CBSE Class 10 Mathematics Applications of Trigonometry MCQs Set 05

Answer: (c) \(75\sqrt{3}\) m

Explanation:
1. \(\tan 30^\circ = \frac{75}{OB}\), so \(\frac{1}{\sqrt{3}} = \frac{75}{OB}\).
2. \(OB = 75\sqrt{3}\) m.

Answer: (a) \(45^\circ\)

Explanation:
1. Height \(=\) shadow, so \(\tan\theta = \frac{AB}{BC} = 1\).
2. So \(\theta = 45^\circ\).

Answer: (d) 15 m

Explanation:
1. \(\tan 45^\circ = \frac{h}{15}\), and \(\tan 45^\circ = 1\).
2. So \(h = 15\) m.

Answer: (a) 75 m

Explanation:
1. \(\cot\theta = \frac{8}{15}\), so \(\csc^2\theta = 1 + \frac{64}{225} = \frac{289}{225}\), giving \(\csc\theta = \frac{17}{15}\).
2. \(\sin\theta = \frac{15}{17}\). In the triangle, \(\sin\theta = \frac{\text{height}}{85}\).
3. So height \(= 85 \times \frac{15}{17} = 75\) m.

Answer: (a) 12 m

Explanation:
1. Difference in heights \(= 20 - 14 = 6\) m.
2. \(\sin 30^\circ = \frac{6}{\text{wire}}\), so \(\frac{1}{2} = \frac{6}{\text{wire}}\).
3. Wire \(= 12\) m.

Answer: (b) \(45^\circ\)

Explanation:
1. Height above the eye \(= 22 - 1.5 = 20.5\) m.
2. \(\tan\theta = \frac{20.5}{20.5} = 1\), so \(\theta = 45^\circ\).

Teacher's Note:
1. The stem is reconstructed from the source's working, which uses 22 − 1.5 = 20.5 m.

Answer: (b) \(\sqrt{ab}\)

Explanation:
1. \(\tan\theta = \frac{h}{a}\) and \(\tan(90^\circ - \theta) = \cot\theta = \frac{h}{b}\).
2. Multiplying: \(\frac{h}{a} \times \frac{h}{b} = 1\), so \(h^2 = ab\) and \(h = \sqrt{ab}\).

Teacher's Note:
1. Options (c) and (d) are reconstructed from the extraction. Please check against the printed book.

Answer: (a) \(14.64\) m

Explanation:
1. Distance from the point to the building \(= 20\) m, since \(\tan 45^\circ = 1\).
2. \(\tan 60^\circ = \frac{20 + x}{20}\), so \(x = 20(\sqrt{3} - 1) = 20 \times 0.732 = 14.64\) m.