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

• Nov 23rd 2010, 07:02 AM
Veronica1999
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?

• Nov 23rd 2010, 07:12 AM
emakarov
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).
• Nov 23rd 2010, 07:31 AM
Veronica1999
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?
• Nov 23rd 2010, 07:48 AM
emakarov
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.
• Nov 23rd 2010, 08:42 AM
Soroban
Hello, Veronica1999!

Did you make a sketch?

Quote:

$\text{How many triangles have area 10 and vertices}}$
$\text{at }A(-5,0),\;B(5,0),\;C(5\cos\theta, 5\sin\theta)}\text{ for some angle }\theta?$
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.