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 $NP$ bisects $\angle MNQ$, then $m\angle MNQ = 2m\angle PNQ$. Given $m\angle MNQ=(8x + 12)^{\circ}$ and $m\angle PNQ = 78^{\circ}$, we have the equation $8x+12=2\times78$.

Step2: Solve for $x$

First, simplify the right - hand side of the equation: $2\times78 = 156$. So, $8x+12 = 156$. Subtract 12 from both sides: $8x=156 - 12=144$. Then divide both sides by 8: $x=\frac{144}{8}=18$.

Step3: Use vertical - angle property

$\angle RNM$ and $\angle PNQ$ are vertical angles. Vertical angles are equal, so $m\angle RNM=m\angle PNQ$. Given $m\angle RNM=(3y - 9)^{\circ}$ and $m\angle PNQ = 78^{\circ}$, we have the equation $3y-9 = 78$.

Step4: Solve for $y$

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

Answer:

$x = 18$, $y = 29$