# Thread: triangle with vertices (-5,0), (5,0) and (5cos A , 5sin A )

1. ## triangle with vertices (-5,0), (5,0) and (5cos A , 5sin A )

How many triangles have area 10 and vertices at (-5,0), (5,0) and (5cosA, 5sinA) for some angle A?

I know that the height of the triangle has to be 2.
I recently studied trig and know how to graph functions.
Is that enough trig to solve this?

2. Since, as you say, the height is 2, the third vertex lies on one of the lines y = 2 or y = -2. Also, the point (5cosA, 5sinA) for every A lies on the circle of radius 5 with the center (0, 0).

3. I think I understand. Am I supposed to know that (5cosA, 5 sinA) lies on a circle because cos squared A + sin squared A=1?

4. This is one reason. In fact, even without the use of Pythagorean theorem, $\cos\alpha$ is defined (at least it was when I studied it) as the x-coordinate of a point on a unit circle that is rotated $\alpha$ radians counterclockwise from (1, 0) . Similarly, $\sin\alpha$ is the y-coordinate. So a point $(\cos\alpha,\sin\alpha)$ lies on a unit circle by definition.

5. Hello, Veronica1999!

Did you make a sketch?

$\text{How many triangles have area 10 and vertices}}$
$\text{at }A(-5,0),\;B(5,0),\;C(5\cos\theta, 5\sin\theta)}\text{ for some angle }\theta?$
Code:
                |
* * *
*     |     *  C
*       |       o
*        |  5  * :*
|   *   :h
*         | * @   : *
- - o - - - - + - - - + o - -
A*         |         *B
|
*        |        *
*       |       *
*     |     *
* * *
|

$\,C$ is on a circle with radius 5.

$\Delta ABC$ has base 10 and height $\,h$
. . where: . $h \:=\:5\sin\theta \:=\:\pm2$

Hence: . $\sin\theta \:=\:\dfrac{\pm2}{5} \quad\Rightarro\quad \theta \:=\:\arcsin(\pm0.4)$

$\text{Therefore: }\:\theta \;\approx\;\begin{Bmatrix}23.6^o \\ 156.4^o \\ 203.6^o \\ 336.4^o \end{Bmatrix}$

. . There are four triangles.