Question : Find the angular elevation of the Sun when the shadow of a 15 metres long pole is $\frac{15}{\sqrt{3}}$ metres.
Option 1: $45^\circ$
Option 2: $60^\circ$
Option 3: $30^\circ$
Option 4: $90^\circ$
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Correct Answer: $60^\circ$
Solution : Let $\theta$ be the angular elevation of the Sun. Here, $\tan\theta = \frac{AB}{BC} = (15×\frac{\sqrt{3}}{15}) = \sqrt{3}$ metres ⇒ $\tan\theta = \sqrt{3}=\tan60^\circ$ ⇒ $\theta = 60^\circ$ $\therefore$ The angular elevation of the Sun is $60^\circ$. Hence, the correct answer is $60^\circ$.
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Question : If the angle of elevation of the sun decreases from $45^\circ$ to $30^\circ$, then the length of the shadow of a pillar increases by 60 m. The height of the pillar is:
Question : If the elevation of the Sun changes from 30° to 60°, then the difference between the lengths of shadows of a pole 15 metres high, is:
Question : If the length of the shadow of a vertical pole is $\sqrt{3}$ times the height of the pole, the angle of elevation of the sun is:
Question : The shadow of a tower when the angle of elevation of the sun is 45°, is found to be 10 metres longer than when it was 60°. The height of the tower is:
Question : Find the value of the given expression. $\frac{4}{3} \tan^2 45^{\circ}+3 \cos^2 30^{\circ}-2 \sec^2 30^{\circ}-\frac{3}{4} \cot^2 60^{\circ}$
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