Trigonometric Equations Question 119
Question: A tower is situated on horizontal plane. From two points, the line joining three points passes through the base and which are $ a $ and $ b $ distance from the base. The angle of elevation of the top are $ \alpha $ and $ 90{}^\circ -\alpha $ and $ \theta $ is that angle which two points joining the line makes at the top, the height of tower will be
[UPSEAT 1999]
Options:
A) $ \frac{a+b}{a-b} $
B) $ \frac{a-b}{a+b} $
C) $ \sqrt{ab} $
D) $ {{(ab)}^{1/3}} $
Show Answer
Answer:
Correct Answer: C
Solution:
- Let there are two points C and D on horizontal line passing from point B of the base of the tower AB. The distance of these points are b and a from B respectively.
$ \therefore BD=aandBC=b $ . $ \because $ line CD, on the top of tower A subtends an angle $ \theta $ ,
$ \therefore \angle CAD=\theta $ According to question, on point C and D, the elevation of top are $ \alpha $ and $ 90^{o}-\alpha $ .
$ \therefore \angle BCA=,\alpha $ and $ \angle BDA= $ $ 90^{o}-\alpha $ . In $ \Delta ABC, $ $ AB=BC\tan \alpha $ $ =b\tan \alpha $ ?.. (i) and in $ \Delta ABD $ , AB = BD $ \tan (90^{o}-\alpha ) $ $ =a\cot \alpha $ ?..(ii) Multiplying equation (i) and (ii), $ {{(AB)}^{2}}=(b\tan \alpha )(a\cot \alpha )=ab $ ,
$ \therefore ,AB=\sqrt{(ab)} $ .
BETA
