Trigonometry & Inverse Trigonometry
Heights and Distances
Grade 11
Question:
<p>Consider a triangular plot ABC with sides AB = 7 m, BC = 5 m and CA = 6 m. A vertical lamp-post at the mid-point D of AC subtends an angle 30° at B. The height (in m) of the lamp-post is</p>
<p>(a) \(\frac{21}{3}\)</p>
<p>(b) \(2\sqrt{21}\)</p>
<p>(c) \(7\sqrt{3}\)</p>
<p>(d) \(\frac{21}{2}\)</p>
Step-by-Step Solution
Key Concept: Use Apollonius theorem to find the median length, then apply trigonometry to find the height from the angle subtended.
<p><strong>Step 1:</strong> By Apollonius theorem, the length of median BD from B to midpoint D of AC is:</p><p>$$BD = \frac{1}{2}\sqrt{2a^2 + 2c^2 - b^2}$$</p><p>where $c = AB = 7$, $a = BC = 5$, and $b = CA = 6$</p><p><strong>Step 2:</strong> Calculate BD:</p><p>$$BD = \frac{1}{2}\sqrt{2(25) + 2(49) - 36} = \frac{1}{2}\sqrt{50 + 98 - 36} = \frac{1}{2}\sqrt{112} = \frac{1}{2} \cdot 4\sqrt{7} = 2\sqrt{7}$$</p><p><strong>Step 3:</strong> Let ED = h be the height of the lamp post. In triangle BDE, the angle subtended is 30°:</p><p>$$\tan 30° = \frac{h}{BD} = \frac{h}{2\sqrt{7}}$$</p><p><strong>Step 4:</strong> Solve for h:</p><p>$$\frac{1}{\sqrt{3}} = \frac{h}{2\sqrt{7}}$$</p><p>$$h = \frac{2\sqrt{7}}{\sqrt{3}} = \frac{2\sqrt{7} \cdot \sqrt{3}}{3} = \frac{2\sqrt{21}}{3}$$</p><p>∴ Answer is (a).</p>
Correct Answer: a