Question

1. graph δ abc and its perpendicular bisectors, and then find the coordinates of its circumcenter.

a (2, 7), b(10, 7) and c(10, 3)

1. circle p is inscribed in δ abc. if m∠bac = 41° and m∠abc = 85°, find m∠bcp and m∠apc.
2. if qn = 5x + 36 and qm = 2x + 51, find qo.

Explanation:

Step1: Analyze the properties of angle bisectors and incenter

Since \( P \) is the incenter of \( \triangle ABC \), it is the intersection of angle bisectors. So \( BP \) bisects \( \angle ABC \), and \( CP \) bisects \( \angle ACB \), \( AP \) bisects \( \angle BAC \).

Given \( m\angle BAC = 41^\circ \), then \( m\angle PAC=\frac{1}{2}m\angle BAC=\frac{41^\circ}{2} = 20.5^\circ \).

Given \( m\angle ABC = 85^\circ \), then \( m\angle PBC=\frac{1}{2}m\angle ABC=\frac{85^\circ}{2}=42.5^\circ \).

Step2: Find \( m\angle ACB \)

In \( \triangle ABC \), the sum of interior angles is \( 180^\circ \). So \( m\angle ACB=180^\circ - m\angle BAC - m\angle ABC \).

Substitute \( m\angle BAC = 41^\circ \) and \( m\angle ABC = 85^\circ \):

\( m\angle ACB=180^\circ - 41^\circ - 85^\circ=54^\circ \).

Then \( m\angle BCP=\frac{1}{2}m\angle ACB=\frac{54^\circ}{2} = 27^\circ \).

Step3: Find \( m\angle APC \)

In \( \triangle APC \), we know \( m\angle PAC = 20.5^\circ \) and \( m\angle ACP=m\angle BCP = 27^\circ \).

Using the angle - sum property of a triangle (\( m\angle PAC + m\angle ACP+m\angle APC = 180^\circ \)):

\( m\angle APC=180^\circ-(m\angle PAC + m\angle ACP) \)

Substitute \( m\angle PAC = 20.5^\circ \) and \( m\angle ACP = 27^\circ \):

\( m\angle APC=180^\circ-(20.5^\circ + 27^\circ)=180^\circ - 47.5^\circ = 132.5^\circ \)

Answer:

\( m\angle BCP = 27^\circ \), \( m\angle APC=132.5^\circ \)