JEE Mains · Maths · STD 11 - Trigonometrical equations
The angle of elevation of the summit of a mountain from a point on the ground is \(45^{\circ}\). After climding up one \(km\) towards the summit at an inclination of \(30^{\circ}\) from the ground, the angle of elevation of the summit is found to be \(60^{\circ} .\) Then the height (in \(km\) ) of the summit from the ground is
- A \(\frac{1}{\sqrt{3}-1}\)
- B \(\frac{1}{\sqrt{3}+1}\)
- C \(\frac{\sqrt{3}-1}{\sqrt{3}+1}\)
- D \(\frac{\sqrt{3}+1}{\sqrt{3}-1}\)
Answer & Solution
Correct Answer
(A) \(\frac{1}{\sqrt{3}-1}\)
Step-by-step Solution
Detailed explanation
\(\sin 30^{\circ}=x \Rightarrow x=\frac{1}{2}\) \(\cos 30^{\circ}=z \Rightarrow z=\frac{\sqrt{3}}{2}\) \(\tan 45^{\circ}=\frac{ h }{ y + z } \Rightarrow h = y + z\) \(\tan 60^{\circ}=\frac{ h - x }{ y } \Rightarrow \tan 60^{\circ}=\frac{ h - x }{ h - z }\)…
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