Question

Jacob is standing 112 m from a building. The angle of elevation from where he is standing on the ground to the top of the building is 62°.

How tall is the building, to the nearest tenth of a meter?

Answer 1

$Sin\frac{Opposite}{Hypotenuse}Cos\frac{Adjacent}{Hypotenuse}Tan\frac{Opposite}{Adjacent}$

The hypotenuse is the longest side of the triangle.

The opposite side is the side across from the point/angle.

The adjacent side is the side next to the point/angle.

You need to find "X". So you can do:

tan 62° = opposite / adjacent

tan 62° = $\frac{X}{112}$      Multiply 112 on both sides

112(tan 62°) = X

210.6 = X

210.6 m

Answer 2

The height of  the building is 210.6 metres.

Given,

The distance of Jacob from the building is 112 metres.

The angle of elevation to the top of the building is 62°.

We have to find the height of the building.

Since Jacob is 112 m away from the base of the building, the height of the building is perpendicular to the base making it a right triangle with one angle 62°.

By using trigonometric ratios, we get

$tan62^\circ=\dfrac{\rm Perpendicular}{\rm Base}$

$1.8807=\dfrac{\rm Height \ of \ the\ building}{112}$

$\rm Height \ of\ the\ building=1.88\times 112$

$\rm Height \ of\ the\ building=210.56$

Hence the height of the building is 210.6 metres.

For more details on trigonometric ratios follow the link:

https://brainly.com/question/1201366