A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130º. Find ∠BAC.
Solution:
Given, ABCD is a quadrilateral
Quadrilateral ABCD is inscribed in a circle
AB is the diameter of the circle
∠ADC = 130º
We have to find ∠BAC
Given, ABCD is a cyclic quadrilateral as it is inscribed in a circle with center O
We know that the sum of the opposite sides of a cyclic quadrilateral is 180 degrees.
∠ADC + ∠ABC = 180º
130 + ∠ABC = 180º
∠ABC = 180º - 130º
∠ABC = 50º ------------------------ (1)
We know that the angle subtended by a diameter to the circle is a right angle
So, ∠ACB = 90º ------------------ (2)
In triangle ABC,
By angle sum property of a triangle,
∠BAC + ∠ACB + ∠ABC = 180º
From (1) and (2),
∠BAC + 90º + 50º = 180º
∠BAC + 140º = 180º
∠BAC = 180º - 140º
Therefore, ∠BAC = 40º
✦ Try This: ABCD is a cyclic quadrilateral in which BC||AD, ∠ADC = 110° and ∠BAC = 50°. Find ∠DAC.
☛ Also Check: NCERT Solutions for Class 9 Maths Chapter 10
NCERT Exemplar Class 9 Maths Exercise 10.3 Problem 17
A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130º. Find ∠BAC.
Summary:
A quadrilateral ABCD is inscribed in a circle such that AB is a diameter and ∠ADC = 130º. ∠BAC = 40º
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