# height of a mountain with different elevations

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• May 4th 2010, 06:41 PM
mikeyb
height of a mountain with different elevations
hello everybody,

i got a height of a mountain problem but with a little twist. thanks for any help in advance.

Determine the elevation (height above sea level) of the top of a mountain, Mount 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 A, with an elevation of 1240.8 feet above sea level, the angle of elevation to the top of Mount C is 16.8 degrees. From the top of Mount B, with and elevation of 1575.2 feet above sea level, the angle of elevation to the top of Mount C is 19.3 degrees. The distance along the ground from Mount A to Mount B is measured at 1742.6 feet. Find the elevation of Mount C above sea level.
• May 4th 2010, 08:19 PM
sa-ri-ga-ma
Quote:

Originally Posted by mikeyb
hello everybody,

i got a height of a mountain problem but with a little twist. thanks for any help in advance.

Determine the elevation (height above sea level) of the top of a mountain, Mount 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 A, with an elevation of 1240.8 feet above sea level, the angle of elevation to the top of Mount C is 16.8 degrees. From the top of Mount B, with and elevation of 1575.2 feet above sea level, the angle of elevation to the top of Mount C is 19.3 degrees. The distance along the ground from Mount A to Mount B is measured at 1742.6 feet. Find the elevation of Mount C above sea level.

Let h be the height of the mountain C.
Angle of elevation of C from B is 19.3 degrees. x is the distance between B and C. So
tan(19.3) = (h-1575.2)/x.....(1)
Angle of elevation of C from A is 16.8 degrees.So
tan(16.8) = (h-1240.8)/(1742.6 + x)......(2)
Substitute the value of x from equation 1 in eq.(2) and solve for h.
• May 4th 2010, 08:38 PM
Soroban
Hello, mikeyb!

I'll set it up for you . . .

Quote:

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*
.
• May 5th 2010, 05:00 AM
mikeyb
Thank you both guys :) I really appreciate your help!
• May 5th 2010, 08:19 AM
mikeyb
I dont know why but I cant solve this problem....

y = (y/tan 19.3)tan 16.8 + 1742.6tan 16.8 - 334.4
Brainfart to the extreme....
• May 5th 2010, 08:53 AM
sa-ri-ga-ma
Quote:

Originally Posted by mikeyb
I dont know why but I cant solve this problem....

y = (y/tan 19.3)tan 16.8 + 1742.6tan 16.8 - 334.4
Brainfart to the extreme....

tan16.8/tan19.3 = 0.86
y(1 - 0.86) = 0.14y. = 191.5
So y = ....?