Hello, mikeyb!

I'll set it up for you . . .

Determine the elevation (height above sea level) of the top of Mount $\displaystyle C.$

You take two measurements from the tops of two mountains with known elevations.

All three mountain tops are in a line (A, then B, then C),

and all three mountain tops lie in the same plane.

From the top of Mount $\displaystyle A$, with an elevation of 1240.8 feet above sea level,

the angle of elevation to the top of Mount $\displaystyle C$ is 16.8°.

From the top of Mount $\displaystyle B$, with and elevation of 1575.2 feet above sea level,

the angle of elevation to the top of Mount $\displaystyle C$ is 19.3°.

The distance along the ground from Mount $\displaystyle A$ to Mount $\displaystyle B$ is 1742.6 feet.

Find the elevation of Mount $\displaystyle C$ above sea level. Code:

o C
* |
* * |
* * | y
* * |
* * 19.8° |
* B o - - - - - - o S
* 16.8° | | 334.4
A o - - - - - - - - + - - - - - - o T
| | |
| | |
1240.8| | | 1240.8
| | |
| | |
* - - - - - - - - * - - - - - - *
P 1742.6 Q x R

Mount $\displaystyle A$ is: .$\displaystyle AP \:=\:1240.8 \:=\:TR, \quad \angle CAT = 16.8^o$

Mount $\displaystyle B$ is: .$\displaystyle BQ \,=\,1575.2 \quad\Rightarrow\quad ST \:=\:334.4,\quad \angle CBS = 19.8^o$

Mount $\displaystyle C$ is: .$\displaystyle CR \:=\:CS + ST + TR \;=\;y + 334.4 + 1240.8$ .[1]

$\displaystyle PQ \:=\:1742.6 \quad\text{ Let }x \:=\:QR$

In right triangle $\displaystyle CTA\!:\;\;\tan16.8^o \:=\:\frac{y+334.4}{x + 1742.6}$

. . . . . . $\displaystyle y \;=\;x\tan16.8^o + 1742.6\tan16.8^o - 334.4 $ .[2]

In right triangle $\displaystyle CSB\!:\;\;\tan19.8^o \:=\:\frac{y}{x} \quad\Rightarrow\quad x \:=\:\frac{y}{\tan19.8^o}$ .[3]

Substitute [3] into [2]: .$\displaystyle y \;=\;\left(\frac{y}{\tan19.8^o}\right)\tan16.8^o + 1742.6\tan16.8^o - 334.4 $

Solve for $\displaystyle y$ . . . Substitute into [1].

Edit: Too slow again . . . Sa-ri-ga-ma beat me to it . . . *sigh*

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