# Triangle ABC

Triangle ABC has side lengths m-1, m-2, m-3. What has to be m to be triangle

a) rectangular
b) acute-angled?

Result

m(a) =  6
m(b) =  7

#### Solution:

$(m-1)^2 = (m-2)^2 +(m-3)^2; m>3 \ \\ m^2-2m+1 = m^2-4m+4+m^2-6m+9 \ \\ m^2-8m+12 = 0 \ \\ \ \\ m_{1,2} = \dfrac{ -b \pm \sqrt{ D } }{ 2a } = \dfrac{ 8 \pm \sqrt{ 16 } }{ 2 } \ \\ m_{1,2} = \dfrac{ 8 \pm 4 }{ 2 } \ \\ m_{1,2} = 4 \pm 2 \ \\ m_{1} = 6 \ \\ m_{2} = 2 \ \\ \ \\ m>3 \ \\ m(a) = 6 \ \\$

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$b) \ m>6$

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