Question

People from two houses 9,000 feet apart are looking up at a plane that is BETWEEN them. House 1 is west of the plane has an angle of elevation of 17.5 degrees and house 2 is east of the plane and has an angle of elevation that is 18.8 degrees. How far is the plane from House 1?

Answer 1

The distance between the plane and house 1 is 4899.2 feet

How to determine the distance between the plane and house 1?

See attachment for illustration of the question, where x represents the distance between the plane and house 1

Start by calculating the angle A using:

A = 180 - 17.5 - 18.8 ----- (angles in a triangle theorem)

This gives

A = 143.7

The distance x is then calculated using the following sine theorem

a/sin(A) = b/sin(B) = c/sin(C)

Using the parameters in the diagram, we have:

x/sin(18.8) = 9000/sin(143.7)

Multiply both sides by sin(18.8)

x= sin(18.8) * 9000/sin(143.7)

Evaluate

x = 4899.2

Hence, the distance between the plane and house 1 is 4899.2 feet

Read more about angle of elevation at:

brainly.com/question/88158

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