Question

An vulture is perched 40 ft up in a tree and looks down at an angle of depression of a 35? angle and spots roadkill. How far is the roadkill from the vulture? Round to the nearest tenth

Answer 1

69.7 feet.

Explanation:

Let x represent the distance between vulture and roadkill.

We have been given that a vulture is perched 40 ft up in a tree and looks down at an angle of depression of a 35 and spots roadkill.

We can see from the attachment that vulture, roadkill and angle of depression forms a right triangle with respect to ground, where, x is hypotenuse and 40 ft is opposite side.

`\text{sin}=\frac{\text{Opposite}}{\text{Hypotenuse}}`

`\text{sin}(35^(\circ))=(40)/(x)`

`x=\frac{40}{\text{sin}(35^(\circ))}`

`x=(40)/(0.573576436351)`

`x=69.7378718`

`x\approx 69.7`

Therefore, the roadkill is 69.7 feet away from the vulture.

Answer 2

69.7 ft

Explanation:

we know that

The function sine of angle of 35 degrees is equal to divide the opposite side to the angle of 35 degrees (the height of the vulture in a tree) by the hypotenuse ( the distance from the vulture to the roadkill)

Let

z -----> the distance from the vulture to the roadkill

sin(35°)=40/z

z=40/sin(35°)=69.7 ft