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You observe a plane approaching overhead and assume that its speed is 500 miles per hour. The angle of elevation of the plane is 15° at one time and 56° one minute later. Approximate the altitude of the plane. (Round your answer to two decimal places.)

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tochinwachukwu33

Answer:

The altitude of the plane is 2.7244 miles

Explanation:

Consider the sketch shown below.

XY = distance traveled by the airplane after 1 minute

recall distance = speed × time

XY = 500 ×[tex]\frac{1}{60}[/tex] =8.33 miles

From trigonometry laws,

cot 15 = [tex]\frac{AX}{AB}[/tex]

cot 56 = [tex]\frac{AY}{AB}[/tex]

cross multiplying,

AX = AB cot 15

AY = AB cot 56

AX - AY = AB cot 15 - AB cot 56

Recall that AX - AY = XY

Hence,

XY = AB (cot 15 - cot 56)

XY = AB (3.73205 - 0.67451)

XY = 3.05754 AB

AB = [tex]\frac{XY}{3.05754}[/tex] = [tex]\frac{8.33}{3.05754}[/tex] = 2.7244 miles

∴ The altitude of the plane is 2.7244 miles

Ver imagen tochinwachukwu33
abiolataiwo2015

Based on the information given the altitude of the plane is 2.73 miles.

Altitude of the plane

Let A represent the house

Let B represent the initial

Let C represent the later position of plane after one minutes

Consider BP and CQ perpendiculars to the ground

In right angle triangle APB

Angle APB = 90

BP=H

Angle BAP = 15

AP=H×COT 15

In right angle triangle AQC

Angle AQC = 90

CQ=H

ANGLE CAQ = 56

AQ=H×COT 56

Speed of plane=500 mph

Hence,

BC=500/60=QP=AP-AQ=H×(cot 15 - cot 56)

H= 500/(60×(cot 15 - cot 56 )

H=500/(60×(3.7321-0.67450852)

H=500/(60×3.05759148)

H=500/183.456

H =2.725 miles

H =2.73 miles (Approximately)

Inconclusion  the altitude of the plane is 2.73 miles.

Learn more about  the altitude of the plane here:https://brainly.com/question/24587018

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