NCERT Solutions for Class 10 Maths Some Applications of Trigonometry — Chapter 9 Exercise 9.1 | 2026-27

Key Concept: The rope is the hypotenuse of a right triangle. The pole is the perpendicular (opposite side). The angle at the ground is 30°.

Step 1: Let the height of the pole be \( h \) metres. The rope length (hypotenuse) = 20 m. Angle with ground = 30°.

Step 2: Using sin ratio (opposite/hypotenuse):

\[ \sin 30° = \frac{h}{20} \]

Step 3: Substitute \( \sin 30° = \frac{1}{2} \):

\[ \frac{1}{2} = \frac{h}{20} \]
\[ h = 20 \times \frac{1}{2} = 10 \text{ m} \]

Key Concept: The broken part of the tree forms the hypotenuse, the standing part is the perpendicular, and the distance on the ground is the base. Total height = standing part + broken part (length).

Step 1: Let the standing part of the tree = \( h_1 \) m and the broken part = \( l \) m. The distance on the ground = 8 m. Angle = 30°.

Step 2: Using tan for the standing part:

\[ \tan 30° = \frac{h_1}{8} \]
\[ \frac{1}{\sqrt{3}} = \frac{h_1}{8} \]
\[ h_1 = \frac{8}{\sqrt{3}} = \frac{8\sqrt{3}}{3} \text{ m} \]

Step 3: Using cos for the broken part (hypotenuse):

\[ \cos 30° = \frac{8}{l} \]
\[ \frac{\sqrt{3}}{2} = \frac{8}{l} \]
\[ l = \frac{16}{\sqrt{3}} = \frac{16\sqrt{3}}{3} \text{ m} \]

Step 4: Total height of the tree:

\[ H = h_1 + l = \frac{8\sqrt{3}}{3} + \frac{16\sqrt{3}}{3} = \frac{24\sqrt{3}}{3} = 8\sqrt{3} \text{ m} \]

Step 1: Let height of tower = \( h \) m. Horizontal distance = 30 m. Angle of elevation = 30°.

Step 2: Using tan:

\[ \tan 30° = \frac{h}{30} \]
\[ \frac{1}{\sqrt{3}} = \frac{h}{30} \]
\[ h = \frac{30}{\sqrt{3}} = \frac{30\sqrt{3}}{3} = 10\sqrt{3} \text{ m} \]

Step 1: Height = 60 m (opposite side). String = hypotenuse = \( l \). Angle = 60°.

\[ \sin 60° = \frac{60}{l} \]
\[ \frac{\sqrt{3}}{2} = \frac{60}{l} \]
\[ l = \frac{60 \times 2}{\sqrt{3}} = \frac{120}{\sqrt{3}} = \frac{120\sqrt{3}}{3} = 40\sqrt{3} \text{ m} \]

Key Concept: Since the boy is 1.5 m tall, the effective height of the building from his eye level = 30 − 1.5 = 28.5 m.

Step 1: Let the initial distance from the building = \( d_1 \) and final distance = \( d_2 \).

Step 2: At angle 30°:

\[ \tan 30° = \frac{28.5}{d_1} \Rightarrow \frac{1}{\sqrt{3}} = \frac{28.5}{d_1} \Rightarrow d_1 = 28.5\sqrt{3} \text{ m} \]

Step 3: At angle 60°:

\[ \tan 60° = \frac{28.5}{d_2} \Rightarrow \sqrt{3} = \frac{28.5}{d_2} \Rightarrow d_2 = \frac{28.5}{\sqrt{3}} = \frac{28.5\sqrt{3}}{3} = 9.5\sqrt{3} \text{ m} \]

Step 4: Distance walked = \( d_1 – d_2 \):

\[ = 28.5\sqrt{3} – 9.5\sqrt{3} = 19\sqrt{3} \text{ m} \]

Step 1: Let horizontal distance from the point to the building = \( d \) m. Height of building = 20 m. Let height of tower = \( h \) m.

Step 2: Angle of elevation to bottom of tower (top of building) = 45°:

\[ \tan 45° = \frac{20}{d} \Rightarrow 1 = \frac{20}{d} \Rightarrow d = 20 \text{ m} \]

Step 3: Angle of elevation to top of tower = 60°. Total height = 20 + h:

\[ \tan 60° = \frac{20 + h}{d} = \frac{20 + h}{20} \]
\[ \sqrt{3} = \frac{20 + h}{20} \]
\[ 20 + h = 20\sqrt{3} \]
\[ h = 20\sqrt{3} – 20 = 20(\sqrt{3} – 1) \text{ m} \]

Step 1: Let height of pedestal = \( h \) m. Height of statue = 1.6 m. Total height = \( h + 1.6 \) m. Let horizontal distance = \( d \) m.

