Question
Write the correct answer in the following:
ABCD is a cyclic quadrilateral such that AB is a diameter of the circle circumscribing it and $\angle\text{ADC}=140^\circ,$ then $\angle\text{BAC}$ is equal to:
  1. 80°.
  2. 50°.
  3. 40°.
  4. 30°.

Answer

  1. 50°.
Solution:
Given, ABCD is a cyclic quadrilateral and $\angle\text{ADC}=140^\circ.$

We know that, sum of the opposite angles in a cyclic quadrilateral is 180°.
$\angle\text{ADC}+\angle\text{ABC}=180^\circ$
$\Rightarrow\ 140^\circ+\angle\text{ABC}=180^\circ$
$\Rightarrow\ \angle\text{ABC}=180^\circ-140^\circ$
$\therefore\angle\text{ABC}=40^\circ$
Since, $\angle\text{ACB}$ is an angle in a semi-circle.
$\therefore\angle\text{ACB}=90^\circ$
In $\triangle\text{ABC},\ \ \angle\text{BAC}+\angle\text{ACB}+\angle\text{ABC}=180^\circ$ [by angle sum property of a triangle]
$\Rightarrow\angle\text{BAC}+90^\circ+40^\circ=180^\circ$
$\Rightarrow\angle\text{BAC}=180^\circ-130^\circ=50^\circ$

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