Question

The shadow of a tower standing on a level ground is found to be 40 m longer when the Sun’s altitude is 30° than when it was 60°. Find the height of the tower and the length of the original shadow. (use \(\sqrt{ 3}\) = 1.73)

Answer 1

- Let h be the height of the tower and x be the length of the original shadow.
- From the first situation (altitude 30°):

\[ \tan 30^\circ = \frac{h}{x + 40} \]

\[ \frac{1}{\sqrt{3}} = \frac{h}{x + 40} \implies h = \frac{x + 40}{\sqrt{3}} \]

- From the second situation (altitude 60°):

\[ \tan 60^\circ = \frac{h}{x} \]

\[ \sqrt{3} = \frac{h}{x} \implies h = \sqrt{3}x \]

- Equating the two expressions for h:

\[ \frac{x + 40}{\sqrt{3}} = \sqrt{3}x \]

Solving:

\[ x + 40 = 3x \implies 40 = 2x \implies x = 20 \]

- Therefore, the height of the tower is:

\[ h = \sqrt{3} \times 20 = 20\sqrt{3} \approx 34.64 \, \text{m} \]

- The length of the original shadow is x = 20 m.