Answer:
Option D. 25636 ft
Explanation:
From the given right triangle,
Observer is at the point O and he observes the plane flying at point P.
Angel of elevation of the plane is 55° and altitude of the plane was 21000 ft.
To get the distance OP we will apply Sine rule in the given right triangle,
Sin 55 =
![(h)/(x)](https://img.qammunity.org/2021/formulas/mathematics/high-school/85k7ugs5opfb9zrfoa4pemd0xkmigww42f.png)
Sin 55 =
![(21000)/(x)](https://img.qammunity.org/2021/formulas/mathematics/middle-school/yjqnyrlsouy3qyk9vjsuu7lqt87i9a5x7n.png)
x =
![\frac{21000}{\text{sin}55}](https://img.qammunity.org/2021/formulas/mathematics/middle-school/uex88xsv3maef9zehi4cf0xlxii2v7lurt.png)
x = 25636.27
x ≈ 25636 ft
Therefore, distance between the plane and observer is 25636 ft.
Option D. is the answer