Trigonometric Equations Question 132
Question: The angle of elevation of the top of a tower from a point A due south of the tower is $ \alpha $ and from a point B due east of the tower is $ \beta $ . If AB =d, then the height of the tower is
[Roorkee 1979; Kurukshetra CEE 1998]
Options:
A) $ \frac{d}{\sqrt{{{\tan }^{2}}\alpha -{{\tan }^{2}}\beta }} $
B) $ \frac{d}{\sqrt{{{\tan }^{2}}\alpha +{{\tan }^{2}}\beta }} $
C) $ \frac{d}{\sqrt{{{\cot }^{2}}\alpha +{{\cot }^{2}}\beta }} $
D) $ \frac{d}{\sqrt{{{\cot }^{2}}\alpha -{{\cot }^{2}}\beta }} $
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Answer:
Correct Answer: C
Solution:
- $ OB=h\cot \beta ,,OA=h\cot \alpha $ $ h^{2}=\frac{d^{2}}{{{\cot }^{2}}\beta +{{\cot }^{2}}\alpha } $
$ \Rightarrow $ $ \Delta =( \frac{\sqrt{3}a^{2}}{4} )=3\sqrt{3}cm^{2},,\therefore ,R=\frac{abc}{4\Delta }=2cm $
BETA
