Question

A boy 1.5 m tall is standing 12 m away from a church building which has a tower on top of its roof. The top of the cross on the tower is 14.6 m away from the boy’s head (eyes). If the boy has to raise his eyes through an angle of 31° in order to see the top of the roof, calculate,Correct to the nearest degree, the angle through which the boy must raise his eyes to see the top of the cross on the tower.

Correct to one decimal place, the height of the top of the cross from ground.
Correct to one decimal place the height of the church building.​

Answer 1

1.
`35^(o)`

2. 9.8 m

3. 8.7 m.

Explanation:

1. Let the angle be represented by θ, so that:

Cos θ =
`(adjacent)/(hypotenuse)`

=
`(12)/(14.6)`

Cos θ = 0.82192

θ =
`Cos^(-1)` 0.8219

=
`34.73^(o)`

θ =
`35^(o)`

The angle through which the boy must raise his eyes to see the top of the cross on the tower is
`35^(o)`.

2. Let the height from the boy's head to the top of the cross be represented by y, so that:

Sin
`34.73^(o)` =
`(y)/(14.6)`

y = Sin
`34.73^(o)` x 14.6

= 0.5697 x 14.6

y = 8.32 m

The height of the top of the cross from the ground = 1.5 + 8.32

= 9.82 m

The height of the top of the cross from the ground is 9.8 m.

3. Let the height from the boy's head to the top of the building be represented by x, so that:

Tan 31 =
`(opposite)/(adjacent)`

=
`(x)/(12)`

x = Tan 31 x 12

= 0.6009 x 12

x = 7.21

The height of the church building = 1.5 + 7.21

= 8.71

The height of the church building is 8.7 m.

The sketch to the question is attached to this answer.