}{3} = \frac{1}{t} \implies t = 3 \text{ s}$$The time taken by the car to reach the foot of the tower from point $D$ is $3 \text{ seconds}$.

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