Question

sides ab and DC of a cyclic quadrilateral are produced to meet at a point P on the sides AD and BC are produced to meet at a point Q angle ADC is equal to 75 degree and Angle BPC is equals to 30 degree
calculate
1. Angle B A D using property exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.
2. angle CQD using angle sum property of a triangle.
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Answer 1

SOLUTION:

□ABCD is cyclic quadrilateral.

∠ADC ≅∠CBP (Exterior angle property of cyclic quadrilateral).

∠CBP= 75° .........(i)

In △CBP,

∠CBP=75° [ From (i) ]

∠CBP=50° [Given]

∠BCP=55° [ Remaining angle]

∠BCP≅∠BAD (Exterior angle property of cyclic quadrilateral).

∠BAD=55°.

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Answer 2

Side AB abd DC of a cyclic quadrilateral are produced to meet at point P on the sides AD and BC are produced to meet at a point Q angle ADC =75° and angle BPC =50°.

1)Angle BAD=55° using property exterior angle of a cyclic quadrilateral is equal to its interior opposite angle.

2) Angle CQD=20° using angle sum property of a triangle.

Stepwise explanation is given below:

- ABCD is cyclic quadrilateral.

∠ADC =∠CBP (Exterior angle property of cyclic quadrilateral).

∠CBP= 75° .........(i)

- In △CBP,

∠CBP=75° [ From (i) ]

∠BPC=50° [Given]

∠BCP=55° [ angle sum property ]

∠BCP=∠BAD (Exterior angle property of cyclic quadrilateral is equal to its interior opposite angle.).

∠BAD=55°.......(2)

- We know that ABP is a straight line. So,

∠ABQ+∠QBP=180°

∠ABQ+75°=180°

∠ABQ=180°-75°=105°......(3)

- In ΔABQ

∠ABQ+∠BQA+∠QAB=180...[ by angle sum property of a triangle ].

105+∠BQA+55=180° ......[from (2) and (3)].

∠BQA=180-160=20°