Question

given: ( p ) is the incenter of ( \triangle mno ). ( overline{pm} ), ( overline{pn} ), and ( overline{po} ) are angle bisectors.
( py = 23mathrm{mm} ), ( po = 52mathrm{mm} ), ( mangle zmp = 30^{circ} ), ( mangle mon = 40^{circ} )
what is the length of ( overline{px} )?
what is the measure of ( angle pmx )?
what is the measure of ( angle pox )?
what is the length of ( overline{xo} )?
(hint: use the pythagorean theorem.)

Explanation:

Step1: Find the length of \( \overline{PX} \)

Since \( P \) is the in - center of \( \triangle MNO \), the distances from the in - center to the sides of the triangle are equal. So \( PX = PY\). Given \( PY = 23\mathrm{mm}\), then \( PX=23\mathrm{mm}\).

Step2: Find the measure of \( \angle PMX \)

Given \( \angle ZMP = 30^{\circ}\), and \( \overline{PM}\) is an angle bisector. So \( \angle PMX=\angle ZMP = 30^{\circ}\).

Step3: Find the measure of \( \angle POX \)

Given \( \angle MON = 40^{\circ}\), and \( \overline{PO}\) is an angle bisector. By the definition of an angle bisector, \( \angle POX=\frac{1}{2}\angle MON\). So \( \angle POX = 20^{\circ}\).

Step4: Find the length of \( \overline{XO}\)

We know that \( PO = 52\mathrm{mm}\) and \( PX = 23\mathrm{mm}\). In right - triangle \( PXO\), using the Pythagorean theorem \(a^{2}+b^{2}=c^{2}\), where \(c = PO\), \(a = PX\), and \(b = XO\). Then \(XO=\sqrt{PO^{2}-PX^{2}}\). Substitute \(PO = 52\) and \(PX = 23\) into the formula:

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Answer:

The length of \( \overline{PX}\) is \(23\mathrm{mm}\).
The measure of \( \angle PMX\) is \(30^{\circ}\).
The measure of \( \angle POX\) is \(20^{\circ}\).
The length of \( \overline{XO}\) is \(5\sqrt{87}\mathrm{mm}\approx46.64\mathrm{mm}\).