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°
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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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