Section A (1 Mark each)
Q1. (b) 10√3 m
\(\tan(30^\circ) = \text{height}/30 \Rightarrow \text{height} = 30 \times (1/\sqrt{3}) = 10\sqrt{3}\) m.
Q2. (b) 15/2 m
The angle with the wall is 60°, so the angle with the ground is 30°. Let h be the height. \(\sin(30^\circ) = h/15 \Rightarrow h = 15 \times (1/2) = 7.5\) m.
Q3. (a) 75√3 m
Angle of elevation is 30°. \(\tan(30^\circ) = 75/\text{distance} \Rightarrow \text{distance} = 75 / (1/\sqrt{3}) = 75\sqrt{3}\) m.
Q4. (b) 45°
\(\tan(\theta) = \text{height}/\text{distance}\). Since height = distance, \(\tan(\theta) = 1 \Rightarrow \theta = 45^\circ\).
Section B (2 Marks each)
Q5. Height of the chimney.
Height of chimney above eye level = \(28.5 \times \tan(45^\circ) = 28.5\) m. (1 Mark)
Total height = \(28.5 + 1.5 = 30\) m. 30 m (1 Mark)
Q6. Length of the wire.
Difference in height = \(20 - 14 = 6\) m. (1 Mark)
\(\sin(30^\circ) = 6/\text{length} \Rightarrow \text{length} = 6 / (1/2) = 12\) m. 12 m (1 Mark)
Q7. Width of the river.
Width = \(3/\tan(30^\circ) + 3/\tan(45^\circ) = 3\sqrt{3} + 3 = 3(\sqrt{3}+1)\) m. 3(√3 + 1) m (2 Marks)
Section C (3 Marks each)
Q8. Height of the building.
Let d be the distance between building and tower. \(\tan(60^\circ) = 50/d \Rightarrow d = 50/\sqrt{3}\) m. (1.5 Marks)
Height of building = \(d \times \tan(30^\circ) = (50/\sqrt{3}) \times (1/\sqrt{3}) = 50/3\) m. 50/3 m (1.5 Marks)
Q9. Distance between the two ships.
Distance of first ship \(d_1 = 75/\tan(45^\circ) = 75\) m. (1 Mark)
Distance of second ship \(d_2 = 75/\tan(30^\circ) = 75\sqrt{3}\) m. (1 Mark)
Distance between them = \(d_2 - d_1 = 75(\sqrt{3}-1)\) m. 75(√3 - 1) m (1 Mark)
Section D (5 Marks)
Q10. Distance travelled by the balloon.
Height of balloon from girl's eyes = \(88.2 - 1.2 = 87\) m. (1 Mark)
Initial distance \(d_1 = 87/\tan(60^\circ) = 87/\sqrt{3} = 29\sqrt{3}\) m. (2 Marks)
Final distance \(d_2 = 87/\tan(30^\circ) = 87\sqrt{3}\) m. (1 Mark)
Distance travelled = \(d_2 - d_1 = 87\sqrt{3} - 29\sqrt{3} = 58\sqrt{3}\) m. 58√3 m (1 Mark)
Section E (Case Study - 4 Marks)
Q11. Case Study:
(i) A figure showing a vertical lighthouse and two ships on opposite sides.
(ii) Distance of ship at 45°: \(d_1 = 100/\tan(45^\circ) = 100\) m. 100 m (1 Mark)
(iii) Distance of ship at 30°: \(d_2 = 100/\tan(30^\circ) = 100\sqrt{3}\) m. 100√3 m (1 Mark)
(iv) Distance between ships = \(d_1 + d_2 = 100 + 100\sqrt{3} = 100(1+\sqrt{3})\) m. 100(1 + √3) m (2 Marks)