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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