In $\Delta ABC$, let side $a = 8\sqrt2$, side $b = 8$ and $\measuredangle B = \frac{\pi }{6}$ radians. Find $\measuredangle A$

Question

I have tried using the trig difference formulas but I ultimately arrive at $\tan(\measuredangle A - 30) = \frac{1}{2\sqrt2}$ and don't know how to proceed from there. Is there an easier way to do this question?

Answer 1

Use the law of sines: $$\frac{\sin(A)}{a}=\frac{\sin(B)}{b}$$ where three of the four quantities above are known.

Answer 2

Use the Cosine Rule formula $\cos B=\dfrac{c^2+a^2-b^2}{2ca}$ to get the value of $c$.

and then use the formula $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ to find the $\measuredangle A$