JEE Mains · Maths · STD 11 - Trigonometrical equations
The angle of elevation of a jet plane from a point \(A\) on the ground is \(60^{\circ}\). After a flight of \(20\, seconds\) at the speed of \(432\, km / hour\), the angle of elevation changes to \(30^{\circ}\). If the jet plane is flying at a constant height, then its height is ..... \(m.\)
- A \(1800 \sqrt{3}\)
- B \(3600 \sqrt{3}\)
- C \(2400 \sqrt{3}\)
- D \(1200 \sqrt{3}\)
Answer & Solution
Correct Answer
(D) \(1200 \sqrt{3}\)
Step-by-step Solution
Detailed explanation
\(\tan 60^{\circ}=\frac{ h }{ y }\) \(\sqrt{3}=\frac{ h }{ y } \Rightarrow h =\sqrt{3} y \quad \ldots \ldots .(1)\) \(\tan 30^{\circ}=\frac{ h }{ x + y }\) \(\frac{1}{\sqrt{3}}=\frac{ h }{ x + y } \Rightarrow \sqrt{3} h = x + y \quad \ldots \ldots .(2)\) Speed…
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