Hello, Sarah!

Jennifer and Alex were flying a hot air balloon when they decided to calculate

the straight line distance on the ground from Mathville to Trigtown.

From a vertical height of 340m, they measured a 2° angle of depression to Mathville,

and a 3° angle of depression to Trigtown.

The angle between these two sight lines, from the balloon, to the two towns was 80°.

Find the distance along the ground from Mathville to Trigtown to the nearest tenth of a metre. I won't even try to type a 3-D diagram for this problem . . .

Consider the triangle involving the balloon and Mathville. Code:

B
* - - - - - - - - -
| * 2°
| 88° *
340 | *
| *
| *
* - - - - - - - - - - - *
G a M

The balloon is at B directly above point G on the ground; Mathville is at M.

The angle of depression is 2°, hence: __/__ GBM = 88°.

. . And we have: .tan 88° = a/340 . → . a .= .340·tan88° .≈ .9736.326

Consider the triangle involving the balloon and Trigtown. Code:

B
- - - - - - - - *
3° * |
* 87° |
* | 340
* |
* |
* - - - - - - - - - - - *
T b G

The balloon is at B direcly above point G on the ground; Trigtown is at T.

The angle of depression is 3°, hence: __/__TBG = 87°.

. . And we have: .tan 87° = b/340 . → . b .= .340·tan87° .≈ .6487.586

Looking down at the ground, we have this triangle: Code:

M
*
* *
9736.326 * *
* *
* *
* 80° *
G * * * * * * * T
6487.586

We can apply the Law of Cosines to determine side MT.

. . MT˛ .= .9736.326˛ + 6487.586˛ - 2(9736.326)(6487.586)cos 80°

I'll let you crank it out . . .