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if in triangle ABC a=6,b=3 and cos(A-B)=4/5 then its area is

HEMANTH , 7 Years ago
Grade 9
anser 1 Answers
Arun

Last Activity: 7 Years ago

Dear Hemant
 
cos (A-B) = (1 -tan²(A-B)/2)/(1+tan²(A-B)/2) = 4/5
On cross multiplication and then solving
 
tan²(A-B)/2 = 1/9
tan(A-B)/2 = 1/3
 
 
Hence
(a-b) cot(C/2)/(a+b) = 1/3
On solving
 
Cot(c/2) = 1
C/2 = 45°
C = 90°
Hence it is a right angle triangle
 
 
Now area = (1/2)*3*6 = 9 square units
 
 
Regards
Arun (askIITians forum expert)

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