Question

(i)
Fig. 10.45
The diagonals of a ||gm ABCD intersect at P. If ZBPC = 85° and CBDC = 65°, find ZPAB.​

Answer 1

The angle PAB=20 degrees.

By definition of parallelogram

AB || CD, BC || 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 + ZAPB = 180

ZAPB = 180 – 85 = 95°

ZABP = LBDC = 65

By using Alternate Interior angles theorem

In triangle APB

ZABP+ ZAPB+ ZPAB = 180°

Using triangle angles sum property

Substitute the values then, we get

65 + 95 + ZPAP = 180°

160 + ZPAB = 180

ZPAB = 180 160 20°

Hence, the angle PAB=20 degrees.

Answer 2

Answer:

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