3 Hard Trigonometry Questions

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• Oct 2nd 2009, 03:04 PM
yeah:)
3 Hard Trigonometry Questions
Here are the questions (image below). Visual representations of the answers would be much appreciated!

http://farm3.static.flickr.com/2539/...49cdae3b_o.jpg
• Oct 2nd 2009, 04:21 PM
Soroban
Hello, yeah!

#2 took a lot of work . . . #3 is similar.
Maybe someone else can find a more direct solution.

Quote:

2. The angle of elevation of the top of a tower is 38° from a point A south of it.
The angle of elevation of the top of the tower is 29° from a point B east of it.
Find the height of the tower if the distqance AB is 50m.

Code:

                  P                   o                 *|  *                 * |    *               *  |        *               *  |h          *             *    |              *             *    |                *           *      |        x      29° *           *      o  *  *  *  *  *  o B         *38°  *  Q              *         *  * y        *  50       * *    *     A o
The tower is: $h = PQ.$
$\angle PAQ = 38^o,\;\angle PBQ = 29^o$
$\angle AQB = 90^o,\;AB = 50\text{m}$
Let: $x = QB,\;y = QA$

In right triangle $PQA\!:\;\tan38^o \:=\:\frac{h}{y} \quad\Rightarrow\quad h \:=\:y\tan38^o$ .[1]

In right triangle $PQB\!:\;\;\tan29^o \:=\:\frac{h}{x} \quad\Rightarrow\quad h \:=\:x\tan29^o$ .[2]

Equate [1] and [2]: . $y\tan38 \:=\:x\tan29 \quad\Rightarrow\quad y \:=\:\frac{\tan29}{\tan38}\,x$ .[3]

In right triangle $AQB\!:\;\;x^2+y^2 \:=\:50^2$

Substitute [3]: . $x^2 + \left(\frac{\tan29}{\tan38}\,x\right)^2 \:=\:2500 \quad\Rightarrow\quad x^2 + \frac{\tan^229}{\tan^238}\,x^2 \:=\:2500$

. . $x^2\left(1 + \frac{\tan^229}{\tan^238}\right) \:=\:2500 \quad\Rightarrow\quad x^2\left(\frac{\tan^238 + \tan^229}{\tan^238}\right) \:=\:2500$

. . $x^2 \:=\:\frac{2500\tan^238}{\tan^238 + \tan^229} \quad\Rightarrow\quad x\;=\;\frac{50\tan38}{\sqrt{\tan^238 + \tan^229}}$

Substitute into [2]: . $h \;=\;\frac{\tan29}{\tan38}\cdot\frac{50\tan38}{\sq rt{\tan^238 + \tan^229}}$

. . Grab your calculator . . .