Question

find all the angles of the parallelogram ABCD given in the following figure. if angle ADB =3x, angle BCD =2x and angle ABD =4x ​

Answer 1

Answer:

Refer Figure

Given :

∠ABD=2x

∠ADB=3x

∠APB=4x

∠CBQ=7x

Solution:

∠ABD=∠ACD=2x (Angles subtended by a chord in the same segment)

∠ADB=∠ACB=3x (Angles subtended by a chord in the same segment)

∠APB=∠DPC=4x (Opposite angles)

In straight angle, ABQ,

∠PBC=180°−∠ABD−∠QBC

∠PBC=180°−2x−7x=180°−9x

In △CPD

∠CDP=180°−∠CPD−∠ACD

∠CDP=180°−4x−2x=180°−6x

In quad ABCD

∠ABC+∠ADC=180° (sum of opposite angles of a cycli quadrilateral)

2x+180°−9x+3x+180°−6x=180°

10x=180°

x=18°

∠BCD=∠BCA+∠DCA=3x+2x=5x=5×18°=90°

∠BCD=90°

Answer 2

Answer:

This is wrong the question is different and your answer is differnet, lol