Question

The diagonal of a parallelogram ABCD intersects at p if angle BPC equal 85 degree and Angle bdc E equal 65 degree find angle PAB

Answer 1

Answer:

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Answer 2

The angle PAB=20 degrees.

Step-by-step explanation:

By definition of parallelogram

$AB\parallel CD,BC\parallel AD$

$AB=CD,BC=AD$

Angle BPC=85 degrees

Angl BDC=65 degrees

Angle BPC+angle APB=180 degrees (linear angles)

Using linear angle property

Substitute the value

$85+\angle APB=180$

$\angle APB=180-85=95^{\circ}$

$\angle ABP=\angle BDC=65$

By using Alternate Interior angles theorem

In triangle APB

$\angle ABP+\angle APB+\angle PAB=180^{\circ}$

Using triangle angles sum property

Substitute the values then, we get

$65+95+\angle PAB=180^{\circ}$

$160+\angle PAB=180$

$\angle PAB=180-160=20^{\circ}$

Hence, the angle PAB=20 degrees.

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