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sir i have a question: If in triangle ABC, (tanA-tanB)/(tanA+tanB) = (c-b)/c, prove that A= 60 degree.

Asked by 12th August 2012, 2:51 PM
Answered by Expert
Answer:

(tanA-tanB?/(tanA+tanB) = (c-b)/c

sin(A-B)/sin(A+B)=(c-b)/c

sin(A-B)/sin(A+B) =(sinC-sinB)/sinC

sin(A-B)/sinC =(sinC-sinB)/sinC   (Since, sin(A+B)=sinC)

sin(A-B)=sinC-sinB

sinB = sin(A+B)-sin(A-B) = 2sinBcosA

2cosA=1

cosA=1/2

Thus, angle A = 60 degrees

Answered by Expert 12th August 2012, 6:48 PM
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