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17. if np bisects ∠mnq, m∠mnq = 8x + 12, m∠pnq = 78°, and m∠rnm = 3y - …

Question

  1. if np bisects ∠mnq, m∠mnq = 8x + 12, m∠pnq = 78°, and m∠rnm = 3y - 9, find the values of x and y.

Explanation:

Step1: Use angle - bisector property

Since $\overrightarrow{NP}$ bisects $\angle MNQ$, then $m\angle MNP=m\angle PNQ$. And $m\angle MNQ = 8x + 12$, $m\angle PNQ=78^{\circ}$. So $8x + 12=2\times78$.

Step2: Solve the equation for $x$

First, simplify the right - hand side of the equation: $8x+12 = 156$. Then subtract 12 from both sides: $8x=156 - 12=144$. Divide both sides by 8: $x=\frac{144}{8}=18$.

Step3: Use vertical - angle property

$\angle RNM$ and $\angle PNQ$ are vertical angles, so $m\angle RNM=m\angle PNQ$. Given $m\angle RNM = 3y-9$ and $m\angle PNQ = 78^{\circ}$, we set up the equation $3y-9 = 78$.

Step4: Solve the equation for $y$

Add 9 to both sides of the equation: $3y=78 + 9=87$. Divide both sides by 3: $y=\frac{87}{3}=29$.

Answer:

$x = 18$, $y = 29$