# ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to:

a. 80º

b. 50º

c. 40º

d. 30º

**Solution:**

It is given that

ABCD is a cyclic quadrilateral

∠ADC = 140º

Sum of the opposite angles in a cyclic quadrilateral is 180º

∠ADC + ∠ABC = 180º

Substituting the values

140 + ∠ABC = 180º

∠ABC = 180º - 140º

∠ABC = 40º

∠ACB is an angle in a semi circle

∠ACB = 90º

In triangle ABC

Using the angle sum property

∠BAC + ∠ACB + ∠ABC = 180º

Substituting the values

∠BAC + 90º + 40º = 180º

By further calculation

∠BAC + 130º = 180º

∠BAC = 180 - 130

∠BAC = 50º

Therefore, ∠BAC is equal to 50º.

**✦ Try This:** ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 120º, then ∠BAC is equal to

**☛ Also Check: **NCERT Solutions for Class 9 Maths Chapter 10

**NCERT Exemplar Class 9 Maths Exercise 10.1 Problem 8**

## ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to: a. 80º, b. 50º, c. 40º, d. 30º

**Summary:**

ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and ∠ADC = 140º, then ∠BAC is equal to 50º

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