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: Solve for \(x\)

Since \(\overline{NP}\) 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 \(8x+12=2\times78\).
First, simplify the right - hand side: \(2\times78 = 156\).
Then, solve the equation \(8x+12 = 156\) for \(x\). Subtract 12 from both sides: \(8x=156 - 12\), so \(8x=144\). Divide both sides by 8: \(x=\frac{144}{8}=18\).

Step2: Solve for \(y\)

\(\angle MNQ\) and \(\angle RNM\) are supplementary (they form a linear pair). So \(m\angle MNQ+m\angle RNM = 180^{\circ}\).
We know \(m\angle MNQ = 156^{\circ}\) (from Step 1) and \(m\angle RNM=(3y - 9)^{\circ}\).
Set up the equation \(156+(3y - 9)=180\).
Simplify the left - hand side: \(3y+147 = 180\).
Subtract 147 from both sides: \(3y=180 - 147\), so \(3y = 33\).
Divide both sides by 3: \(y=\frac{33}{3}=11\).

Answer:

\(x = 18\), \(y = 11\)