Question : ABCD is a cyclic quadrilateral and BC is the diameter of the related circle on which A and D also lie. $\angle \mathrm{BCA}=19°$ and $\angle \mathrm{CAD}=32°$. What is the measure of $\angle \mathrm{ACD}$?
Option 1: 41°
Option 2: 38°
Option 3: 40°
Option 4: 39°
Correct Answer: 39°
Solution :
Given,
$\angle{CAD}=32°$ and $\angle{BCA}=19°$
We know, that the angle formed by the diameter is 90°.
⇒ $\angle{BAC}=90°$
The sum of the opposite angle of a quadrilateral = 180°.
⇒ $\angle{CAD}+\angle{BAC}+\angle{BCA}+\angle{ACD}=180°$
⇒ $32°+90°+19°+\angle{ACD}=180°$
⇒ $141°+\angle{ACD}=180°$
⇒ $\angle{ACD}=180°-141°$
⇒ $\angle{ACD}=39°$
Hence, the correct answer is 39°.
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