InterviewSolution
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InterviewSolution
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If the angles `alpha, beta, gamma` of a triangle satisfy the relation, `sin((alpha-beta)/(2)) + sin ((alpha -gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)`, then Triangle isA. acute angledB. right angled but not isoscelesC. isoscelesD. isosceles right angled |
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Answer» Correct Answer - C We have `sin((alpha - beta)/(2)) + sin ((alpha - gamma)/(2)) + sin((3alpha)/(2)) = (3)/(2)` `therefore sin (((pi-(beta+gamma))-beta)/(2))+ sin (((pi-(gamma +beta))-gamma)/(2)) + sin((3pi-3(beta+gamma))/(2)) = (3)/(2)` `therefore cos((2beta + gamma)/(2)) + cos((2gamma + beta)/(2)) - cos((3(beta+gamma))/(2)) = (3)/(2)` `therefore 2cos ((3)/(4)(beta + gamma)) cos ((beta-gamma)/(4)) +1 - 2cos^(2)((3)/(4)(beta +gamma)) = (3)/(2)` `therefore 4 cos^(2) ((3)/(4) (beta + gamma)) - 4cos((3)/(4) (beta + gamma)) cos ((beta-gamma)/(4)) +1 =0`...(1) Above equation is quadratic in `cos((3)/(4)(beta +gamma))` Since `cos((3)/(4)(beta+gamma))` is a real number, Discriminant `D ge 0` `therefore 16 cos^(2)((beta-gamma)/(4)) -16 ge 0` `rArr cos ^(2)((beta -gamma)/(4)) ge 1` `rArr cos^(2)((beta -gamma)/(4)) =1 ` `rArr beta = gamma ` From equation (1) for `beta = gamma`, we get `[ 2 cos((3)/(4)(beta+gamma)) -1]^(2) =0` `rArr 2cos""(3)/(4) (beta + gamma) = 1` `rArr cos""(3)/(4) (beta +gamma) = (1)/(2)` `rArr (3)/(4) (beta +gamma) = 60^(@)` `therefore beta +gamma = 80^(@)` `therefore alpha = 100^(@)` `therefore beta = 40^(@), gamma = 40^(@)` |
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