Question 58

ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and $$\angle$$ADC = $$129^\circ$$. Then, $$\angle$$BAC is equal to:

Solution

As per the given question,

It is given that $$\angle ADC=129^\circ$$

As per the diagram, ABCD is a cyclic quadrilateral. So $$\angle ADC+\angle ABC=180$$

Now, $$\angle ABC=180^\circ-129^\circ=51^\circ$$

Now, In $$\triangle ACB$$

$$\angle ACB=90^\circ$$ (angle at the circumference with in half circle)

So, $$\angle BAC=90^\circ-51^\circ=39^\circ$$

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