Question

what are the angle measures of triangle abc? m∠a = 30°, m∠b = 60°, m∠c = 90° m∠a = 60°, m∠b = 90°, m∠c = 30° m∠a = 90°, m∠b = 30°, m∠c = 60° m∠a = 90°, m∠b = 60°, m∠c = 30°

Explanation:

Step1: Check the Pythagorean theorem

In a right - triangle, \(a^{2}+b^{2}=c^{2}\). Let \(a = 12\), \(b=24\), \(c = 12\sqrt{3}\).
\(12^{2}+24^{2}=144 + 576=720\), \((12\sqrt{3})^{2}=144\times3 = 432\) (Wrong). Let \(a = 12\), \(b = 12\sqrt{3}\), \(c = 24\).
\(12^{2}+(12\sqrt{3})^{2}=144+432 = 576\), \(24^{2}=576\). So \(\angle A=90^{\circ}\) (since \(BC^{2}=AB^{2}+AC^{2}\), by the Pythagorean theorem converse).

Step2: Use the sine function

We know that \(\sin B=\frac{AC}{BC}\). Given \(AC = 12\sqrt{3}\), \(BC = 24\).
\(\sin B=\frac{12\sqrt{3}}{24}=\frac{\sqrt{3}}{2}\). So \(m\angle B = 60^{\circ}\) (since \(\sin60^{\circ}=\frac{\sqrt{3}}{2}\)).

Step3: Use the angle - sum property of a triangle

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

Answer:

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