Question 111

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and angle $$ADC = 130^\circ$$. Then angle BAC is equal to:

Solution

Given,

$$ \angle ADC=130^{\circ\ }$$

we know that,

sum of opposite angle of cyclic quadrilateral=$$180^{\circ\ }$$

$$\angle ADC+\angle ABC=180^{\circ\ }$$

$$130^{\circ\ }+\angle ABC=180^{\circ\ }$$

$$\angle ABC=50^{\circ\ }$$

$$In\ \triangle ABC,$$

$$\angle ABC=50^{\circ\ }, \angle BCA=90^{\circ\ }$$

$$\angle BAC=40^{\circ\ }$$

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