Need help with inscribed angles If M
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Given:
In the given circle O, BC is diameter, OA is radius, DC is a chord parallel to chord BA and [tex]m\angle BCD=30^\circ[/tex].
To find:
The [tex]m\angle AOB[/tex].
Solution:
If a transversal line intersect two parallel lines, then the alternate interior angles are congruent.
We have, DC is parallel to BA and BC is the transversal line.
[tex]\angle OBA\cong \angle BCD[/tex] [Alternate interior angles]
[tex]m\angle OBA=m\angle BCD[/tex]
[tex]m\angle OBA=30^\circ[/tex]
In triangle AOB, OA and OB are radii of the circle O. It means OA=OB and triangle AOB is an isosceles triangle.
We know that base angles of an isosceles triangle are congruent.
[tex]\angle OAB\cong \angle OBA[/tex] [Base angles of an isosceles triangle]
[tex]m\angle OAB=m\angle OBA[/tex]
[tex]m\angle OAB=30^\circ[/tex]
In triangle AOB,
[tex]m\angle OAB+m\angle OBA+m\angle AOB=180^\circ[/tex]
[tex]30^\circ+30^\circ+m\angle AOB=180^\circ[/tex]
[tex]60^\circ+m\angle AOB=180^\circ[/tex]
[tex]m\angle AOB=180^\circ-60^\circ[/tex]
[tex]m\angle AOB=120^\circ[/tex]
Therefore, the measure of angle AOB is 120 degrees.