Trigonometric Equations Question 271
Question: The angular elevation of a tower CD at a point A due south of it is $ 60^{o} $ and at a point B due west of A, the elevation is $ 30^{o} $ . If AB = 3 km, the height of the tower is
[MP PET 1998]
Options:
A) $ 2\sqrt{3},km $
B) $ 2\sqrt{6},km $
C) $ \frac{3\sqrt{3}}{2}km $
D) $ \frac{3\sqrt{6}}{4}km $
Show Answer
Answer:
Correct Answer: D
Solution:
- From $ \Delta CDA,,x=h\cot 60^{o} $ = $ \frac{h}{\sqrt{3}} $ From $ \Delta ,CDB,,y=h\cot 30^{o}=\sqrt{3}h $ From $ \Delta ABC $ , by Pythagoras theorem, $ x^{2}+3^{2}=y^{2} $ $ c_2 $ $ ( \frac{h}{\sqrt{3}} )+3^{2}={{(\sqrt{3}h)}^{2}}\Rightarrow h=\frac{3\sqrt{6}}{4} $ km.
BETA
