Question

drag the correct expressions into the boxes to correctly complete the solution. show hints m∠bcd = 1/2 = 1/2 = 1/2 = 160° 360° (260° - 100°) (m⌢bad - m⌢bd) (260° + 100°) 80° 180° (m⌢bad + m⌢bd)

Explanation:

Step1: Recall the inscribed - angle formula

The measure of an inscribed angle is half the difference of the measures of the intercepted arcs. If the arcs are $\overset{\frown}{BAD}$ and $\overset{\frown}{BD}$, then $m\angle BCD=\frac{1}{2}(m\overset{\frown}{BAD}-m\overset{\frown}{BD})$.

Step2: Find the measures of the arcs

Assume that the measure of the major - arc $\overset{\frown}{BAD} = 260^{\circ}$ and the measure of the minor - arc $\overset{\frown}{BD}=100^{\circ}$. Then $m\overset{\frown}{BAD}-m\overset{\frown}{BD}=(260^{\circ}-100^{\circ})$.

Step3: Calculate the measure of the angle

$m\angle BCD=\frac{1}{2}(260^{\circ}-100^{\circ})$. First, calculate $260 - 100=160$. Then $\frac{1}{2}\times160^{\circ}=80^{\circ}$.

Answer:

$m\angle BCD=\frac{1}{2}(m\overset{\frown}{BAD}-m\overset{\frown}{BD})=\frac{1}{2}(260^{\circ}-100^{\circ})=\frac{1}{2}(160^{\circ}) = 80^{\circ}$
So the blanks should be filled in order with $(m\overset{\frown}{BAD}-m\overset{\frown}{BD})$, $(260^{\circ}-100^{\circ})$, $160^{\circ}$, $80^{\circ}$