Question

If two sides of a triangle are $50$ and $40$ and sine of the included angle is $\sqrt{15}/4$, then the square of the third side is equal to

Answer 1

Given $b=50,c=40,\sin A=\frac{\sqrt{15}}{4}$

$\Rightarrow \cos A=\sqrt{1-{\sin }^{2}A}=\frac{1}{4}$
Now, $a^2=b^2+c^2-2bc\cos A$

$=(50{)}^2+(40{)}^2-2\times 50\times 40\times \left(\frac{1}{4}\right)$

$=3100$