In triangle,
\(\tan 30=\frac{150-h}{60}\)
1/√3 \(=\frac{150-h}{60}\)
20√3 = 150 - h
h = 150- 20√3
h = 150 - 34.64
h = 115.36m (nearly equal to 116m).
So the correct option is (C)
The value of \( \cosec x + \cot x \) is
Evaluate \[ \frac{\cosec^2(\theta) - 1}{\cosec^2(\theta)} - \frac{\sec^2(\theta) - 1}{\sec^2(\theta)} \]