Question

In the given figure (ii), if angle A= 50° and angle DCP = 20°, then find angle BDC and angle DPC?
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Answer 1

$\huge \fbox \purple{❥ Answer}$

Step-by-step explanation:

Angle A+ Angle DCB =180° (opposite side of a cyclic quadrilateral is supplementary)

Therefore angle DCB=120°

Similarly angle ADC = 100°

AP is a straight line angle ADC + PDC=180°(linear pair)

Therefore angle CDA = 80°

Similarly angle PCD=60°

Angle CDA+PCD+DPC=180°(angle sum property of a triangle

Therefore angle DPC=40°

Angle DCB = 120° previously found

Angle ABC =80° given

As DQ is a straight line DCB+BCQ=180°(linear pair)

Therefore angle BCQ=60°

Similarly angle CBQ =100°

In triangle CBQ

angle BCQ+CBQ+BQC=180°(angle sum property)

Therefore angle angle BCQ=20°