Question

practice!

1. if p is the incenter of △jkl, find each missing measure.

a) np = ______
b) nk = ______
c) pk = ______
d) lo = ______

1. if g is the incenter of △abc, find each missing measure.

a) gd = ______
b) bg = ______
c) fc = ______
d) bf = ______

1. if s is the incenter of △pqr, m∠prq = 8x - 10, m∠rpq = 4x - 14, and m∠pqr = 7x - 5, find each missing measure.

a) m∠prq = ______
b) m∠rpq = ______
c) m∠pqr = ______
d) m∠rps = ______
e) m∠pqs = ______
f) m∠prs = ______
g) m∠psr = ______

Explanation:

Step1: Recall triangle angle sum

The sum of angles in a triangle is \(180^\circ\). So, \(m\angle PRQ + m\angle RPQ + m\angle PQR = 180^\circ\). Substitute the given expressions: \((8x - 10)+(4x - 14)+(7x - 5)=180\).

Step2: Solve for \(x\)

Combine like terms: \(8x + 4x + 7x - 10 - 14 - 5 = 180\) → \(19x - 29 = 180\). Add 29 to both sides: \(19x = 209\). Divide by 19: \(x = 11\).

Step3: Find \(m\angle PRQ\)

Substitute \(x = 11\) into \(8x - 10\): \(8(11)-10 = 88 - 10 = 78^\circ\).

Step4: Find \(m\angle RPQ\)

Substitute \(x = 11\) into \(4x - 14\): \(4(11)-14 = 44 - 14 = 30^\circ\).

Step5: Find \(m\angle PQR\)

Substitute \(x = 11\) into \(7x - 5\): \(7(11)-5 = 77 - 5 = 72^\circ\).

Step6: Find \(m\angle RPS\)

The incenter bisects angles, so \(m\angle RPS=\frac{1}{2}m\angle RPQ=\frac{1}{2}(30^\circ)=15^\circ\).

Step7: Find \(m\angle PQS\)

The incenter bisects \(\angle PQR\), so \(m\angle PQS=\frac{1}{2}m\angle PQR=\frac{1}{2}(72^\circ)=36^\circ\).

Step8: Find \(m\angle PRS\)

The incenter bisects \(\angle PRQ\), so \(m\angle PRS=\frac{1}{2}m\angle PRQ=\frac{1}{2}(78^\circ)=39^\circ\).

Step9: Find \(m\angle PSR\)

In \(\triangle PSR\), sum of angles is \(180^\circ\). So, \(m\angle PSR = 180^\circ - m\angle RPS - m\angle PRS = 180 - 15 - 39 = 126^\circ\).

Answer:

a) \(78^\circ\)
b) \(30^\circ\)
c) \(72^\circ\)
d) \(15^\circ\)
e) \(36^\circ\)
f) \(39^\circ\)
g) \(126^\circ\)