Question

When the sun's angle of elevation is 76°, a tree casts an 18-foot shadow on the ground. Find the length of the tree to the nearest tenth of a foot.

Answer 1

56.5 ft

Explanation:

See the attached figure which represents the explanation of the problem.

We need to find the length of the tree to which is the length of AD

From the graph ∠BAC = 90° and ∠ABD = 76°, AB = 18 ft

At ΔABD:

∠BAD = ∠BAC - ∠DAC = 90° - 4° = 86°

∠ADB = 180° - ( ∠BAD + ∠ABD) = 180 - (86+76) = 180 - 162 = 18°

Apply the sine rule at ΔABD

∴
`(AB)/(SinD) =(AD)/(sinB)`

∴ 18/sin 18 = AD/sin 76

∴ AD = 18 * (sin 76)/(sin 18) ≈ 56.5 (to the nearest tenth of a foot)

So, The length of the tree = 56.5 ft.

The answer is 56.5 ft