Question

1. a triangle $\triangle abc$ with $mangle a = 40$, $mangle b = 60$, and $mangle c = 80$ is reflected across $y$-axis to obtain the $\triangle abc$. what is the $mangle c$, in degrees? 10. a figure is shown below. what is the $mangle qsr? p q r 112° 42° s

Explanation:

Step1: Recall reflection property

Reflection is a rigid - motion. Rigid - motions preserve angle measures. So, when \(\triangle ABC\) is reflected across the \(y\) - axis to get \(\triangle A'B'C'\), the angle measures of the two triangles are equal.

Step2: Identify the corresponding angle

Since \(\angle C\) and \(\angle C'\) are corresponding angles in the reflection, \(m\angle C'=m\angle C\). Given \(m\angle C = 80^{\circ}\), then \(m\angle C'=80^{\circ}\).

Step3: For the second problem, use the exterior - angle property

The exterior - angle of a triangle is equal to the sum of the two non - adjacent interior angles. In \(\triangle QSR\), the exterior angle \(\angle PQR = 112^{\circ}\), and one non - adjacent interior angle is \(\angle R=42^{\circ}\). Let \(m\angle QSR=x\). Then, by the exterior - angle property of a triangle, \(112^{\circ}=x + 42^{\circ}\).

Step4: Solve for \(x\)

Subtract \(42^{\circ}\) from both sides of the equation \(112^{\circ}=x + 42^{\circ}\). We get \(x=112^{\circ}-42^{\circ}=70^{\circ}\), so \(m\angle QSR = 70^{\circ}\).

Answer:

1. \(80^{\circ}\)
2. \(70^{\circ}\)