Question

In triangle ABC b=3 and c= 1 and cos(B-C)=3/5 then it’s area is

Answer 1

Given : In triangle ABC b=3 and c= 1 and cos(B-C)=3/5

To find : area of triangle

Solution:

Cos ( B - C)  =  (1  -  Tan²( (B - C)/2) ) /  (1  +  Tan²( (B - C)/2) )   = 3/5

=> 5 - 5  Tan²( (B - C)/2)  =  3 + 3 Tan²( (B - C)/2)

=> 8 Tan²( (B - C)/2) = 2

=>  Tan²( (B - C)/2)  = 1/4

=>  Tan ( (B - C)/2)  = 1/2

Tan ( (B - C)/2)  = ( ( b - c)/(b + c) ) Cot A/2

=> 1/2 = ((3 - 1)/(3 + 1) )Cot A/2

=> Cot A/2  = 1

=> Tan (A/2) = 1

=> A/2 = 45°

=> A = 90°

Hence area of triangle =  (1/2) 3 ( 1)   = 3/2

area of triangle =  3/2 sq units

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