Angle in RT

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

Correct answer:

x =  -90 °

Step-by-step explanation:

$$\begin{gathered} a^2+b^2=c^2 \\ {a}^2+(a+d{)}^2=(a+2d{)}^2 \\ a=1.23 \\ 1.23^2+(1.23+d{)}^2=(1.23+2\cdot d{)}^2 \\ 1.23^2+(1.23+d{)}^2=(1.23+2\cdot d{)}^2 \\ -3d^2-2.46d+1.513=0 \\ 3d^2+2.46d-1.513=0 \\ a=3;b=2.46;c=-1.513 \\ D=b^2-4ac=2.46^2-4\cdot 3\cdot (-1.513)=24.2064 \\ D>0 \\ {d}_{1,2}=\frac{-b\pm \sqrt{D}}{2a}=\frac{-2.46\pm \sqrt{24.21}}{6} \\ {d}_{1,2}=-0.41\pm 0.82 \\ {d}_1=0.41 \\ {d}_2=-1.23 \\ d>0 \\ d=d_2=(-1.23)=-\frac{123}{100}=-1.23 \\ a=1.23 \\ b=a+d=1.23+(-1.23)=0 \\ c=a+2\cdot d=1.23+2\cdot (-1.23)=-\frac{123}{100}=-1.23 \\ x=\frac{180°}{\pi }\cdot \arcsin (a/c)=\frac{180°}{\pi }\cdot \arcsin (1.23/(-1.23))=-90^{\circ } \end{gathered}$$

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