## Trigonometry (11th Edition) Clone

If $a$ is twice as long as $b$, $A$ is not necessarily twice as large as $B$.
The law of sines for a triangle having sides $a, b$ and angles $A, B$: $$\frac{a}{\sin A}=\frac{b}{\sin B}$$ This would mean $$\frac{a}{b}=\frac{\sin A}{\sin B}$$ Since $a$ is twice as long as $b$: $$\frac{a}{b}=\frac{\sin A}{\sin B}=2$$ Therefore, $$\sin A=2\sin B$$ Now this only means the value of $\sin A$ is twice as large as that of $\sin B$, not necessarily angle $A$ is twice as large as angle $B$. To show this, we can take a case where $\sin B=\frac{1}{2}$ and $\sin A=1$. Obviously, $\sin A=2\sin B$ because $1=2\times\frac{1}{2}$ However, $\sin A=1$ refers to $A=90^\circ$, while $\sin B=\frac{1}{2}$ refers to $B=30^\circ$. Yet here we can see that $A=3B$. Therefore, if $a$ is twice as long as $b$, only the value of $\sin A$ is necessarily twice as large as that of $\sin B$. $A$ is not necessarily twice as large as $B$.