Question

Find the angle of elevation of the Sun, when the length of the shadow of a tree is \( \sqrt{3} \) times the height of the tree.

• A. \( 30^\circ \)
• B. \( 45^\circ \)
• C. \( 60^\circ \)
• D. \( 90^\circ \)

Hint:

\textbf{Trigonometric Ratios.} Remember the values of trigonometric ratios for standard angles. For tangent: \( \tan(30^\circ) = \frac{1}{\sqrt{3}} \) \( \tan(45^\circ) = 1 \) \( \tan(60^\circ) = \sqrt{3} \) These values are frequently used in problems involving angles of elevation and depression.

Answer 1

The Correct Option is A

Let the height of the tree be \(h\) and the length of the shadow of the tree be \(s\). According to the problem, the length of the shadow is \( \sqrt{3} \) times the height of the tree: $$ s = \sqrt{3} h $$ Let \( \theta \) be the angle of elevation of the Sun. We can consider a right-angled triangle formed by the tree (perpendicular), its shadow (base), and the line from the top of the tree to the end of the shadow (hypotenuse). The angle of elevation \( \theta \) is the angle between the ground (shadow) and the line of sight to the top of the tree. We can use the tangent function: $$ \tan(\theta) = \frac{\text{height of the tree}}{\text{length of the shadow}} = \frac{h}{s} $$ Substitute the given relationship between \(s\) and \(h\): $$ \tan(\theta) = \frac{h}{\sqrt{3} h} $$ $$ \tan(\theta) = \frac{1}{\sqrt{3}} $$ We need to find the angle \( \theta \) whose tangent is \( \frac{1}{\sqrt{3}} \). We know from trigonometric values that: $$ \tan(30^\circ) = \frac{1}{\sqrt{3}} $$ Therefore, the angle of elevation of the Sun is \( 30^\circ \).