Question

1. A plane is flying at an altitude of 13000 feet. The angle of depression from the plane to a control tower is 20 degrees. The control tower is 150 feet high . What is the horizontal distance from the plane to the tower? Round to the nearest foot.

2. Arianna is standing on top of a 60 foot building and spots spots her friend Rita, across the street. Arianna's eye height is 5 feet. If Rita is 37 feet away from the building, what is angle of depression from Arianna to Rita. Round to the nearest degree.

Answer 1

Answer:

see explanation

Step-by-step explanation:

There is a right triangle formed between the height above the tower to the plane, the horizontal and the line from the tower to the plane.

The height above the tower to the plane = 13000 - 150 = 12850 ft

The angle between the plane and the tower is 20° , angle of depression.

Using the tangent ratio in the right triangle formed, with the adjacent side being the horizontal ( hor ) distance, then

tan20° = $\frac{opposite}{adjacent}$ = $\frac{12850}{hor}$ ( multiply both sides by hor)

hor × tan20° = 12850 ( divide both sides by tan20° )

hor = $\frac{12850}{tan20}$ ≈ 35305 ft ( to the nearest foot )

(2)

Arianna is 60 + 5 = 65 ft above the ground ( to her eye level )

Using the right triangle formed with legs 65 and 37  and angle of depression x, then

tanx° = $\frac{opposite}{adjacent}$ = $\frac{37}{65}$ , thus

x = $tan^{-1}$ ($\frac{37}{65}$ ) ≈ 30° ( to the nearest degree )

Answer 2

Answer:  1) 35,305 ft          2) 60°

Step-by-step explanation:

Note that angle of depression (outside of the triangle) is EQUAL to the angle of elevation (inside the triangle measured from the ground).  See images

$1)\quad \tan \theta=\dfrac{opposite}{adjacent}\\\\\\.\qquad \tan 20^o=\dfrac{13,000-150}{x}\\\\\\.\qquad \tan 20^o=\dfrac{12,850}{x}\\\\\\.\qquad x=\dfrac{12,850}{\tan 20^o}\\\\\\.\qquad x=\large\boxed{35,305}$

$2)\quad \tan \theta=\dfrac{opposite}{adjacent}\\\\\\.\qquad \tan \theta=\dfrac{60+5}{37}\\\\\\.\qquad \tan \theta=\dfrac{65}{37}\\\\\\.\qquad \theta=\tan^{-1}\bigg(\dfrac{65}{37}\bigg)\\\\\\.\qquad \theta=\large\boxed{60^o}$