the diagonal of a parallelogram ABCD intersect at p if angle BPC is equal to 85 degree and angle BDC us equal to 65 degree find the angle PAB
Answers
The angle PAB=20 degrees.
Step-by-step explanation:
By definition of parallelogram
AB\parallel CD,BC\parallel ADAB∥CD,BC∥AD
AB=CD,BC=ADAB=CD,BC=AD
Angle BPC=85 degrees
Angl BDC=65 degrees
Angle BPC+angle APB=180 degrees (linear angles)
Using linear angle property
Substitute the value
85+\angle APB=18085+∠APB=180
\angle APB=180-85=95^{\circ}∠APB=180−85=95
∘
\angle ABP=\angle BDC=65∠ABP=∠BDC=65
By using Alternate Interior angles theorem
In triangle APB
\angle ABP+\angle APB+\angle PAB=180^{\circ}∠ABP+∠APB+∠PAB=180
∘
Using triangle angles sum property
Substitute the values then, we get
65+95+\angle PAB=180^{\circ}65+95+∠PAB=180
∘
160+\angle PAB=180160+∠PAB=180
\angle PAB=180-160=20^{\circ}∠PAB=180−160=20
∘
Hence, the angle PAB=20 degrees.
Step-by-step explanation:
The angle PAB=20 degrees.
Step-by-step explanation:
By definition of parallelogram
Angle BPC=85 degrees
Angl BDC=65 degrees
Angle BPC+angle APB=180 degrees (linear angles)
Using linear angle property
Substitute the value
By using Alternate Interior angles theorem
In triangle APB
Using triangle angles sum property
Substitute the values then, we get
Hence, the angle PAB=20 degrees.