Mathematics
In the given figure, O is center of the circle, AB || DC and ∠ACD = 32°, ∠DAB is equal to :
122°
148°
90°
none of the above
![In the given figure, O is center of the circle, AB || DC and ∠ACD = 32°, ∠DAB is equal to : Circles, Concise Mathematics Solutions ICSE Class 10.](https://cdn1.knowledgeboat.com/img/cm10/q1-e-ex-17-a-circles-maths-concise-icse-class-10-solutions-837x861.png)
Circles
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Answer
Join DA.
From figure,
![In the given figure, O is center of the circle, AB || DC and ∠ACD = 32°, ∠DAB is equal to : Circles, Concise Mathematics Solutions ICSE Class 10.](https://cdn1.knowledgeboat.com/img/cm10/q1-e-ex-17-a-circles-maths-answer-concise-icse-class-10-solutions-830x862.png)
In △DAC,
∠DAC = 90° (Angle in semicircle is a right angle)
As, alternate angles are equal.
∴ ∠CAB = ∠ACD = 32°
From figure,
∠DAB = ∠DAC + ∠CAB = 90° + 32° = 122°.
Hence, Option 1 is the correct option.
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