Question

parallel lines and triangle angle sums

1. use the diagrams to fill in the blanks.

interior angle measures
exterior angle measure
m∠a + m∠b + m∠c = 180°
m∠dca = m∠a + m∠b
+ m∠b + = 180°
m∠dca = +
m∠b + = 180°
m∠dca =
m∠b =

Explanation:

Step1: Find angle \(B\) in the first - triangle

We know that for a triangle, the sum of interior angles is \(180^{\circ}\). In \(\triangle ABC\) with \(m\angle A = 54^{\circ}\) and \(m\angle C=48^{\circ}\), we use the formula \(m\angle A + m\angle B + m\angle C=180^{\circ}\). So, \(m\angle B=180^{\circ}-m\angle A - m\angle C\).

$$m\angle B = 180^{\circ}-54^{\circ}-48^{\circ}=78^{\circ}$$

Step2: Find the exterior - angle measure in the second - triangle

For the second triangle, we know that the exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles. Given \(m\angle A = 59^{\circ}\) and \(m\angle B = 40^{\circ}\), and \(m\angle DCA=m\angle A + m\angle B\).

$$m\angle DCA=59^{\circ}+40^{\circ}=99^{\circ}$$

Answer:

In the first triangle: \(m\angle B = 78^{\circ}\). In the second triangle: \(m\angle DCA = 99^{\circ}\)