Question

use the diagram at the right. solve for x. find the angle measures to check your work. m∠aob = 4x - 2, m∠boc = 5x + 10, m∠cod = 2x + 14
understanding the problem

1. so, m∠aob =

o m<cod
o m<cob
o m<abo
o m<boc
o m<boa
o m<dco
o m<ocd

Explanation:

1. First, assume that \(\angle AOB+\angle BOC+\angle COD = 180^{\circ}\) (assuming they form a straight - line angle. If it is a full - circle angle, the sum is \(360^{\circ}\), but without further information, we assume a straight - line angle for simplicity).

• Set up the equation: \((4x - 2)+(5x + 10)+(2x+14)=180\).
• Combine like terms: \(4x+5x + 2x-2 + 10+14 = 180\).
• Simplify the left - hand side: \(11x+22 = 180\).
• Subtract 22 from both sides: \(11x=180 - 22\).
• Calculate: \(11x = 158\).
• Solve for \(x\): \(x=\frac{158}{11}\approx14.36\).

1. Then, find \(m\angle AOB\):

• Substitute \(x\) into the formula for \(m\angle AOB\). Since \(m\angle AOB = 4x-2\), then \(m\angle AOB=4\times\frac{158}{11}-2=\frac{632}{11}-2=\frac{632 - 22}{11}=\frac{610}{11}\approx55.45^{\circ}\).

The question also asks to match \(m\angle AOB\) with the given angle - naming options. Since \(m\angle AOB\) can also be written as \(m\angle BOA\), the answer for the multiple - choice part is \(m\lt BOA\).

Step1: Set up the angle - sum equation

\((4x - 2)+(5x + 10)+(2x+14)=180\)

Step2: Combine like terms

\(11x+22 = 180\)

Step3: Solve for \(x\)

\(x=\frac{158}{11}\)

Step4: Find \(m\angle AOB\)

\(m\angle AOB = 4x-2=4\times\frac{158}{11}-2=\frac{610}{11}\)

Answer:

\(m\lt BOA\)