A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depression. the local elementary school is directly to the west of the helicopter at a 62° angle of depression. the distance between the high school and the elementary school is 5 miles. find the distance from the helicopter to the high school. round your answer to the nearest tenth of a mile.

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Mathematics

11 August, 02:22

A helicopter is flying above a town. the local high school is directly to the east of the helicopter at a 20° angle of depression. the local elementary school is directly to the west of the helicopter at a 62° angle of depression. the distance between the high school and the elementary school is 5 miles. find the distance from the helicopter to the high school. round your answer to the nearest tenth of a mile.

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Orlando · Answer 1

Answer: 4.5 miles

Explanation:

When you draw the situation you find two triangles.

1) Triangle to the east of the helicopter

a) elevation angle from the high school to the helicopter = depression angle from the helicopter to the high school = 20°

b) hypotensue = distance between the high school and the helicopter

c) opposite-leg to angle 20° = heigth of the helicopter

d) adyacent leg to the angle 20° = horizontal distance between the high school and the helicopter = x

2) triangle to the west of the helicopter

a) elevation angle from elementary school to the helicopter = depression angle from helicopter to the elementary school = 62°

b) distance between the helicopter and the elementary school = hypotenuse

c) opposite-leg to angle 62° = height of the helicopter

d) adyacent-leg to angle 62° = horizontal distance between the elementary school and the helicopter = 5 - x

3) tangent ratios

a) triangle with the helicpoter and the high school

tan 20° = Height / x ⇒ height = x tan 20°

b) triangle with the helicopter and the elementary school

tan 62° = Height / (5 - x) ⇒ height = (5 - x) tan 62°

c) equal the height from both triangles:

x tan 20° = (5 - x) tan 62°

x tan 20° = 5 tan 62° - x tan 62°

x tan 20° + x tan 62° = 5 tan 62°

x (tan 20° + tan 62°) = 5 tan 62°

⇒ x = 5 tant 62° / (tan 20° + tan 62°)

⇒ x = 4,19 miles

=> height = x tan 20° = 4,19 tan 20° = 1,525 miles

4) Calculate the hypotenuse of this triangle:

hipotenuese ² = x² + height ² = (4.19) ² + (1.525) ² = 19.88 miles²

hipotenuse = 4.46 miles

Rounded to the nearest tenth = 4.5 miles

That is the distance between the helicopter and the high school.