Question

The angle of elevation of a tower, 45 m high from a ball on the ground 52 degree (a) how far is the ball from the foot of the tower (b) how much further away should the ball be moved from the foot of the tower so that the angle of elevation is halved

Answer 1

The distance of the ball from the foot of the tower is : 35.18m

The ball would be moved 57.2m away from the foot of the tower for the Angle of elevation to be halved.

What is angle of elevation?

Angle of elevation is the angle formed between the horizontal and the  line of view from the vertical.

Analysis:

The height of the tower and the distance of the ball from the foot of the tower form a right angle triangle.

so we use trigonometry.

a) let distance of the ball from foot of tower be x.

  so that, tan 52 = 45/x

  x = 45/tan52

  x = 45/1.279 = 35.18m

b) let the distance of the ball in the new position from the foot of the tower be y.

if the angle of elevation is halved, then new angle is 52/2 = 26°

 tan 26 = 45/y

y = 45/tan26 = 45/0.487 = 92.4m

distance moved from old position to new position = 92.4 - 35.18 = 57.2m

In conclusion, the distance of the ball from the foot of the tower and the distance the ball should move to make its elevation 26° are 35.18m and 57.2m respectively.

Learn more about angle of elevation: brainly.com/question/88158

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