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 feet, what is the distance (x) from the plane (P) to the observer (O)? A right triangle is shown with angle O marked 42 degrees, hypotenuse marked x, and height marked 15,000 feet. 10,035 feet
16,648 feet
20,188 feet
22,417 feet

8 2/3 is the answer

step-by-step explanation: just use a calculator

m∠cdf = 43° , m∠gdf = 114°, m∠gde = 23°, m∠edh = 23° , m∠cdh = 157°

step-by-step explanation:

since, it is given that de bisects the angle gdh ⇒ de divides the angle gdh into two equal angles.

⇒ m∠gde = m∠edh

⇒ 8·x   - 1 = 6·x + 5

⇒ x = 3

so, m∠gde = 8·x - 1 = 23°

and m∠edh = 23°

now, it is given ∠cde is an straight angle so, m∠cde = 180°

⇒ m∠cdf + m∠fdg + m∠gde = 180°

⇒ 43° + m∠fdg + 23° = 180°

⇒ m∠fdg = 114°

again, m∠cde = m∠cdh + m∠edh = 180°

m∠cdh + 23° = 180°

⇒ m∠cdh = 157°

d is the answer

step-by-step explanation: