Question

AB is parallel to CD. If BCD= 100° and BAC= 40°, calculate
(i) CAD (ii) CBD (iii) BCA
​

Answer 1

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:)

Answer 2

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