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?
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?
This is one reason. In fact, even without the use of Pythagorean theorem, $\displaystyle \cos\alpha$ is defined (at least it was when I studied it) as the x-coordinate of a point on a unit circle that is rotated $\displaystyle \alpha$ radians counterclockwise from (1, 0) . Similarly, $\displaystyle \sin\alpha$ is the y-coordinate. So a point $\displaystyle (\cos\alpha,\sin\alpha)$ lies on a unit circle by definition.
Hello, Veronica1999!
Did you make a sketch?
$\displaystyle \text{How many triangles have area 10 and vertices}}$
$\displaystyle \text{at }A(-5,0),\;B(5,0),\;C(5\cos\theta, 5\sin\theta)}\text{ for some angle }\theta?$Code:| * * * * | * C * | o * | 5 * :* | * :h * | * @ : * - - o - - - - + - - - + o - - A* | *B | * | * * | * * | * * * * |
$\displaystyle \,C$ is on a circle with radius 5.
$\displaystyle \Delta ABC$ has base 10 and height $\displaystyle \,h$
. . where: .$\displaystyle h \:=\:5\sin\theta \:=\:\pm2$
Hence: .$\displaystyle \sin\theta \:=\:\dfrac{\pm2}{5} \quad\Rightarro\quad \theta \:=\:\arcsin(\pm0.4)$
$\displaystyle \text{Therefore: }\:\theta \;\approx\;\begin{Bmatrix}23.6^o \\ 156.4^o \\ 203.6^o \\ 336.4^o \end{Bmatrix}$
. . There are four triangles.