Question : ABCD is a cyclic quadrilateral such that AB is the diameter of the circle circumscribing it and $\angle A D C=148^{\circ}$. What is the measure of the $\angle BAC$?
Option 1: $32^{\circ}$
Option 2: $45^{\circ}$
Option 3: $58^{\circ}$
Option 4: $60^{\circ}$
Correct Answer: $58^{\circ}$
Solution :
ABCD is a cyclic quadrilateral.
$\angle ADC = 148^\circ$
The angle formed by drawing lines from the ends of the diameter of a circle to its circumference forms a right angle.
The sum of opposite angles of the cyclic quadrilateral is 180$^\circ$
In quadrilateral ABCD,
$\angle ADC+\angle CBA = 180^\circ$
⇒ $148^\circ+\angle CBA = 180^\circ$
⇒ $\angle CBA = 180^\circ - 148^\circ$
⇒ $\angle CBA = 32^\circ$
In $\triangle ABC$,
$\angle BAC + 90^\circ + \angle CBA = 180^\circ$
⇒ $\angle BAC + 90^\circ + 32^\circ = 180^\circ$
⇒ $\angle BAC + 122^\circ = 180^\circ$
⇒ $\angle BAC = 180^\circ - 122^\circ$
⇒ $\angle BAC = 58^\circ$
$\therefore$
The measure of the ∠BAC is 58°
Hence, the correct answer is 58$^\circ$
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