# CHALLENGE - 8th Grade Math Help ??Challenge : The angle of depression from an airplane to the outskirts of a city measures 5.8 (degrees Tangent) The airplane is flying 6 miles above the ground at a...

CHALLENGE - 8th Grade Math Help ??

Challenge : The angle of depression from an airplane to the outskirts of a city measures 5.8 (degrees Tangent) The airplane is flying 6 miles above the ground at a speed of 200 mi/h. How much time will elapse before the airplane begins passing over the city?

*print*Print*list*Cite

Note that the angle of depression refers to the angle between the horizontal line and the line of sight. And it is located below the horizontal line. So, the triangle formed when the angle of depression is `5.8^o` is :

Note that the red dot indicates the position of the plane when the angle of depression is 5.8. And the yellow dot indicates its position when it is directly above the city.

Since the plane is flying 6 miles above the ground, the height of the triangle is y=6.

Then, let x be the base of the triangle. This is also the distance travelled by the plane from the red dot to yellow dot in the graph.

To solve for x, use the tangent function.

`tan theta=y/x`

Substitute `theta =5.8^o` and `y=6` .

`tan5.8^o=6/x`

Multiply both sides by x.

`x*tan5.8^o=6/x*x`

`x tan5.8^o=6`

Then, divide both sides by `tan 5.8^o` .

`(xtan5.8^o)=6/(tan5.8^o)`

`x=6/(tan5.8^o)`

`x=59.07`

Hence, the plane has travelled 59.07 miles in order to be directly above the city starting from the red dot in the graph.

Next, determine the time it takes for the plane to travel 59.07 miles.

To do so, apply the formula:

`s=d/t`

where s -speed, d-distance travelled and t-time elapsed.

Substitute s= 200 mi/h and d=59.07 miles to the formula.

`200=59.07/t`

Multiply both sides by t.

`t*200=59.07/t*t`

`200t = 59.07`

Divide both sides by 200.

`(200t)/200=59.07/200`

`t=59.07/200`

`t=0.3`

**Hence, it takes the plane 0.3 hours to pass over the city.**