# Thread: 3 Hard Trigonometry Questions

1. ## 3 Hard Trigonometry Questions

Here are the questions (image below). Visual representations of the answers would be much appreciated!

2. Hello, yeah!

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

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: $\displaystyle h = PQ.$
$\displaystyle \angle PAQ = 38^o,\;\angle PBQ = 29^o$
$\displaystyle \angle AQB = 90^o,\;AB = 50\text{m}$
Let: $\displaystyle x = QB,\;y = QA$

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

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

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

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

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

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

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

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

. . Grab your calculator . . .