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Part (1)Part (2)at Bconsidur moment equilibrium abaut point A;\begin{array}{c}\left(+\sum M_{A}=0^{\circ}\right. \\+\quad\left(R_{B} \sin 40^{\circ}\right) \times 7+\left(R_{B} \cos 40^{\circ}\right) \times 1-(125 \times 4) \\-(600 \times 9)=0 \\1.40 R_{B}=5900 \quad 4214.28 l b \\R_{B}=4\end{array}Part (3) \rightarrow \Sigma f_{x}=0^{*},A_{x}+R_{B} \cos 40^{\circ}=0^{\circ}Part (4)\begin{array}{l}+\uparrow \delta f_{y}=0 \\A_{y}+R_{B} \sin 40^{\circ}-125-600=0\end{array}Part (5)Rolsing equation (1)\begin{array}{l}A_{x}=-R_{B} \cos 40^{\circ}=-(4214.28) \cos 40^{\circ} \\(\rightarrow) A_{x}=-3228.33 \mathrm{lb} \\(\leftarrow) A_{x}=3228.33 \mathrm{lb} \\\end{array}Raluing equation (11)total reaction at A.\begin{aligned}R_{A} & =\sqrt{A_{x^{2}}+A_{y}^{2}}- \\& =\sqrt{(3228.33)^{2}+(1983.88)^{2}}\end{aligned}eaction at R_{A}=3789.18 \mathrm{lb} Ansreaction at B ;, R_{B}=4214.28 \mathrm{lb} Ans ...