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

In a right - triangle, \(a^{2}+b^{2}=c^{2}\). Let \(a = 6\), \(b=6\sqrt{3}\), \(c = 12\).
\(a^{2}=6^{2}=36\), \(b^{2}=(6\sqrt{3})^{2}=36\times3 = 108\), \(c^{2}=12^{2}=144\).
Since \(36 + 108=144\), by the Pythagorean theorem, \(\angle A=90^{\circ}\).

Step2: Use the sine function

The sine function is defined as \(\sin C=\frac{\text{opposite}}{\text{hypotenuse}}\).
In \(\triangle ABC\), for \(\angle C\), the opposite side to \(\angle C\) is \(AB = 6\) and the hypotenuse \(BC = 12\).
So \(\sin C=\frac{AB}{BC}=\frac{6}{12}=\frac{1}{2}\).
Since \(\sin30^{\circ}=\frac{1}{2}\), then \(m\angle C = 30^{\circ}\).

Step3: Use the angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). Let \(m\angle A = 90^{\circ}\), \(m\angle C=30^{\circ}\).
Using the formula \(m\angle A+m\angle B+m\angle C = 180^{\circ}\), we substitute the known values: \(90^{\circ}+m\angle B + 30^{\circ}=180^{\circ}\).
Solving for \(m\angle B\): \(m\angle B=180^{\circ}-(90^{\circ}+30^{\circ})=60^{\circ}\).

Answer:

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