Question

what are the angle measures in triangle abc?
$m\angle a = 90^{\circ}, m\angle b = 30^{\circ}, m\angle c = 60^{\circ}$
$m\angle a = 60^{\circ}, m\angle b = 90^{\circ}, m\angle c = 30^{\circ}$
$m\angle a = 90^{\circ}, m\angle b = 60^{\circ}, m\angle c = 30^{\circ}$
$m\angle a = 60^{\circ}, m\angle b = 30^{\circ}, m\angle c = 90^{\circ}$

Explanation:

Step1: Check the Pythagorean theorem

For a triangle with sides \(a = 6\), \(b=6\sqrt{3}\), \(c = 12\).
Check if \(a^{2}+b^{2}=c^{2}\).
\(a^{2}=6^{2}=36\), \(b^{2}=(6\sqrt{3})^{2}=36\times3 = 108\), \(a^{2}+b^{2}=36 + 108=144\), and \(c^{2}=12^{2}=144\). So \(\angle A=90^{\circ}\) (since it is opposite the longest side \(BC\)).

Step2: Use trigonometric ratios

We know that \(\sin C=\frac{AB}{BC}\) (sine ratio: \(\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}\)).
Since \(AB = 6\) and \(BC=12\), \(\sin C=\frac{6}{12}=\frac{1}{2}\).
We know that if \(\sin\theta=\frac{1}{2}\), then \(\theta = 30^{\circ}\). So \(m\angle C = 30^{\circ}\).

Step3: Use the angle - sum property of a triangle

Since the sum of angles in a triangle is \(180^{\circ}\), and \(\angle A = 90^{\circ}\), \(\angle C=30^{\circ}\).
Let \(\angle B=x\), then \(90^{\circ}+x + 30^{\circ}=180^{\circ}\).
\(x=180^{\circ}-(90^{\circ}+ 30^{\circ})=60^{\circ}\).

Answer:

\(m\angle A = 90^{\circ},m\angle B = 60^{\circ},m\angle C = 30^{\circ}\) (the third option)