Question

A cliff forms an angle of x∘ with a lake. If you stand on the edge of the cliff and throw a rock at a distance of 48 ft to the water, the rock will strike the water about 5 ft from the base of the cliff. What is measure of the angle that the cliff forms with the lake? Round your answer to the nearest whole degree.

Answer 1

Answer:

∅=6°

Step-by-step explanation:

we have that the distance from the edge of the cliff to the lake is 48 feet, and when the stone is thrown into the lake it collides with the water 5 feet from the base of the cliff so you must find the distance that the stone travels from the edge of the sewer up to 5 feet beyond its base (hypotenuse) and then the angle that forms it (slope)

Hypotenuse (c)

$c^{2}=a^{2}+b^{2}\\c^{2}=48ft^{2}+5ft^{2}\\c^{2}=2304ft^{2} +25ft^{2}=2329ft^{2}\\c=\sqrt{2329ft^{2}}=48.25ft$:

Slope ∅

Sin∅ = 5ft/48.25ft = 0.1036

∅=arcsin(0.1036) ≅ 5.9°, but how should we round to the nearest integer

, then:

∅=6°