Angle in RT

Determine the size of the smallest internal angle of a right triangle whose sides constitutes sizes consecutive members of arithmetic progressions.

Result

x =  -90 °

Solution:

$a^2 + b^2=c^2 \ \\ a^2 + (a+d)^2=(a+2d)^2 \ \\ a=1.23 \ \\ 1.23^2 + (1.23+d)^2=(1.23+2*d)^2 \ \\ \ \\ 1.23^2 + (1.23+d)^2=(1.23+2 \cdot \ d)^2 \ \\ -3d^2 -2.46d +1.513=0 \ \\ 3d^2 +2.46d -1.513=0 \ \\ \ \\ a=3; b=2.46; c=-1.513 \ \\ D=b^2 - 4ac=2.46^2 - 4\cdot 3 \cdot (-1.513)=24.2064 \ \\ D>0 \ \\ \ \\ d_{1,2}=\dfrac{ -b \pm \sqrt{ D } }{ 2a }=\dfrac{ -2.46 \pm \sqrt{ 24.21 } }{ 6 } \ \\ d_{1,2}=-0.41 \pm 0.82 \ \\ d_{1}=0.41 \ \\ d_{2}=-1.23 \ \\ \ \\ \text{ Factored form of the equation: } \ \\ 3 (d -0.41) (d +1.23)=0 \ \\ d>0 \ \\ d=d_{2}=(-1.23)=- \dfrac{ 123 }{ 100 }=-1.23 \ \\ a=1.23 \ \\ b=a+d=1.23+(-1.23)=0 \ \\ c=a+2 \cdot \ d=1.23+2 \cdot \ (-1.23)=- \dfrac{ 123 }{ 100 }=-1.23 \ \\ x=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(a/c)=\dfrac{ 180^\circ }{ \pi } \cdot \arcsin(1.23/(-1.23))=-90=-90 ^\circ$

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