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
PQRS is acyclic quadrilateral sum of its opposite angles is 180o
So ‹P +‹R = 4b + 20-4b=180 i.e b-a = 40
Solving we get a = 15
b= 25
4b + 20 = 4(25) + 20 = 120°
3b– 5 = 3(25) – 5 = 70°
–4a = –4(–15) = 60°
–7a + 5 = –7(–15) + 5 = 110°
Answered by | 14th Jul, 2010, 02:34: PM
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