JEE Mathematics Heights and Distances MCQs Set 01

Answer: (c) 6√3 ft

Explanation:
1. Let the distance be d and the elevation θ. The top is 18 ft above his eye and the base 6 ft below.
2. tan θ = 18/d and tan(90° − θ) = cot θ = 6/d.
3. Multiplying: 1 = 108/d², so d = 6√3 ft.

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Answer: (b) 7√3 m

Explanation:
1. Distance moved by time t is \( \int_0^t(2t+3)dt=t^2+3t \): 4 m at t = 1 and 18 m at t = 3.
2. With distance d: \( h+4=d\tan30^\circ=\frac d{\sqrt3} \) and \( h+18=\sqrt3\,d \).
3. Subtract: \( 14=d\left(\sqrt3-\frac1{\sqrt3}\right)=\frac{2d}{\sqrt3} \), so d = 7√3 m.

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Answer: (c) HP

Explanation:
1. With circumradius R, cot α = R/h₁, cot β = R/h₂, cot γ = R/h₃.
2. These are in AP, so 1/h₁, 1/h₂, 1/h₃ are in AP.
3. Hence h₁, h₂, h₃ are in HP.

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Answer: (b) ½ a tan α · cosec A

Explanation:
1. The foot of the hill is the circumcentre, with circumradius R = a/(2 sin A).
2. Height = R tan α = \( \frac12a\tan\alpha\,\csc A \).

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Answer: (a) \( \frac35 \)

Explanation:
1. Arc length \( =10\times\frac{216\pi}{180}=12\pi \) equals the base circumference \( 2\pi r \), so r = 6 cm.
2. The slant height is 10 cm, hence sin θ = r/l = 6/10 = 3/5.

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Answer: (d) \( \frac{3\sqrt3}{2\pi}A \)

Explanation:
1. The pole stands at the centre of the hexagon; elevation does not matter for the area.
2. Circumradius R satisfies \( A=\pi R^2 \) and the hexagon area is \( \frac{3\sqrt3}2R^2=\frac{3\sqrt3}{2\pi}A \).

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Answer: (c) 60°

Explanation:
1. AC subtends 90° at P and O is the midpoint of AC, so PO = OA = h.
2. Side AB = OA√2 = h√2 and PA = PB = \( \sqrt{h^2+h^2}=h\sqrt2 \).
3. Triangle PAB is equilateral, so the angle is 60°.

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Answer: (a) \( 81\sqrt2\ \text{m}^2 \)

Explanation:
1. Diagonal = 9 cot 30° = 9√3 m. The other corner is at distance 9 cot 45° = 9 m.
2. Other side = \( \sqrt{(9\sqrt3)^2-9^2}=\sqrt{162}=9\sqrt2 \) m.
3. Area = 9 × 9√2 = 81√2 m².

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Answer: (a) \( \frac52 \) m

Explanation:
1. Take the lamp-post as origin. The ray from its top (0, 6) through the wall top (2, 4) has equation y = 6 − x.
2. The man's head (height 1.5) is on the ray when x = 4.5 m from the lamp-post.
3. Distance from the wall = 4.5 − 2 = 2.5 m.

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Answer: (c) \( \frac{16}{25} \)

Explanation:
1. Each string has length \( \sqrt{3^2+4^2}=5 \) cm. Consecutive attachment points are 90° apart on the circle, so the chord is \( 3\sqrt2 \).
2. By the cosine rule \( \cos\theta=\frac{25+25-18}{2\cdot25}=\frac{16}{25} \).

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Answer: (c) either 3 : 1 or 1 : 3

Explanation:
1. Let the man's height be h and his distance from the shorter post be x. The posts rise h and 3h above his eyes; the posts are 4h apart.
2. Right angle gives \( \frac hx\cdot\frac{3h}{4h-x}=1 \), so \( x^2-4hx+3h^2=0 \).
3. So x = h or x = 3h, giving the ratio 1 : 3 or 3 : 1.

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Answer: (c) 1 : √3

Explanation:
1. Let the pillar be h, the distance d and the angle subtended by the pillar θ, so tan θ = h/d = t. The flagstaff then also subtends θ, so the top is at angle 2θ.
2. tan 2θ = 3h/d = 3t, so \( \frac{2t}{1-t^2}=3t \), giving \( t^2=\frac13 \).
3. Hence h : d = 1 : √3.

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