JEE Mathematics Heights and Distances MCQs Set 03

Answer: (a) 60

Explanation:
1. \(\tan\theta=\frac{6}{2\sqrt3}=\sqrt3\), so \(\theta=60^\circ\).

Teacher's Note:
1. \( \tan \) of the sun's elevation is height divided by shadow length.
2. Recall \( \tan 60 \) degrees = \( sqrt3 \).

Answer: (d) \(30(\sqrt3+1)\text{ m}\)

Explanation:
1. Height \(h\): shadows are \(h\) and \(h\sqrt3\).
2. So \(h(\sqrt3-1)=60\) and \(h=\frac{60}{\sqrt3-1}=30(\sqrt3+1)\).

Teacher's Note:
1. Shadow length = \( h \cot \) of the elevation.
2. Rationalise the denominator.

Answer: (a) \(\frac{5h}3\)

Explanation:
1. The tower top makes angle \(\alpha\) with \(\tan\alpha=\frac12\).
2. Equal angles means the pole subtends \(\alpha\), so the pole top has elevation \(2\alpha\): \(\tan2\alpha=\frac43=\frac{h+H}{2h}\).
3. So \(H=\frac{8h}3-h=\frac{5h}3\).

Teacher's Note:
1. Equal angles subtended by the tower and the pole mean the top of the pole is at elevation \( 2 \alpha \).
2. Use \( \tan 2 \alpha = 2 \tan \alpha / (1 - \tan^2 \alpha) \).

Answer: (a) \(48\text{ m}\)

Explanation:
1. The window sees the house from \( 36 m \) below its top and \( 64 m \) above its foot.
2. A right angle gives \(d^2=64\times36\), so \( d=48 m \).

Teacher's Note:
1. The angles of elevation and depression are complementary when the subtended angle is 90 degrees.
2. \( d^2 \) equals the product of the two heights.

Answer: (b) 346.4 m

Explanation:
1. Distances from the foot are \(300\cot60^\circ=100\sqrt3\) and \(300\cot30^\circ=300\sqrt3\).
2. The difference is \( 200\sqrt3\approx346.4 m \).

Teacher's Note:
1. Horizontal distance = height times \( \cot \) of the depression angle.
2. Subtract the two distances.

Answer: (c) 1.5 min

Explanation:
1. With tree height \(h\), the car's distances are \(h\sqrt3\) and \(\frac h{\sqrt3}\).
2. It covers \(\frac{2h}{\sqrt3}\) in \( 3 \min \), and the remaining \(\frac h{\sqrt3}\) takes half of that, \( 1.5 \min \).

Teacher's Note:
1. The speed is constant, so time is proportional to distance.
2. Remaining distance is half of the distance covered.

Answer: (c) \((150-20\sqrt3)\text{ m}\)

Explanation:
1. The drop is \( 60\tan30^\circ=20\sqrt3 m \).
2. So the first tower is \( 150-20\sqrt3 m \) high.

Teacher's Note:
1. Height difference = horizontal distance times \( \tan \) of the depression angle.
2. \( 60/sqrt3 = 20 sqrt3 \).

Answer: (b) \(240\sqrt3\)

Explanation:
1. Horizontal distances: \(\cot60^\circ=\frac1{\sqrt3}\) and \(\cot30^\circ=\sqrt3\) km.
2. Distance covered \(=\frac2{\sqrt3}\) km in \( 10 s \), so speed \(=\frac2{\sqrt3}\times360=240\sqrt3\) km/h.

Teacher's Note:
1. Take the aeroplane as approaching and then receding in the same line; the horizontal distance changes by \( \cot30 - \cot60 \).
2. Convert \( 10 s \) to hours by multiplying by 360.