# How do I draw the diagram for this trig. problem???

• Aug 12th 2009, 01:08 AM
xwrathbringerx
How do I draw the diagram for this trig. problem???
Hi

The elevation of a hill at a place P due east of it is 48 degrees and at a place due south of P the elevation is 30 degrees. If the distance from P to Q is 500 m, find the height of the hill.

Could someone please show me how to draw a diagram of this trig. problem?

Thanx
• Aug 12th 2009, 02:39 AM
bruxism
I hope this came out clear enough for you.

The problem with this question is that you have to think in 3 dimensions instead of two.

The picture I've drawn has replaced "the hill" from the question with "a flagpole". Same ideas obviously, just easier to draw.

If this image doesn't make much sense to you, let me know and i'll give it another shot with an isometric view instead of an oblique view like the one I've drawn.
• Aug 12th 2009, 07:03 AM
Soroban
Hello, xwrathbringerx!

Quote:

The elevation of a hill at a place $P$ due east of it is 48°
and at a place due south of $P$ the elevation is 30°.
If the distance from $P$ to $Q$ is 500 m, find the height of the hill.

Standing on the ground and looking north, we have this view:
Code:

    A *       | *       |42°*     h |    *       |      *       |        *       |      48° *     B * - - - - - - * P
The hill is: . $h = AB.$
$\angle APB = 48^o \quad\Rightarrow\quad \angle BAP = 42^o$
. . Hence: . $BP = h\tan42^o$

Standing on the ground and looking southwest, we have this view:
Code:

                    * A                   * |                 *60°|               *    | h             *      |           *        |         * 30°      |     Q * - - - - - - * B
The hill is: . $h = AB.$
$\angle AQB = 30^o \quad\Rightarrow\quad \angle QAB = 60^o$
. . Hence: . $BQ = h\tan60^o$

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

          h·tan42°     B * - - - - - - * P         *          |           *        |             *      | 500     h·tan60°  *    |                 *  |                   * |                     * Q
Using Pythagorus: . $\left(h\tan60^o\right)^2 \;=\;\left(h\tan42^o\right)^2 + 500^2 \quad\Rightarrow\quad h^2\tan^2\!60^o \;=\;h^2\tan^2\!42^o + 500^2$

. . $h^2\tan^2\!60^o - h^2\tan^2\!42^o \:=\:500^2 \qquad\Rightarrow\qquad h^2\left(\tan^2\!60^o - \tan^2\!42^o\right) \:=\:500^2$

. . $h^2 \;=\;\frac{500^2}{\tan^2\!60^o - \tan^2\!42^o} \qquad\Rightarrow\qquad h \;=\;\frac{500}{\sqrt{\tan^2\!60^i - \tan^2\!42^o}}$ . . . . There!