Question

point o is the incenter of δabc. what is m∠qbo? (3x − 15)° (2x + 6)° (4x − 12)° 5° 9° 12° 24°

Explanation:

Step1: Recall Incenter Property

The incenter is the intersection of angle bisectors, so \(AO\) bisects \(\angle BAC\). Thus, \(2x + 6 = 4x - 12\).

Step2: Solve for \(x\)

\[

$$\begin{align*} 2x + 6 &= 4x - 12\\ 6 + 12 &= 4x - 2x\\ 18 &= 2x\\ x &= 9 \end{align*}$$

\]

Step3: Find \(m\angle QBO\)

Substitute \(x = 9\) into \(3x - 15\):
\[
3(9) - 15 = 27 - 15 = 12
\]
Wait, no—wait, \(BO\) is an angle bisector? Wait, no, \(AO\) is the angle bisector, but \(BO\) is also an angle bisector? Wait, no, the incenter is the intersection of angle bisectors, so \(AO\) bisects \(\angle BAC\), \(BO\) bisects \(\angle ABC\), and \(CO\) bisects \(\angle ACB\). But in the diagram, \(AQO\) and \(ASO\) are right angles, so \(AO\) is the angle bisector, so we found \(x = 9\). Then, the angle at \(B\) is bisected by \(BO\), but wait, the expression for \(\angle QBO\) is \(3x - 15\)? Wait, no, maybe I misread. Wait, the angle at \(A\) is bisected by \(AO\), so \(2x + 6 = 4x - 12\), solving gives \(x = 9\). Then, the angle \(\angle QBO\) is \(3x - 15\), so substituting \(x = 9\), we get \(3(9) - 15 = 27 - 15 = 12\)? Wait, but let's check again. Wait, maybe the angle \(\angle QBO\) is half of some angle? No, the problem gives \(\angle QBO\) as \(3x - 15\), and we found \(x = 9\), so \(3(9) - 15 = 12\), which is one of the options (12°).

Answer:

12°