Question

problem 21: (first taught in lesson 30) in △abc, m∠a = 120 and m∠b is twice m∠c. find m∠b and m∠c. after you enter your answer press go. m∠b = m∠c =

Explanation:

Step1: Use the angle - sum property of a triangle

The sum of the interior angles of a triangle is \(180^{\circ}\). So, \(m\angle A+m\angle B + m\angle C=180^{\circ}\). Given \(m\angle A = 120^{\circ}\), then \(120^{\circ}+m\angle B + m\angle C=180^{\circ}\), which simplifies to \(m\angle B + m\angle C=180^{\circ}- 120^{\circ}=60^{\circ}\).

Step2: Set up an equation based on the relationship between \(\angle B\) and \(\angle C\)

Let \(m\angle C=x\). Since \(m\angle B\) is twice \(m\angle C\), then \(m\angle B = 2x\). Substitute into \(m\angle B + m\angle C=60^{\circ}\), we get \(2x+x=60^{\circ}\).

Step3: Solve for \(x\)

Combine like terms: \(3x = 60^{\circ}\). Divide both sides by 3: \(x=\frac{60^{\circ}}{3}=20^{\circ}\).

Step4: Find \(m\angle B\)

Since \(m\angle B = 2x\) and \(x = 20^{\circ}\), then \(m\angle B=2\times20^{\circ}=40^{\circ}\).

Answer:

\(m\angle B = 40^{\circ}\), \(m\angle C = 20^{\circ}\)