Question

determining angle/side relationships
what are the angle measures in triangle abc?
options:

• ( mangle a = 90^circ ), ( mangle b = 30^circ ), ( mangle c = 60^circ )
• ( mangle a = 60^circ ), ( mangle b = 90^circ ), ( mangle c = 30^circ )
• ( mangle a = 90^circ ), ( mangle b = 60^circ ), ( mangle c = 30^circ )
• ( mangle a = 60^circ ), ( mangle b = 30^circ ), ( mangle c = 90^circ )

triangle abc has sides: ab = 6 in., ac = ( 6sqrt{3} ) in., bc = 12 in.

Explanation:

Step1: Check if it's a right triangle

First, we can use the Pythagorean theorem to check if triangle \( ABC \) is a right triangle. The Pythagorean theorem states that for a right triangle with legs \( a \), \( b \) and hypotenuse \( c \), \( a^{2}+b^{2}=c^{2} \). Let's check the sides: \( AB = 6 \) in, \( AC=6\sqrt{3}\) in, \( BC = 12 \) in.

Calculate \( AB^{2}+AC^{2}=(6)^{2}+(6\sqrt{3})^{2}=36 + 36\times3=36+108 = 144\). And \( BC^{2}=(12)^{2}=144\). Since \( AB^{2}+AC^{2}=BC^{2}\), triangle \( ABC \) is a right triangle with the right angle at \( A \) (because \( AB \) and \( AC \) are the legs, and \( BC \) is the hypotenuse), so \( m\angle A = 90^{\circ}\).

Step2: Use trigonometric ratios to find other angles

We can use the sine ratio to find the angles. For angle \( B \), \( \sin B=\frac{AC}{BC}=\frac{6\sqrt{3}}{12}=\frac{\sqrt{3}}{2} \). We know that \( \sin60^{\circ}=\frac{\sqrt{3}}{2} \), so \( m\angle B = 60^{\circ} \).

For angle \( C \), since the sum of angles in a triangle is \( 180^{\circ} \), \( m\angle C=180^{\circ}-m\angle A - m\angle B=180^{\circ}-90^{\circ}-60^{\circ}=30^{\circ} \).

Answer:

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