Question

Please help!!! A plane takes off from the airport and climbs at a steady rate. If the plane travelled
5.4km after take-off and gained 1120m of elevation (vertical gain), what angle did the plane take off at?

Answer 1

Answer:

$12.0^\circ$

Step-by-step explanation:

Think of a right triangle. It may help you to draw it.

Start at a point. Draw a point and label it point A. That is where the plane takes off from. Now draw a diagonal segment tilted up to the right and stop at a point and label it B. Segment AB is the path of the airplane. Now draw a segment down  vertically to a point you will label C, so that points A and C are on the same horizontal line. Connect points A and C.

You now have a right triangle. Angle C is the right angle. AB is the hypotenuse and is 5.4 km. That is the actual distance the plane traveled. BC is 1.12 km (since 1120 m = 1.12 km) and is a leg of the right triangle. The angle you are looking for is angle A.

For angle A, BC is the opposite leg. AB is the hypotenuse.

The trig ratio that relates the opposite leg to the hypotenuse is the sine.

$\sin A = \dfrac{opp}{hyp}$

$\sin A = \dfrac{1.12~km}{5.4~km}$

$\sin A = 0.2074$

$A = \sin^{-1} 0.2074$

$A = 12.0^\circ$

Answer: 12.0 degrees