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if np bisects ∠mnq, m∠mnq=(8x + 12)°, m∠pnq = 78°, and m∠rnm=(3y - 9)°,…

Question

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 $\angle MNQ$, we have $m\angle MNP=m\angle PNQ$. So, $8x + 12=78$.

Step2: Solve for $x$

Subtract 12 from both sides of the equation $8x + 12=78$: $8x=78 - 12$, so $8x = 66$. Then divide both sides by 8: $x=\frac{66}{8}=\frac{33}{4}=8.25$.

Step3: Use vertical - angle property

$\angle RNM$ and $\angle MNQ$ are vertical angles. So $m\angle RNM=m\angle MNQ = 2\times78^{\circ}=156^{\circ}$. Also, $\angle RNM$ and $3y-9$ are vertical angles, so $3y-9 = 156$.

Step4: Solve for $y$

Add 9 to both sides of the equation $3y-9 = 156$: $3y=156 + 9=165$. Then divide both sides by 3: $y = 55$.

Answer:

$x = 8.25,y = 55$