# Question #3365f

The required angle is ${59.04}^{0} \left(2 \mathrm{dp}\right)$
Let "p" be height where ladder rests on wall and "b" be the base distance of the ladder from wall at ground, then $\tan \theta = \frac{p}{b} = \frac{50}{30} = \frac{5}{3} \therefore \theta = {\tan}^{-} 1 \left(\frac{5}{3}\right) = {59.04}^{0} \left(2 \mathrm{dp}\right)$[Ans]