InterviewSolution
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InterviewSolution
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There exists triangle ABC satisfyingA. `tan A + tan B + tan C = 0`B. `(sin A)/(2) = (sin B)/(3) = (sin C)/(7)`C. `(a + b)^(2) = c^(2) + ab and sqrt2 (sinA + cos A) = sqrt3`D. `sin A + sin B = (sqrt3 + 1)/(2), cos A cos B = (sqrt3)/(4) = sin A sin B` |
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Answer» Correct Answer - C::D (1) `tan A + tan B + tan C = tan A tan B tan C = 0` Therefore, either A or B or C = 0. This is not possible in a triangle (2) By sine rule, `(a)/(2) = (b)/(3) = (c)/(7) = lamda`(say) `a + b = 5 lamda, c = 7 lamda` i.e., `a + b lt c` This is not possible in a triangle, as the sum of two sides is greater than the third. (3) Given that `(a + b)^(2) = c^(2) + ab` or `a^(2) + b^(2) - c^(2) + ab = 0` or `2ab cos C + ab = 0` or `cos C = -(1)/(2) or angleC = 120^(@)` Also `sqrt2 (sin A + cos A) = sqrt3`(given) or `sin (A + 45^(@)) = (sqrt3)/(2)` or `A = 45^(@) = 60^(@) " " (A + 45^(@) = 120^(@) " out as " angle C )` or `A = 15^(@) and "hence " B = 45^(@)` `:. Delta` is possible (4) Given that `cos a cos B = (sqrt3)/(4) = sin A sin B` `rArr cos (A -B) = (sqrt3)/(2)` or `A - B = (pi)/(6)` and `cos (A +B) = 0` or `A +B = (pi)/(2)` `rArr A = 60^(@), B = 30^(@)` Hence, `sin A + sin B = (sqrt3+1)/(2)`. This is possible in a triangle |
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