Question

In the figure, ABCD is a parallelogram in which angle DAP = 20° , angle BAP = 40° and angle ABP = 80° . Find angle APD and angle BPC.

Most satisfying answer will mark as brainliest. ​

Answer 1

Solution:

Given:

=> ∠ABP = 80°

=> ∠BAP = 40°

To Find:

=> ∠APD

=> ∠BPC

So,

=> ∠APD = ∠PAB            (Alternate interior angle)

=> ∠APD = 40°

We know that sum of all sides of angle is 180°. So,

=> ∠PAB + ∠ABP + ∠BPA = 180°

=> 40° + 80° + ∠BPA = 180°

=> 120° + ∠BPA = 180°

=> ∠BPA = 180° - 120°

=> ∠BPA = 60°

Now,

=> ∠APD + ∠BPA + ∠BPC = 180°

=> 40° + 60° + ∠BPC = 180°

=> 100° + ∠BPC = 180°

=> ∠BPC = 180° - 100°

=> ∠BPC = 80°

So,

=> ∠APD = 40°

=> ∠BPC = 80°

Answer 2

Given :-

ABCD is a parallelogramGiven:

ABP = 80°

BAP = 40°

To Find :-

∠APD &∠BPC

Solution :-

∠APD = ∠PAB [Alternate interior angle]

∠APD = 40°

∠PAB + ∠ABP + ∠BPA = 180° [ ASP of traingle]

40° + 80° + ∠BPA = 180°

120° + ∠BPA = 180°

∠BPA = 180° - 120°

∠BPA = 60°

Similarly with other traingle

We get,

∠APD + ∠BPA + ∠BPC = 180°

40° + 60° + ∠BPC = 180°

100° + ∠BPC = 180°

∠BPC = 180° - 100°

∠BPC = 80°