Solve the triangle using the law of Cosine

Question

$a = 20$, $c= 20$, $B = 30^\circ$, $\cos 30^\circ = \frac{\sqrt{3}}{2}$

Using the law of cosine: $b^2 = a^2 + c^2 - 2ac\cos(B)$

\begin{align*} b^2 & = 20^2 + 20^2 -2(20)(20)\frac{\sqrt{3}}{2}\\ b^2 & = 800 - 800\frac{\sqrt{3}}{2}\\ b^2 & = -800\frac{\sqrt{3}}{2} + 800\\ b & = \sqrt{-400\sqrt{3}+800} \end{align*} (given that it's a triangle I only used the positive result)

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Now I have $a = 20$, $c = 20$, $b=\sqrt{-400\sqrt{3} + 800}$, $B = 30^\circ$, $\sin 30^\circ = \frac{1}{2}$

I can take it from here then use law of Sines to find the rest: $\frac{\sin A}{a} = \frac{\sin B}{b} = \frac{\sin C}{c}$

$$\frac{\sin A}{20} = \frac{\frac{1}{2}}{\sqrt{-400\sqrt{3} + 800}} = \frac{\sin C}{20}$$

I could do the math here and find the rest but if this were given as a quiz and I am supposed to finish it in a short amount of time, would there be any tricks I could use to get results quicker?

Answer 1

Since $a=c=20$, why don't you use the fact that the triangle is an isosceles triangle? You immediately know that the angles of the triangle are $30^\circ,75^\circ,75^\circ$.