Question

The angle of depression from an approaching airplane to an aircraft carrier is 52°. If the plane is 700 feet above sea level, how far does it have to fly to reach the carrier?

Answer 1

sohcahtoa, you can do sin52=700/x, multiply all by x, divide by sin52 so 700/sin52 = about 888.31 feet

Answer 2

The distance that is required to fly to reach the carrier should be considered as the 888.31 feet.

Calculation of the distance;

Since

The angle of depression from an approaching airplane to an aircraft carrier is 52°. And, the plane is 700 feet above sea level

So here we assume the distance be x

So, the following equation should be applied

$sin \ 52=700/x$

x = 888.31 feet

Hence, The distance that is required to fly to reach the carrier should be considered as the 888.31 feet.

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