in the given figure ,Ab is the diameter of the circle and angle Cba is 50° then measure of angle Cdb is equal to 40°. Prove it
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ACB=90° (Angle subtended on semi-circle)
In, △ACB,
∠CAB=180°-(∠ACB+∠CBA)
=180°-(90°+70°)
=20°
Now,
∠DCA+∠ACB+∠BAC+∠DAC=180° [Sum of opp. anglesof cyclic quadrilateralis supplementary.]
∠DCA+90°+20°+30°=180°
∠ACD=180°-40°
∠ACD=40°
Hence, 40° is the correct answer.
Answered by
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Answer:
ACB=90° (Angle subtended on semi-circle)
In, △ACB,
∠CAB=180°-(∠ACB+∠CBA)
=180°-(90°+70°)
=20°
Now,
∠DCA+∠ACB+∠BAC+∠DAC=180° [Sum of opp. anglesof cyclic quadrilateralis supplementary.]
∠DCA+90°+20°+30°=180°
∠ACD=180°-40°
∠ACD=40°
Hence, 40° is the correct answer.
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