Given:
AD is diameter of the circle, AB is the tangent, and measure of arc ADC is 228 degrees.
To find:
The [tex]m\angle CAB[/tex] and [tex]m\angle CAD[/tex].
Solution:
AD is diameter of the circle. So, the measure of arc AD is 180 degrees.
[tex]m(arcADC)=m(arcAD)+m(arcDC)[/tex]
[tex]228^\circ=180^\circ+m(arcDC)[/tex]
[tex]228^\circ-180^\circ+=m(arcDC)[/tex]
[tex]48^\circ+=m(arcDC)[/tex]
The measure inscribed angle is half of the corresponding subtended arc.
[tex]m\angle CAD=\dfrac{1}{2}\times m(arcDC)[/tex]
[tex]m\angle CAD=\dfrac{1}{2}\times 48^\circ[/tex]
[tex]m\angle CAD=24^\circ[/tex]
AB is the tangent. So, [tex]m\angle BAD=90^\circ[/tex] because radius is perpendicular on the tangent and the point of tangency.
[tex]m\angle BAD=m\angle CAB+m\angle CAD[/tex]
[tex]90^\circ=m\angle CAB+24^\circ[/tex]
[tex]90^\circ -24^\circ=m\angle CAB[/tex]
[tex]66^\circ=m\angle CAB[/tex]
Therefore, [tex]m\angle CAB=66^\circ[/tex] and [tex]m\angle CAD=24^\circ[/tex].