Step 2: Angle of elevation to top of pedestal = 45°:

\[ \tan 45° = \frac{h}{d} \Rightarrow 1 = \frac{h}{d} \Rightarrow d = h \]

Step 3: Angle of elevation to top of statue = 60°:

\[ \tan 60° = \frac{h + 1.6}{d} = \frac{h + 1.6}{h} \]
\[ \sqrt{3} = \frac{h + 1.6}{h} \]
\[ \sqrt{3}\,h = h + 1.6 \]
\[ h(\sqrt{3} – 1) = 1.6 \]
\[ h = \frac{1.6}{\sqrt{3} – 1} = \frac{1.6(\sqrt{3} + 1)}{(\sqrt{3} – 1)(\sqrt{3} + 1)} = \frac{1.6(\sqrt{3} + 1)}{2} = 0.8(\sqrt{3} + 1) \text{ m} \]

Step 1: Let height of building = \( h \) m, height of tower = 50 m, horizontal distance between them = \( d \) m.

Step 2: From the foot of the tower, angle of elevation to top of building = 30°:

\[ \tan 30° = \frac{h}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{d} \Rightarrow d = h\sqrt{3} \]

Step 3: From the foot of the building, angle of elevation to top of tower = 60°:

\[ \tan 60° = \frac{50}{d} \Rightarrow \sqrt{3} = \frac{50}{d} \Rightarrow d = \frac{50}{\sqrt{3}} \]

Step 4: Equate the two expressions for \( d \):

\[ h\sqrt{3} = \frac{50}{\sqrt{3}} \]
\[ h = \frac{50}{\sqrt{3} \times \sqrt{3}} = \frac{50}{3} \text{ m} \]

Step 1: Let height of each pole = \( h \) m. Let the point be at distance \( x \) m from the first pole, so distance from second pole = \( (80 – x) \) m.

Step 2: Angle of elevation to first pole = 60°:

\[ \tan 60° = \frac{h}{x} \Rightarrow \sqrt{3} = \frac{h}{x} \Rightarrow h = x\sqrt{3} \quad \text{…(i)} \]

Step 3: Angle of elevation to second pole = 30°:

\[ \tan 30° = \frac{h}{80 – x} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{80 – x} \Rightarrow h = \frac{80 – x}{\sqrt{3}} \quad \text{…(ii)} \]

Step 4: From (i) and (ii):

\[ x\sqrt{3} = \frac{80 – x}{\sqrt{3}} \]
\[ 3x = 80 – x \]
\[ 4x = 80 \Rightarrow x = 20 \text{ m} \]

Step 5: Height: \( h = 20\sqrt{3} \) m. Distance from second pole = 80 − 20 = 60 m.

Step 1: Let height of tower AB = \( h \) m. Let width of canal = \( d \) m (distance from point C directly opposite to the foot of tower). Point D is 20 m farther from C.

Step 2: From point C, angle of elevation = 60°:

\[ \tan 60° = \frac{h}{d} \Rightarrow \sqrt{3} = \frac{h}{d} \Rightarrow h = d\sqrt{3} \quad \text{…(i)} \]

Step 3: From point D (distance = d + 20), angle of elevation = 30°:

\[ \tan 30° = \frac{h}{d + 20} \Rightarrow \frac{1}{\sqrt{3}} = \frac{h}{d + 20} \Rightarrow h = \frac{d + 20}{\sqrt{3}} \quad \text{…(ii)} \]

Step 4: From (i) and (ii):

\[ d\sqrt{3} = \frac{d + 20}{\sqrt{3}} \]
\[ 3d = d + 20 \]
\[ 2d = 20 \Rightarrow d = 10 \text{ m} \]

Step 5: Height: \( h = 10\sqrt{3} \) m.

Key Concept: The observer is at height 7 m. The angle of depression to the foot of the tower means the foot of the tower is at ground level, and the observer looks down at 45°.

Step 1: Let height of cable tower = \( H \) m. Horizontal distance between building and tower = \( d \) m.

Step 2: Angle of depression to foot of tower = 45°. The observer is 7 m high, foot of tower is at ground level:

\[ \tan 45° = \frac{7}{d} \Rightarrow 1 = \frac{7}{d} \Rightarrow d = 7 \text{ m} \]

Step 3: Angle of elevation to top of tower from top of building = 60°. The top of tower is \( (H – 7) \) m above the observer’s level:

\[ \tan 60° = \frac{H – 7}{d} = \frac{H – 7}{7} \]
\[ \sqrt{3} = \frac{H – 7}{7} \]
\[ H – 7 = 7\sqrt{3} \]
\[ H = 7 + 7\sqrt{3} = 7(1 + \sqrt{3}) \text{ m} \]

Step 1: Height of lighthouse = 75 m. Let distance of nearer ship = \( d_1 \) m, farther ship = \( d_2 \) m.

