InterviewSolution
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InterviewSolution
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The sides of a triangle are `x^2+x+1, 2x+1 and x^2-1.` Prove that the greatest angle is `120^0` |
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Answer» Let `a = x^(2) + x + 1, b = 2x + 1, and c = x^(2) - 1`. First, we have to decide which side is the greatest. We know that in a triangle, the length of each side is greater than zero. Therefore, we have `b = 2x + 1 gt 0 and c = x^(2) - 1 0`. Thus, `x gt -(1)/(2) and x^(2) gt 1` `rArr x gt - (1)/(2) and x lt -1 " or " x gt 1` `rArr x gt 1` `a = x^(2) + x + 1 = (x + (1)/(2))^(2) + ((3)/(4))` is always positive. Thus, all sides a, b, and c are positive when `x gt 1`. Now `x gt 1 " or" x^(2) gt x` or `x^(2) + x + 1 gt 2x + 1 rArr a gt b` Also, when `x gt 1` `x^(2) + x + 1 gt x^(2) - 1 rArr a gt c` Thus, `a = x^(2) + x + 1` is the greatest side and the angle A opposite to this side is the greatest angle `:. cos A = (b^(2) + c^(2) -a^(2))/(2bc)` `= ((2x + 1)^(2) + (x^(2) - 1)^(2) - (x^(2) + x + 1)^(2))/(2(2x + 1) (x^(2) -1))` `= (-2x^(3) - x^(2) + 2x + 1)/(2(2x^(3) + x^(2) - 2x -1)) = -(1)/(2) = cos 120^(@)` `rArr A = 120^(@)` |
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