cyclic quadrilateral

Asked by fathima12 | 10th Jun, 2010, 12:57: PM

Expert Answer:

Dear Student
Following is the solution of your question
PQRS is  acyclic quadrilateral sum of its opposite angles is 180o
Angle P must be 4b+20
So ‹P +‹R = 4b + 20-4b=180 i.e b-a = 40
‹Q + ‹ S = 3b -5  -7a + 5=180
3b-7a = 180
Solving we get a = 15
b= 25
Angles will be

 4b + 20 = 4(25) + 20 = 120°

 3b– 5 = 3(25) – 5 = 70°

 –4a = –4(–15) = 60°

 –7a + 5 = –7(–15) + 5 = 110°

Regards
Team Topperlearning

Answered by  | 14th Jul, 2010, 02:34: PM

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