Question

question 10(multiple choice worth 1 points)
(07.01 mc)
in circle a, bd is a diameter, and the measure of cb is 54°. what is the measure of ∠dbc?
o 54°
o 36°
o 27°
o 63°

Explanation:

Step1: Recall the property of a semicircle

Since \(BD\) is a diameter, \(\angle BCD = 90^{\circ}\) (angle inscribed in a semicircle).

Step2: Use the angle - sum property of a triangle

In \(\triangle BCD\), we know that the sum of angles in a triangle is \(180^{\circ}\). Let \(\angle DBC=\theta\). We know that \(\angle BCD = 90^{\circ}\) and the measure of arc \(CB\) is \(54^{\circ}\), and the measure of an inscribed angle is half the measure of its intercepted arc. But we can also use the angle - sum formula.
We have \(\angle BDC=\frac{1}{2}\text{arc}(BC)\) (inscribed - angle theorem). Since \(\text{arc}(BC) = 54^{\circ}\), \(\angle BDC = 27^{\circ}\).
Using the angle - sum property of a triangle \(\angle DBC+\angle BDC+\angle BCD=180^{\circ}\). Substituting \(\angle BCD = 90^{\circ}\) and \(\angle BDC = 27^{\circ}\) into the formula: \(\theta+27^{\circ}+90^{\circ}=180^{\circ}\).

Step3: Solve for \(\angle DBC\)

\(\theta=180^{\circ}-(90^{\circ} + 27^{\circ})\)
\(\theta = 63^{\circ}\)

Answer:

\(63^{\circ}\)