Question

consider ⊙b in the figure. what is m∠abc? m∠abc = select choice degrees

Explanation:

Step1: Recall the angle - sum property of a triangle

In \(\triangle BCD\), the sum of interior angles of a triangle is \(180^{\circ}\). But we also know that \(\angle DBC\) and \(\angle ABC\) are supplementary (a straight - line angle). First, note that in \(\triangle BCD\), we have an exterior - angle relationship. The exterior angle of a triangle is equal to the sum of the two non - adjacent interior angles. Here, \(\angle ABC\) is an exterior angle of \(\triangle BCD\) with respect to the non - adjacent interior angles \(\angle BDC\) and \(\angle BCD\). Since \(BD = BC\) (radii of the same circle \(\odot B\)), \(\angle BDC=\angle BCD = 2x - 8\). And the central angle \(\angle DBC = 3x\).

Step2: Use the angle - sum property of a triangle

In \(\triangle BCD\), we know that the sum of interior angles \(\angle BDC+\angle BCD+\angle DBC = 180^{\circ}\). But we can also use the exterior - angle property. The exterior angle \(\angle ABC\) of \(\triangle BCD\) is given by \(\angle ABC=\angle BDC+\angle BCD\). Since \(\angle BDC = 2x - 8\) and \(\angle BCD = 2x - 8\), then \(\angle ABC=(2x - 8)+(2x - 8)=4x-16\). Also, we know that the central angle \(\angle DBC = 3x\) and \(\angle ABC + \angle DBC=180^{\circ}\) (a straight - line angle). Substituting \(\angle ABC = 4x - 16\) and \(\angle DBC = 3x\) into \(\angle ABC+\angle DBC = 180^{\circ}\), we get \((4x - 16)+3x=180\).

Step3: Solve the equation for \(x\)

Combine like terms: \(4x+3x-16 = 180\), which simplifies to \(7x-16 = 180\). Add 16 to both sides of the equation: \(7x=180 + 16=196\). Divide both sides by 7: \(x=\frac{196}{7}=28\).

Step4: Find the measure of \(\angle ABC\)

Substitute \(x = 28\) into the expression for \(\angle ABC\). \(\angle ABC=4x-16\). So, \(\angle ABC=4\times28-16\). First, calculate \(4\times28 = 112\), then \(112-16 = 96\).

Answer:

\(96\)