Answer:
[tex]\angle COD =\frac{90+x}{2} [/tex] in term of x
Step-by-step explanation:
Given that AB is tangent to a circle with center O and radius of OC=OB
D is the mid-point of the chord BC and D is 90
Here, Angle BAC = x.
From figure,
AC is a straight line.
we can write as
[tex]\angle AOB+\angle BOD +\angle COD =180[/tex]
Since D is the mid-point of the chord BC
We know,
Angle opposite to sides are congruent
[tex]\angle BOD=\angle COD[/tex]
So,
[tex]\angle AOB+\angle BOD +\angle COD =180[/tex]
[tex]\angle AOB +2\angle COD =180[/tex]
Now, In triangle AOB
[tex]\angle ABO = 90[/tex]
Therefore,
[tex]\angle AOB+\angle ABO+\angle BAO=180[/tex]
[tex]\angle AOB=90-x[/tex]
Therefore.
[tex]\angle AOB +2\angle COD =180[/tex]
[tex](90-x) +2\angle COD =180[/tex]
[tex]-x+2\angle COD =90[/tex]
[tex]2\angle COD =90+x[/tex]
[tex]\angle COD =\frac{90+x}{2} [/tex]
Thus, [tex]\angle COD =\frac{90+x}{2} [/tex]
