Question

A helicopter is flying at an altitude of 1200°m. The angle of elevation from the tower on the ground to the plane measures 36°. How far is the building to the plane?

Answer 1

`"Distance between the plane and the building is " = color(blue)(2041.56 " m"`

Explanation:

![https://www.slideshare.net/BVCMDP/mat2793-angle-of-elevation]()
Given `BC = " Altitude " = 1200 m, " Angle of elevation " = 36^@`

To find `"Distance AC "`

ABC is a right triangle with AC the hypotenuse. Applying Pythagoras theorem,

`AC = x = (BC) / sin X = 1200 / sin 36 = 2041.56 " m"`

Answer 2

Please read the explanation.

Explanation:

Assumptions:

1. Angle of elevation from the tower on the ground to the plane: `rArr` angle of elevation is from the foot of the tower `/_OFH=36^@`.

2. How far is the building to the plane?: `rArr` horizontal distance from the foot of the tower `bar (OF)`.

Please refer to the diagram (NOT drawn to scale) demonstrating the assumptions used:

`bar (OH) = 1200` meters

`/_HOF = 90^@`

`/_OFH = 36^@`

We must find the magnitude of `bar (OF)`.

For the `/_ OFH`,

`bar (OH)` is the Opposite Side.

`bar (OF)` is the Adjacent Side.

Since we must find the magnitude of `bar (OF)`, we use the reatio for `tan(theta)`

`tan(theta) = tan(36^@) =` Opposite Side`/`Adjacent Side

`rArr tan(36^@) = (OH)/(OT)`

`rArr tan(36^@) = 1200/(OT)`

`rArr OT * tan(36^@) = 1200`

`rArr OT = 1200/tan(36^@)`

Using the calculator, `tan(36^@) ~~ 0.72654`

`rArr OT = 1200/0.72654~~1651.664`

Hence, `bar (OT) ~~ 1651.67` meters.

I have used a software to construct the problem for visual verification:

Scale Factor : 1 unit on the graph = 100 meters

All calculations are carried out using available functions in the software.