IMO Maths Olympiad Class 10 Chapter 09 Some Applications of Trigonometry MCQs with Answers

Answer: (b) (i) 36.6 m, (ii) 100 m

Explanation:
1. (i) Shadow at 45° is 50 cot 45° = 50 m. Shadow at 30° is 50 cot 30° = \( 50\sqrt{3} \) m.
2. So x = \( 50\sqrt{3} - 50 = 50(1.732-1) \) = 36.6 m.
3. (ii) Height = 100 × tan 45° = 100 m.

Teacher's Note:
1. A shorter shadow means a higher Sun, so the larger angle gives the shorter shadow.
2. At 45° the height equals the shadow length.

Answer: (a) 300 cm

Explanation:
1. From the figure, AB = 90 cm, BC = AD = 60 cm, ∠AQD = 30° and ∠APB = 30°.
2. In right △ADQ: sin 30° = \( \dfrac{AD}{AQ} \), so AQ = 2 × 60 = 120 cm.
3. In right △ABP: sin 30° = \( \dfrac{AB}{AP} \), so AP = 2 × 90 = 180 cm. Hence AP + AQ = 300 cm.

Teacher's Note:
1. In each right triangle the hypotenuse is opposite the right angle and the side opposite 30° is half of it.
2. Check which side is opposite the marked 30° angle before choosing sin or cos.

Answer: (a) \( \dfrac{h\,(1+\tan\theta)}{(1-\tan\theta)} \)

Explanation:
1. Let the cloud be H above the lake and BC the horizontal distance. The cloud is H − h above the observer, and its reflection is H + h below him.
2. So \( \tan\theta = \dfrac{H-h}{BC} \) and tan 45° = \( \dfrac{H+h}{BC} \), i.e. BC = H + h.
3. Then H − h = (H + h) tan θ, so \( H = \dfrac{h(1+\tan\theta)}{1-\tan\theta} \).

Teacher's Note:
1. A cloud and its reflection are at the same distance from the surface of the lake.
2. Depth of the reflection below the observer is H + h.

Answer: (c) 23.66 m

Explanation:
1. Let the height be h. Shadows are h (at 45°) and \( \dfrac{h}{\sqrt{3}} \) (at 60°).
2. So \( h - \dfrac{h}{\sqrt{3}} = 10 \), giving \( h = \dfrac{10\sqrt{3}}{\sqrt{3}-1} = 5\sqrt{3}(\sqrt{3}+1) = 15 + 5\sqrt{3} \).
3. h = 15 + 5 × 1.732 = 23.66 m.

Teacher's Note:
1. The same method as the previous shadow problem, with 10 m instead of 15 m.
2. Use the given value of √3 to get a decimal answer.

Answer: (d) \( \dfrac{a}{2\sqrt{2}} \)

Explanation:
1. Let the heights be x and 2x, each at distance \( \dfrac{a}{2} \) from the observer. The angles are θ and 90° − θ, so tan θ₁ × tan θ₂ = 1.
2. \( \dfrac{2x}{a/2}\times\dfrac{x}{a/2} = 1 \), so \( \dfrac{8x^2}{a^2} = 1 \).
3. \( x^2 = \dfrac{a^2}{8} \), so x = \( \dfrac{a}{2\sqrt{2}} \).

Teacher's Note:
1. If two angles are complementary, the product of their tangents is 1.
2. Both horizontal distances equal a/2.

Answer: (b) \( \dfrac{b\cot x - a\cot y}{b-a} \)

Explanation:
1. Let the top be h above the ground, with its vertical projection z to the north of the foot of the tower, so z = h cot θ.
2. From the two stations: a + z = h cot x and b + z = h cot y, so a = h(cot x − cot θ) and b = h(cot y − cot θ).
3. Dividing: \( \dfrac{a}{b} = \dfrac{\cot x - \cot\theta}{\cot y - \cot\theta} \), which gives \( \cot\theta = \dfrac{b\cot x - a\cot y}{b-a} \).

Teacher's Note:
1. Let the vertical height of the top be h and use cot for horizontal distances.
2. Dividing the two equations removes h.

Answer: (d) \( \dfrac{2h\sec x}{\tan y-\tan x} \)

Explanation:
1. Let the cloud be H above the observer's level and CE the horizontal distance. Then CE = \( \dfrac{H}{\tan x} \) and, for the reflection (H + 2h below), CE = \( \dfrac{H+2h}{\tan y} \).
2. Equating, H tan y = (H + 2h) tan x, so \( H = \dfrac{2h\tan x}{\tan y-\tan x} \).
3. Distance = \( \dfrac{H}{\sin x} = \dfrac{2h\tan x}{(\tan y-\tan x)\sin x} = \)\( \dfrac{2h\sec x}{\tan y-\tan x} \).

Teacher's Note:
1. The reflection is as far below the lake surface as the cloud is above it, so it is 2h + H below the observer.
2. The line of sight is the hypotenuse, so divide the height by sin x.

Answer: (d) 819.6 m

Explanation:
1. Take A as the origin with the cliff top at height h. The point reached is 600 m along a 30° slope, so it is 300 m higher and \( 300\sqrt{3} \) m further along the ground.
2. From it the top is at elevation 60°, which makes ∠CAD = ∠ACD = 15°, so the distance to the top is also 600 m.
3. The top is then 300 m above that point plus \( 600\sin 60^\circ = 300\sqrt{3} \) m, so h = 300 + 519.6 = 819.6 m.

Teacher's Note:
1. Angle chasing shows the triangle with the 600 m side is isosceles.
2. The final height is the sum of a vertical rise from each of the two stages.

Answer: (b) 790.56 km/hr

Explanation:
1. At 45° the horizontal distance is 3000 cot 45° = 3000 m, and at 30° it is \( 3000\sqrt{3} \) m.
2. Distance flown = 3000(1.732 − 1) = 2196 m in 10 s.
3. Speed = \( \dfrac{2196}{10}\times\dfrac{18}{5} \) = 790.56 km/hr.

Teacher's Note:
1. Multiply m/s by 18/5 to convert to km/hr.
2. Taking √3 = 1.732 gives the option value; the exact value is 790.61 km/hr.

Answer: (a) 15 m, \( 5\sqrt{3} \) m

Explanation:
1. Let the tall building be h and the distance between the buildings x. The bottom is at depression 60°, so tan 60° = \( \dfrac{h}{x} \) and \( x = \dfrac{h}{\sqrt{3}} \).
2. The top of the small building is h − 10 below, at depression 30°: tan 30° = \( \dfrac{h-10}{x} \), so \( x = \sqrt{3}(h-10) \).
3. \( \sqrt{3}(h-10) = \dfrac{h}{\sqrt{3}} \) gives 3(h − 10) = h, so h = 15 m and x = \( \dfrac{15}{\sqrt{3}} = \)\( 5\sqrt{3} \) m.

Teacher's Note:
1. Depression angles are measured downwards from the horizontal at the top of the tall building.
2. Both angles share the same horizontal distance x.