Question

NO LINKS!!

Two sensors are spaced 700 feet apart along the approach to a small airport. When an aircraft is nearing the airport, the angle of elevation from the first sensor to the aircraft is 20°, and from the second sensor to the aircraft is 15°. Determine how high the aircraft is at this time. ​

Answer 1

711 ft (nearest foot)

Explanation:

Create two equations using trig ratios and the information given (refer to the attached diagram). Equate the equations and solve.

First equation

Let y = horizontal distance between sensor 1 and the aircraft

Let h = height of aircraft above the ground

Using the tangent trig ratio:

`\mathsf{\tan(\theta)=(opposite \ side)/(adjacent \ side)}`

Given:

• angle = 20°
• side opposite the angle = h
• side adjacent the angle = y

`\implies \mathsf{\tan(20)=(h)/(y)}`

Rearrange to make y the subject:

`\implies \mathsf{y=(h)/(\tan(20))}`

Second equation

Let y + 700 = horizontal distance between sensor 2 and the aircraft

Let h = height of aircraft above the ground

Using the tangent trig ratio:

`\mathsf{\tan(\theta)=(opposite \ side)/(adjacent \ side)}`

Given:

• angle = 15°
• side opposite the angle = h
• side adjacent the angle = y + 700

`\implies \mathsf{\tan(15)=(h)/(y+700)}`

Rearrange to make y the subject:

`\implies \mathsf{y=(h)/(\tan(15))-700}`

Now equate the 2 equations and solve for h:

`\implies \mathsf{(h)/(\tan(20))=(h)/(\tan(15))-700}}`

`\implies \mathsf{700=(h)/(\tan(15))-(h)/(\tan(20))}`

`\implies \mathsf{700=(h\tan(20)-h\tan(15))/(\tan(15)\tan(20))}`

`\implies \mathsf{700=(\tan(20)-\tan(15))/(\tan(15)\tan(20))h}`

`\implies \mathsf{h=(700\tan(15)\tan(20))/(\tan(20)-\tan(15))}`

`\implies \mathsf{h=710.9678247...}`

Therefore, the height of the aircraft is 711 ft (nearest foot)