Exercise 363

In $\triangle ABC$, $\sin(2A-B)+\sin(A+B)=2$ and $AB=6$. What is $BC$?

Solution

Each sine is at most $1$, so for the sum to be $2$ both must equal $1$: $\sin(2A-B)=1$ and $\sin(A+B)=1$.
In a triangle, $0^\circ
Adding the two equations gives $3A=180^\circ$, so $A=60^\circ$, $B=30^\circ$ and $C=180^\circ-A-B=90^\circ$.
$AB$ is the hypotenuse, opposite $C=90^\circ$, and $BC$ is opposite $A=60^\circ$. So $BC=AB\sin A=6\cdot\dfrac{\sqrt3}2=\boxed{3\sqrt3}$.