Question

A daredevil decides he wants to jump out of a plane and parachute to the U.S.S. Showboat. He knows he needs to leave the plane at an angle of 58˚ from vertical and he knows he wants to wait until the horizontal distance to the boat is 1000yd. How far away should the plane be from the boat to complete his aerial stunt successfully?

Answer 1

Answer:

1179 yards

Step-by-step explanation:

The question can be described by a right angled triangle.

The angle of depression of the boat from the plane = $90^{o}$ - 58˚

                                                                    = $32^{o}$

Let the distance from the plane to the boat be represented by x.

Thus, applying the appropriate trigonometric function, we have;

Cos θ = $\frac{Adjacent}{Hypotenuse}$

Cos $32^{o}$ = $\frac{1000}{x}$

x = $\frac{1000}{Cos 32^{o} }$

  = $\frac{1000}{0.8481}$

  = 1179.11

x = 1179 yds

The plane should be 1179 yards away from the boat.