A triangle ABC is inscribed in a circle with centre O. AO is produced to meet the circle at K and AD $$\perp$$ BC. If $$\angle$$B = 80$$^\circ$$ and $$\angle$$C = 64$$^\circ$$, then the measure of $$\angle$$DAK is:

As per the given question,

Given that, AD $$\perp$$ BC, $$\angle$$B = 80$$^\circ$$ and $$\angle$$C = 64$$^\circ$$

Now, $$\triangle ADB$$

$$\angle ABD=80$$

$$\angle BAD=90-80=10^\circ$$

In $$\triangle ABC$$ $$\angle BAC=180-80-64=36^\circ$$

Now, In $$\triangle ACK$$ and $$\triangle ACB$$

$$\angle B=\angle K$$ base is same for both triangle in a circle.

Now, in the $$\triangle ACK$$

$$\angle C=90^\circ$$

Hence $$\angle CAK=90-80=10^\circ$$

Now, $$\angle DAK=\angle BAC-\angle BAD-\angle CAK$$

$$\angle DAK=36-10-10=16^\circ$$