Question

m∠aob = 28, m∠boc = 3x - 2, m∠aod = 6x

Explanation:

Step1: Use angle - addition postulate

We know that \(m\angle AOD=m\angle AOB + m\angle BOC\). So, \(6x=28+(3x - 2)\).

Step2: Simplify the right - hand side

Simplify \(28+(3x - 2)\) to get \(3x+26\). So the equation becomes \(6x=3x + 26\).

Step3: Solve for \(x\)

Subtract \(3x\) from both sides: \(6x-3x=3x + 26-3x\), which gives \(3x=26\), then \(x=\frac{26}{3}\).

Step4: Find \(m\angle AOB\)

Given \(m\angle AOB = 28^{\circ}\).

Step5: Find \(m\angle AOD\)

Substitute \(x = \frac{26}{3}\) into \(m\angle AOD=6x\), \(m\angle AOD=6\times\frac{26}{3}=52^{\circ}\).

Step6: Find \(m\angle BOC\)

Substitute \(x=\frac{26}{3}\) into \(m\angle BOC = 3x-2\), \(m\angle BOC=3\times\frac{26}{3}-2=26 - 2=24^{\circ}\).

Step7: Find \(m\angle DOC\)

Since \(m\angle AOD=m\angle AOB+m\angle BOC+m\angle DOC\), then \(m\angle DOC=m\angle AOD-(m\angle AOB + m\angle BOC)\). Substitute the known values: \(m\angle DOC = 52-(28 + 24)=0^{\circ}\).

Answer:

\(x=\frac{26}{3}\)
\(\angle AOB = 28\)
\(\angle AOD = 52\)
\(\angle DOC = 0\)