Solution
\begin{aligned}\\ &\sum{x}=0 & M_A-4.698=0 && M_A=4.698\\ &\sum{M_A}=0 & 3\cdot 4\cdot 2+1.71\cdot 6-4.698\cdot 2\sqrt3-V_B\cdot 8=0 && V_B=2.248\\ &\sum{y}=0 & V_A-3\cdot 4-1.71+2.248=0\\ \\ \\ &Interval I\ \ \ \varphi \in (0;60)\\ \end{aligned}
\begin{aligned}\\ &N(\varphi)=-2.248 cos\varphi+1.71 cos\varphi-4.698 sin\varphi=-0.538 cos\varphi-sin\varphi-4.698 sin\varphi\\ &T(\varphi)=-2.248\cdot sin\varphi+1.71\cdot sin\varphi+4.698 cos\varphi=-0.583 sin\varphi+4.698 cos\varphi\\ &M(\varphi)=2.248\cdot (4-x)-1.71\cdot (2-x)-4.698(y-2\sqrt3)=\\ &=8.992-8.992\cdot cos\varphi-3.42+6.84\cdot cos\varphi-18.792 sin\varphi+16.274=\\ &=21.848-2.152 cos\varphi-18.792 sin\varphi\\ \end{aligned}
\begin{aligned}\\ &Interval III \ \ \ \varphi \in (0;90)\ \ \ Change-coordinate-system\\ \end{aligned}
\begin{aligned}\\ &N(\varphi)=-11.462\cdot cos\varphi -4.698\cdot sin\varphi+3\cdot (4-x) cos\varphi=\\ &=-11.462\cdot cos\varphi-4.698\cdot sin\varphi+12\cdot cos\varphi-12cos^2 \varphi=\\ &=0.538\cdot cos\varphi-4.698\cdot sin\varphi-12\cdot cos^2 \varphi\\ \\ &T(\varphi)=11.462\cdot sin\varphi-4.698\cdot cos\varphi-3\cdot (4-x)\cdot sin\varphi=\\ &=11.462\cdot sin\varphi-4.698\cdot cos\varphi-12\cdot sin\varphi+12\cdot sin\varphi cos\varphi=\\ &=-0.538\cdot sin\varphi-4.698\cdot cos\varphi+12\cdot sin\varphi cos\varphi\\ &M(\varphi)=11.462\cdot (4-x)-4.698y-3\cdot(4-x\cdot \frac{1}{2}(4-x)=\\ &=45.848-45.848\cdot cos\varphi-18.792\cdot sin\varphi-1.5(16-8x+x^2)=\\ &=45.848-45.848\cdot cos\varphi-18.792\cdot sin\varphi-24+48\cdot cos\varphi-24cos^2 \varphi=\\ &21.848+2.152\cdot cos\varphi-18.792\cdot sin\varphi-24\cdot cos^2 \varphi\\ \end{aligned}
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