Question

In the diagram, AB is a tangent to the circle, centre O. D is the mid-point of the chord BC. Given that BAC = x, find COD in the term of x.

Answer 1

`\angle COD =(90+x)/(2)` in term of x

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

`\angle AOB+\angle BOD +\angle COD =180`

Since D is the mid-point of the chord BC

We know,

Angle opposite to sides are congruent

`\angle BOD=\angle COD`

So,

`\angle AOB+\angle BOD +\angle COD =180`

`\angle AOB +2\angle COD =180`

Now, In triangle AOB

`\angle ABO = 90`

Therefore,

`\angle AOB+\angle ABO+\angle BAO=180`

`\angle AOB=90-x`

Therefore.

`\angle AOB +2\angle COD =180`

`(90-x) +2\angle COD =180`

`-x+2\angle COD =90`

`2\angle COD =90+x`

`\angle COD =(90+x)/(2)`

Thus,
`\angle COD =(90+x)/(2)`