Two men on either side of a pole of 30 m high observe its top at angles of elevation α and β respectively. The distance between the two men is 40√3 m and the distance between the first man at A and the pole is 30√3m.
Based on the above information, answer the question:
The value of \(\frac{1}{AB^2}+\frac{1}{BC^2}\)is:
Answer & explanation
Correct answer: option 1
$tanα=\frac{30}{30\sqrt{3}}=\frac{1}{\sqrt{3}}$
$∴α=30°\, or\, \frac{\pi}{6}$
$tanβ=\frac{30}{10\sqrt{3}}=\sqrt{3}$
$β=\frac{\pi}{3}⇒tan(α+β)=tan(\frac{\pi}{2})$ → ∞ (does not exist)
$∠ABC=\frac{\pi}{2}-∞$
$=\frac{\pi}{2}-\frac{\pi}{6}=\frac{\pi}{3}$
$\frac{1}{AB^2}+\frac{1}{BC^2}=\frac{1}{30^2(4)}+\frac{1}{1200}=\frac{1}{3600}+\frac{1}{1200×3}$
$=\frac{4}{3600}=\frac{1}{900}=\frac{1}{BD^2}$
Option 1 is correct.