Trigonometric Equations Question 166
Question: The base of a cliff is circular. From the extremities of a diameter of the base the angles of elevation of the top of the cliff are 30o and $ 60^{o} $ . If the height of the cliff be 500 metres, then the diameter of the base of the cliff is
Options:
A) $ 1000,\sqrt{3},m $
B) $ 2000/\sqrt{3},m $
C) $ 1000/\sqrt{3},m $
D) $ 2000\sqrt{2},m $
Show Answer
Answer:
Correct Answer: B
Solution:
- $ d_2=h\cot 30^{o}=500\sqrt{3},,d_1=\frac{500}{\sqrt{3}} $ Diameter $ D=500\sqrt{3}+\frac{500}{3}\sqrt{3}=\frac{2000}{\sqrt{3}}m $ .
BETA
