A plane rises from take-off and flies at an angle of 10 degrees with the horizontal runway. When it has gained 900 feet, find the distance, to the nearest foot, the plane has flown. An airplane is at vertex B of a right triangle that has a horizontal side AC, a vertical side BC of length 900 feet, and a dashed hypotenuse AB of unknown length c that rises from left to right. Angle C is the right angle and angle A measures 10 degrees. 10 degrees c equals question mark 900 ft A B C

The plane has flown 5,183 feet (approximately)

Step-by-step explanation: First and foremost, the plane is at vertex B of a right angled triangle ABC, and angle A measures 10°. Also the vertical side BC is the distance covered by the plane which in this case is the hypotenuse. The line BC is facing angle A which is the reference angle, hence line BC is the opposite.

With this bit of information we can calculate the distabce flown (hypotenuse) as follows;

Sin A = opposite ÷ hypotenuse

Sin 10 = 900/c

By cross multiplication we now have;

c = 900/Sin 10

c = 900/0.17364817766693

c = 5182.89

c ≈ 5183

Therefore the plane has flown 5,183 feet (approximately)

-2

7 - 9 = -2

reversed:

-2 + 9 = 7