Given:
Consider the below figure attached with this question.
[tex]\angle ABO=20^\circ[/tex]
[tex]\angle ACO=30^\circ[/tex]
To find:
The value of x.
Solution:
Draw a line segment OA as shown below.
In triangle ABO,
[tex]OA=OB[/tex] [Radii of same circle]
[tex]m\angle BAO=m\angle ABO=20^\circ[/tex] [Base angles of an isosceles triangle]
In triangle ACO,
[tex]OA=OC[/tex] [Radii of same circle]
[tex]m\angle CAO=m\angle ACO=30^\circ[/tex] [Base angles of an isosceles triangle]
Now,
[tex]m\angle BAC=m\angle BAO+m\angle CAO[/tex]
[tex]m\angle BAC=20^\circ+30^\circ[/tex]
[tex]m\angle BAC=50^\circ[/tex]
Central angle is always twice of angle subtended by two points on the circle.
[tex]m\angle BOC=2\times m\angle BAC[/tex]
[tex]x=2\times (50^\circ)[/tex]
[tex]x=100^\circ[/tex]
Therefore, the value of x is 100°.