Step 2: For the ship with angle of depression 45° (nearer ship):

\[ \tan 45° = \frac{75}{d_1} \Rightarrow 1 = \frac{75}{d_1} \Rightarrow d_1 = 75 \text{ m} \]

Step 3: For the ship with angle of depression 30° (farther ship):

\[ \tan 30° = \frac{75}{d_2} \Rightarrow \frac{1}{\sqrt{3}} = \frac{75}{d_2} \Rightarrow d_2 = 75\sqrt{3} \text{ m} \]

Step 4: Distance between the two ships:

\[ = d_2 – d_1 = 75\sqrt{3} – 75 = 75(\sqrt{3} – 1) \text{ m} \]

Key Concept: The girl’s eyes are at 1.2 m height. So effective height of balloon from eye level = 88.2 − 1.2 = 87 m.

Step 1: Let initial horizontal distance = \( d_1 \) m, final horizontal distance = \( d_2 \) m.

Step 2: At angle 60°:

\[ \tan 60° = \frac{87}{d_1} \Rightarrow \sqrt{3} = \frac{87}{d_1} \Rightarrow d_1 = \frac{87}{\sqrt{3}} = \frac{87\sqrt{3}}{3} = 29\sqrt{3} \text{ m} \]

Step 3: At angle 30°:

\[ \tan 30° = \frac{87}{d_2} \Rightarrow \frac{1}{\sqrt{3}} = \frac{87}{d_2} \Rightarrow d_2 = 87\sqrt{3} \text{ m} \]

Step 4: Distance travelled by balloon:

\[ = d_2 – d_1 = 87\sqrt{3} – 29\sqrt{3} = 58\sqrt{3} \text{ m} \]

Step 1: Let height of tower = \( h \) m. Initial position of car = A (angle of depression 30°), position after 6 s = B (angle of depression 60°).

Step 2: Distance OA (from foot of tower to A):

\[ \tan 30° = \frac{h}{OA} \Rightarrow OA = h\sqrt{3} \]

Step 3: Distance OB (from foot of tower to B):

\[ \tan 60° = \frac{h}{OB} \Rightarrow OB = \frac{h}{\sqrt{3}} \]

Step 4: Distance covered in 6 seconds = AB = OA − OB:

\[ AB = h\sqrt{3} – \frac{h}{\sqrt{3}} = h\left(\sqrt{3} – \frac{1}{\sqrt{3}}\right) = h \cdot \frac{3 – 1}{\sqrt{3}} = \frac{2h}{\sqrt{3}} \]

Step 5: Speed of car = \( \frac{AB}{6} = \frac{2h}{6\sqrt{3}} = \frac{h}{3\sqrt{3}} \) m/s.

Step 6: Time to travel remaining distance OB from point B:

\[ t = \frac{OB}{\text{speed}} = \frac{\frac{h}{\sqrt{3}}}{\frac{h}{3\sqrt{3}}} = \frac{h}{\sqrt{3}} \times \frac{3\sqrt{3}}{h} = 3 \text{ seconds} \]

Key Concept: Complementary angles means they add up to 90°. If one angle is \( \theta \), the other is \( 90° – \theta \). We use \( \tan\theta \cdot \tan(90°-\theta) = 1 \), i.e., \( \tan\theta \cdot \cot\theta = 1 \).

Step 1: Let height of tower = \( h \) m. Let the angle of elevation from the point 4 m away = \( \theta \), and from the point 9 m away = \( 90° – \theta \).

Step 2: From the point 4 m away:

\[ \tan\theta = \frac{h}{4} \quad \text{…(i)} \]

Step 3: From the point 9 m away:

\[ \tan(90° – \theta) = \frac{h}{9} \Rightarrow \cot\theta = \frac{h}{9} \quad \text{…(ii)} \]

Step 4: Multiply equations (i) and (ii):

\[ \tan\theta \times \cot\theta = \frac{h}{4} \times \frac{h}{9} \]
\[ 1 = \frac{h^2}{36} \]
\[ h^2 = 36 \]
\[ h = 6 \text{ m} \quad (\text{taking positive root since height} > 0) \]

Step 1: \( \cos\theta = \frac{9}{15} = \frac{3}{5} = 0.6 \)

Step 2: \( \theta = \cos^{-1}(0.6) \approx 53.13° \)

Step 1: Let height of tower = \( h \) m. Horizontal distance = \( d \) m.

Step 2: Angle of depression to bottom of tower = 60°:

\[ \tan 60° = \frac{100}{d} \Rightarrow d = \frac{100}{\sqrt{3}} \]

Step 3: Angle of depression to top of tower = 30°. Height difference = \( 100 – h \):

\[ \tan 30° = \frac{100 – h}{d} \Rightarrow \frac{1}{\sqrt{3}} = \frac{(100-h)\sqrt{3}}{100} \]
\[ 100 = \sqrt{3} \cdot (100 – h) \cdot \sqrt{3} = 3(100 – h) \]
\[ 100 = 300 – 3h \Rightarrow 3h = 200 \Rightarrow h = \frac{200}{3} \approx 66.67 \text{ m} \]