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question 3
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which are the correct measures for <ojk, <jko, and <jok?
select one:
a. <ojk = 152°, <jko = 18°, and <jok = 10°
b. <ojk = 102°, <jko = 28°, and <jok = 50°
c. <ojk = 117°, <jko = 36°, and <jok = 37°
d. <ojk = 117°, <jko = 28°, and <jok = 35°

Explanation:

Step1: Use property of parallel - lines

Since $JK\parallel LM$, $\angle JKL+\angle KLM = 180^{\circ}$ (co - interior angles). Given $\angle KLM = 152^{\circ}$, then $\angle JKL=180 - 152=28^{\circ}$.

Step2: Use property of straight - line

Since $\angle PON = 117^{\circ}$, and $\angle PON$ and $\angle JOK+\angle KON$ form a straight - line, assume $\angle KON$ is part of the angle formed by the parallel - line related angles. Also, because of the parallel - line properties in the figure, $\angle OJK$ and $\angle PON$ are corresponding angles (by extending the lines appropriately), so $\angle OJK = 117^{\circ}$.

Step3: Use angle - sum property of a triangle

In $\triangle OJK$, we know that the sum of interior angles of a triangle is $180^{\circ}$. Let $\angle OJK=\alpha = 117^{\circ}$, $\angle JKO=\beta = 28^{\circ}$, then $\angle JOK=\gamma=180-(\alpha+\beta)=180-(117 + 28)=35^{\circ}$.

Answer:

D. $\angle OJK = 117^{\circ}$, $\angle JKO = 28^{\circ}$, and $\angle JOK = 35^{\circ}$