# Trigonometry Question

• Sep 2nd 2012, 11:40 AM
DonManolo
Trigonometry Question
• Sep 3rd 2012, 09:30 PM
kalyanram
Re: Trigonometry Question
Hey Don,
Refer to the figure attached. Choose x-axis along $AB$ and y-axis along $AC$ with the origin at $A$.

$Q,R$ can be interchanged.

Case 1.
$QR$ on sides $AB$ and $AC$ we have $Q=(2\alpha,0)$, $R=(0,2\beta)$. With the additional condition that $0 \le \alpha \le \frac{b}{2}$, $0 \le \beta \le \frac{c}{2}$

Case 2.
$QR$ on sides $AC$ and $BC$ we have the abscissa of $Q$ is $0$ and hence abscissa of $R$ has to be $2\alpha$. Let the ordinate of $Q$ be $\gamma$ then we have ordinate of $R$ has to be $2\beta-\gamma$ and it has to satisfy the line equation $\frac{y}{x-b} = -\frac{c}{b} \implies \gamma = -\frac{c(2\alpha - b)}{b}$ with the additional bound constraints $0 \le \alpha \le \frac{b}{2}$ and $0 \le \gamma \le c$

Case 3.
Can be done on similar lines of Case2.

Now use the rotational matrix to translate to the Rectangular Cartesian Co-ordinates.

~Kalyan.