JEE Mains · Maths · STD 11 - Trigonometrical equations
The angle of elevation of the top of a vertical tower from a point \(P\) on the horizontal ground was observed to be \(\alpha \). After moving a distance \(2\, metres\) from \(P\) towards the foot of the tower, the angle of elevation changes to \(\beta \). Then the height (in metres) of the tower is
- A \(\frac{{2\,\sin \,\alpha \,\sin \,\beta }}{{\sin \,\left( {\beta - \alpha } \right)}}\)
- B \(\frac{{\sin \,\alpha \,\sin \,\beta }}{{\cos \,\left( {\beta - \alpha } \right)}}\)
- C \(\,\,\,\frac{{2\,\sin \,\left( {\beta - \alpha } \right)}}{{\sin \,\alpha \,\sin \,\beta }}\)
- D \(\,\,\,\frac{{\cos \,\left( {\beta - \alpha } \right)}}{{\sin \,\alpha \,\sin \,\beta }}\)
Answer & Solution
Correct Answer
(A) \(\frac{{2\,\sin \,\alpha \,\sin \,\beta }}{{\sin \,\left( {\beta - \alpha } \right)}}\)
Step-by-step Solution
Detailed explanation
Given : In \(\Delta ABP\) \(\tan \alpha =\frac {AB}{PB}\) or \(\frac{{\sin \,\alpha }}{{\cos \,\alpha }} = \frac{h}{{x + 2}}\) \( \Rightarrow \,(x + 2)\sin \,\alpha \, = \,h\,\cos \,\alpha \)…
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