Maths

Let G be a point on the line BC such that EG丄BC.

The required angle = ㄥEAG

Sin ㄥEAG = EG / EA....(☆)

Find EG :

△EBC = (1/2) * 8 * 8 * sin45, also = (1/2) BC * EG

So

8 * 8 * sin45 = BC * EG

64sin45 = 8EG

EG = 8sin45

Find EA :

△ABE is a right-angle △,

EA^2 = AB^2 + EB^2 = 7^2 + 8^2 = 113

EA = √113

Sub to ☆ :

Sin ㄥEAG = 8sin45 / √113

Sin ㄥEAG = 0.532152084

ㄥEAG = 32.15097814 = 32.2(correct to 1 d.p. )