## Trigonometry (11th Edition) Clone

Recall: For an angle $\theta$ whose terminal side contains the point $(x, y)$, we have: $\sin{\theta}=\dfrac{y}{r}$ where $r^2=x^2+y^2$ Thus, with $\sin{\theta}=\frac{3}{4}$, we can assume that $y=3$ and $r=4$. Solve for $x$ using the formula above to obtain: \begin{align*} r^2&=x^2+y^2\\ 4^2&=x^2+3^2\\ 4^2-3^2&=x^2\\ 7&=x^2\\ \pm \sqrt{7}&=x \end{align*} Hence, one triangle whose solution is found from $\sin{\theta}=\dfrac{3}{4}$ is one where $x=\sqrt7$ and $y=3$. Refer to the triangle above.