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 is 60°. Find the height of the tower.
Step-by-Step Solution
Key Concept: Use the definition of tangent in a right‑angled triangle: \(\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}\). For a tower of height \(h\) and its shadow length \(s\), \(\tan\theta = \dfrac{h}{s}\). Set up two equations for the two given altitudes and use the given difference of shadows to solve for \(h\).
1. Let \(h\) be the height of the tower (in metres).\
2. Let \(s_1\) be the length of the shadow when the Sun’s altitude is \(30^{\circ}\) and \(s_2\) be the length when the altitude is \(60^{\circ}\).\
3. From the statement, \(s_1 = s_2 + 40\) (the shadow is 40 m longer at 30°).\
4. Using the definition of tangent:
\[ \tan 30^{\circ} = \frac{h}{s_1} \quad\text{and}\quad \tan 60^{\circ} = \frac{h}{s_2}. \]
5. Substitute the known values of the tangents:
\[ \frac{1}{\sqrt{3}} = \frac{h}{s_1} \quad\Rightarrow\quad h = \frac{s_1}{\sqrt{3}} \]
\[ \sqrt{3} = \frac{h}{s_2} \quad\Rightarrow\quad h = s_2\sqrt{3}. \]
6. Equate the two expressions for \(h\):
\[ \frac{s_1}{\sqrt{3}} = s_2\sqrt{3}. \]
7. Replace \(s_1\) by \(s_2 + 40\):
\[ \frac{s_2 + 40}{\sqrt{3}} = s_2\sqrt{3}. \]
8. Multiply both sides by \(\sqrt{3}\):
\[ s_2 + 40 = 3s_2. \]
9. Solve for \(s_2\):
\[ 3s_2 - s_2 = 40 \Rightarrow 2s_2 = 40 \Rightarrow s_2 = 20\ \text{m}. \]
10. Find the height using \(h = s_2\sqrt{3}\):
\[ h = 20\sqrt{3}\ \text{m}. \]
11. Numerically, \(h \approx 20 \times 1.732 = 34.64\ \text{m}. \]
Correct Answer: 20√3 m (≈ 34.6 m)