AB is parallel to CD. If BCD= 100° and BAC= 40°, calculate
(i) CAD (ii) CBD (iii) BCA
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Answered by
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Step-by-step explanation:
Given,
∠BCD=100 and ∠BAC=40
AB║CD So ,∠BAC=∠ACD=40
∠BAD=180-100=80 (in a cyclic quad sum of opp angles is 180)
∠CAD=∠BAD-40=80-40=40
∴∠CAD=40
∠BCA=∠CAD=40 (alternate interior angles)
∠CBD=180-(∠BCD+∠CDB)
=180-(100+40) (angle subtended by the same chord DC are equal)
∠CBD =180-140=40
hope this helps:)
Answered by
0
Answer:
Step-by-step explanation:
Given,
∠BCD = 100 and ∠BAC = 40
AB║CD
So ,∠BAC = ∠ACD= 40
∠BAD=180-100 = 80 (in a cyclic quad sum of opp angles is 180)
∠CAD=∠BAD-40 = 80-40 = 40
∴∠CAD = 40
∠BCA=∠CAD = 40 (alternate interior angles)
∠CBD = 180-(∠BCD+∠CDB)
= 180-(100+40) (angle subtended by the same chord DC are equal)
∠CBD = 180-140 = 40
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