Trigonometric Equations Question 284
Question: A person is standing on a tower of height $ 15(\sqrt{3}+1),m $ and observing a car coming towards the tower. He observed that angle of depression changes from $ 30^{o} $ to $ 45^{o} $ in 3 sec. What is the speed of the car
[Karnataka CET 1998]
Options:
A) 36 km/hr
B) 72 km/hr
C) 18 km/hr
D) 30 km/hr
Show Answer
Answer:
Correct Answer: A
Solution:
- $ CD=h(\cot 30^{o}-\cot 45^{o}) $ $ 15,(\sqrt{3}+1),[\cot 30^{o}-\cot 45^{o}] $ $ =15,(\sqrt{3}+1),(\sqrt{3}-1) $ = $ 15,(3-1)=30m $ Speed = $ \frac{30}{1000}\times \frac{60\times 60}{3}km/hr $ = $ 36km/hr $ .
BETA
