Thread: Trig word problem

Two surveyors are working together to mark landmarks in the grand canyon. One man is on the south rim of the canyon and measures the angle of depression to an interesting rock formation on the base of the canyon to be 41.162 degrees. The woman on the north rim measures the angle of depression to the same rock formation to be 24.017 degrees. Find the distance from the base of the north rim (woman) to the rock formation if the woman's altitude is 2,000 feet less than the man's and the canyon is 15,840 feet wide at this point.

Thanks for any help.

Hello, ATC!

I assume you made a sketch . . .

Two surveyors are working together to mark landmarks in the Grand Canyon.
The man is on the south rim of the canyon and measures the angle of depression
to an interesting rock formation on the base of the canyon to be 41.162 degrees.
The woman on the north rim measures the angle of depression
to the same rock formation to be 24.017 degrees.
The woman's altitude is 2,000 feet less than the man's.
The canyon is 15,840 feet wide at this point.

Find the distance from the base of the north rim (woman) to the rock formation.

Code:

    M * - - - - - - - P
      | * 41.162o
      |   *
      |     *
      |       *             Q - - - - - - - * W
y+2000|         *               24.017o  *  |
      |           *                   *     |
      |             *              *        | y
      |               *         *           |
      |         41.162o *    * 24.017o      |
      *-------------------*-----------------*
      S     15,840-x      R        x        N
      : - - - - - - - -15,840 - - - - - - - :

The rock formation is at $\displaystyle \,R.$

The woman is at $\displaystyle \,W\!\!:\;W\!N = y$

$\displaystyle \angle QW\!R = 24.017^o = \angle W\!RN$

Let $\displaystyle x = RN \quad\Rightarrow\quad 15,\!840-x = SR$


The man is at $\displaystyle \,M\!:\;MS = y+2000$

$\displaystyle \angle P\!M\!R = 41.162^o = \angle M\!RS$


$\displaystyle \text{In right triangle }W\!N\!R\!:\;\tan24.017^o \:=\:\dfrac{y}{x}$

. . Hence: .$\displaystyle y \:=\:x\tan24.017^o\;\;\bf{[1]}$


$\displaystyle \text{In right triangle }MSR\!:\;\tan41.162^o \:=\:\dfrac{y+2000}{15,\!840 - x}$

. . Hence: .$\displaystyle y \:=\:(15,\!840-x)\tan41.162^o - 2000\;\;\bf{[2]} $


Equate [1] and [2]: .$\displaystyle x\tan24.017^o \;=\;15,\!840\tan41.162^o - x\tan41.162^o - 2000 $

. . . .$\displaystyle x\tan24.017^o + x\tan41.162^o \;=\;15,\!840\tan41.162^o - 2000 $

. . . .$\displaystyle x(\tan24.917^o + \tan41.167^o) \;=\;15,\!840\tan41.162^o - 2000 $

. . . . . . . . . . . . . . . . . . . . . . .$\displaystyle x \;=\;\dfrac{15,\!840\tan41.162^o - 2000}{\tan24.017^o + \tan41.162^o} $

Therefore: .$\displaystyle x \;\approx\;8977$ feet.