Question

An observer (O) spots a plane flying at a 42° angle to his horizontal line of sight. If the plane is flying at an altitude of 15,000 ft., what is the distance (x) from the plane (P) to the observer (O)? (4 points) A right triangle is shown with angle O marked 42 degrees, hypotenuse marked x and the height marked 15000 feet. Select one: a. 10,035 feet b. 16,648 feet c. 20,188 feet d. 22,417 feet

Answer 1

Answer:

D. 22,417 feet

Step-by-step explanation:

Fine the diagram in the attachment for proper elucidation. Using the SOH, CAH, TOA trigonometry identity to solve for the distance (x) from the plane (P) to the observer (O), the longest side x is the hypotenuse and the side facing the angle of elevation is the opposite.

Hypotenuse = x and Opposite = 15,000feet

According to SOH;

$sin 42^0 = \frac{Opposite}{Hypotenuse} \\\\Sin42^0 = \frac{15000}{x}\\ \\x = \frac{15000}{sin42^0}$

$x = \frac{15000}{ 0.6691} \\\\x = 22,417 feet$

Hence the distance (x) from the plane P to the observer O is approximately 22,417 feet