Question

What is the height of the tree to the nearest tenth of a foot?

Answer 1

Answer:

The height of the tree is approximately 21.4 feet

Step-by-step explanation:

We list out the question parameters first as follows;

The distance from the base of the tree where the angle of elevation is measured, d = 10 feet

The angle of elevation to the top of the tree from 10 feet from the base, θ = 65°

Let 'h' represent the height of the tree, then we have;

The line formed by the angle 65° angle, the height of the tree, 'h', and the distance 'd', form a right triangle with 'h' being the opposite leg to the given reference angle, 65°, and 'd' being the adjacent leg

By trigonometric ratio, we have;

$tan(\theta) = \dfrac{Opposite \ leg \ length}{Adjacent\ leg \ length} = \dfrac{h}{d}$

∴ h = d × tan(θ)

Plugging in the given values, we get;

h = 10 feet × tan(65°) = 21 feet $5\frac{11}{32}$ inches

∴  By rounding to the nearest tenth of a foot, the height of the tree, h ≈ 21.4 feet.