Two projectiles are projected at angles `( theta)` and `((pi)/(2) - theta )` to the horizontal respectively with same speed `20 m//s`. One of them rises 10 m higher than the other. Find the angles of projection. (Take `g= 10 m//s^(2)`)

Explanation

To solve the problem, we need to find the angles of projection for two projectiles that are launched at angles \( \theta \) and \( \frac{\pi}{2} - \theta \) with the same speed of \( 20 \, \text{m/s} \). One projectile rises 10 m higher than the other. We will use the formula for the maximum height of a projectile.

### Step-by-step Solution:

1. **Formula for Maximum Height**:
The maximum height \( h \) reached by a projectile is given by the formula:
\[
h = \frac{u^2 \sin^2(\theta)}{2g}
...