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

Answer:

$\angle COD =\frac{90+x}{2}$ 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

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

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.

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

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

$-x+2\angle COD =90$

$2\angle COD =90+x$

$\angle COD =\frac{90+x}{2}$

Thus, $\angle COD =\frac{90+x}{2}$