Question

5 (g.5c) find m∠gmh. a m∠gmh = 130° b m∠gmh = 50° c m∠gmh = 0° d m∠gmh = 180° 6 (g.5b, g.5c) find the supplement of m∠gmh. f ∠hmn g ∠kmn h ∠gmk j ∠hmj 10 (g.4a) name three points

Explanation:

Step1: Observe the protractor measurement

By looking at the protractor - like diagram for $\angle GMH$, the angle measure is $50^{\circ}$. The zero - degree mark starts from the ray $GM$ and the other ray $MH$ intersects the protractor at the $50^{\circ}$ mark.

Step2: Recall the definition of supplementary angles

Two angles are supplementary if their sum is $180^{\circ}$. Let the measure of $\angle GMH=\alpha = 50^{\circ}$, and the measure of its supplement be $\beta$. Then $\alpha+\beta = 180^{\circ}$, so $\beta=180^{\circ}-\alpha$.

Step3: Calculate the supplement

Substitute $\alpha = 50^{\circ}$ into the formula $\beta = 180^{\circ}-\alpha$. We get $\beta=180 - 50=130^{\circ}$. The angle that forms a straight - line with $\angle GMH$ (i.e., is supplementary to it) is $\angle HMN$ since $\angle GMH+\angle HMN = 180^{\circ}$.

Answer:

1. B. $m\angle GMH = 50^{\circ}$
2. F. $\angle HMN$