Let a = 6, b = 3 and `cos (A -B) = (4)/(5)`
Area (in sq. units) of the triangle is equal to

Options

A. 9
B. 12
C. 11
D. 10

Answer

A

Explanation

`cos(A -B) = (4)/(5)`
`rArr (1 - tan^(2).(A-B)/(2))/(1 + tan^(2).(A-B)/(2)) = (4)/(5)`
or `"tan"^(2) (A -B)/(2) = (1)/(9)`
or `"tan" (A -B)/(2) = (1)/(3)`
Now, `tan.(A-B)/(2) = (a-b)/(a+b) "cot"(C)/(2)`
or `(1)/(3) = (6-3)/(6+3) "cot"(C)/(2)`
or `"cot"(C)/(2) = 1 " or " C = (pi)/(2)`
Area of triangle `= (1)/(2) ab sin C = (1)/(2) xx 6 xx 3 xx 1 = 9`
`(a)/(sinA) = (sqrt(a^(2) + b^(2)))/(1)`
or `(6)/(sinA) = sqrt45`
or `sin A = (2)/(sqrt5)`