InterviewSolution
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InterviewSolution
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Let a = 6, b = 3 and `cos (A -B) = (4)/(5)` Area (in sq. units) of the triangle is equal toA. 9B. 12C. 11D. 10 |
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Answer» Correct Answer - A `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)` |
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