Answer:
[tex]\theta_S=\tan ^{-1}(1.85)=61.6069\text{ Deg}[/tex]Explanation: We need to find the resultant vector S that is the sum of the two vectors:
[tex]\begin{gathered} C=C_x+C_y\Rightarrow\lvert C\rvert=\sqrt[]{(C_x)^2+(C_y)^2}=6.28m \\ \theta_c=\tan ^{-1}(\frac{C_y}{C_x})=105^{\circ} \\ \text{ Likewise} \\ D=D_x+D_y\Rightarrow\lvert D\rvert=\sqrt[]{(D_x)^2+(D_y)^2}=3.09m \\ \theta_D=\tan ^{-1}(\frac{D_y}{D_x})=233^{\circ} \\ \text{ We need to find: } \\ \theta_S=\tan ^{-1}(\frac{S_y}{S_x})\rightarrow\text{ ?} \\ S_y\text{ }\rightarrow\text{?} \\ S_z\rightarrow? \\ \end{gathered}[/tex]Next step is fairly easy, we just need to solve simultaneous equations, which will give us components of C D vectors, then finding the direction of vector sum would be the last step
[tex]\begin{gathered} C=C_x+C_y\Rightarrow\lvert C\rvert=\sqrt[]{(C_x)^2+(C_y)^2}=6.28m \\ \theta_c=\tan ^{-1}(\frac{C_y}{C_x})=105^{\circ}\rightarrow\frac{C_y}{C_x}=\tan (105^{\circ})\rightarrow C_y=C_x\tan (105^{\circ})=C_x4.03 \\ \therefore\rightarrow C_y=C_x4.03 \\ \text{ Now we can solve for the vector components from the above equation as:} \\ \sqrt[]{(C_x)^2+(C_y)^2}=6.28m\rightarrow\mleft(C_x\mright)^2+\mleft(C_y\mright)^2=(6.28m)^2 \\ (C_x)^2+(C_y)^2=(6.28m)^2\Rightarrow(C_x)^2(4.03)^2+(C_x)^2=(6.28m)^2 \\ (C_x)^2\lbrack(4.03)^2+1\rbrack=(6.28m)^2\rightarrow17.22(C_x)^2=39.44\rightarrow(C_x)^2=\frac{39.44}{17.22} \\ \therefore\rightarrow \\ C_x=\sqrt[]{\frac{39.44}{17.22}}=1.51m\rightarrow(1) \\ \therefore\rightarrow \\ (C_y)^2=(6.28m)^2-(C_x)^2=39.44m-1.51m=37.93m \\ C_y=\sqrt[]{37.93m}=6.16m\rightarrow(2) \\ \end{gathered}[/tex]Similarly for D same thing as above is done, as shown next:
[tex]\begin{gathered} D=D_x+D_y\Rightarrow\lvert D\rvert=\sqrt[]{(D_x)^2+(D_y)^2}=3.09m \\ \theta_D=\tan ^{-1}(\frac{D_y}{D_x})=233^{\circ}\rightarrow\frac{D_y}{D_x}=\tan (233^{\circ}^{})\rightarrow D_y=D_x\tan (233^{\circ})=D_x0.58 \\ \therefore\rightarrow\text{ }D_y=D_x0.5 \\ \text{Now we can solve for the vector components from the above equation as:} \\ \sqrt[]{(D_x)^2+(D_y)^2}=3.09m\rightarrow(D_x)^2+(D_y)^2=(3.09m)^2 \\ (C_x)^2+(C_y)^2=(3.09m)^2\Rightarrow(D_x)^2(0.58)^2+(D_x)^2=(3.09m)^2 \\ \therefore\rightarrow \\ (D_x)^2\lbrack(0.58)^2+1\rbrack=(3.09m)^2\rightarrow1.34(D_x)^2=9.54\rightarrow(D_x)^2=\frac{9.54}{1.34} \\ \therefore\rightarrow \\ D_x=\sqrt[]{\frac{9.54}{1.34}}=2.67m\Rightarrow(3) \\ (D_y)^2=(3.09m)^2-(D_x)^2=9.55m-7.12m=2.43m \\ D_y=\sqrt[]{2.43m}=1.56m\Rightarrow(4) \end{gathered}[/tex]Now the direction simply is:
[tex]\begin{gathered} \theta_S=\tan ^{-1}(\frac{S_y\text{ }}{S_x_{}})=\tan ^{-1}(\frac{D_y+C_y}{D_x+C_x})=\tan ^{-1}(\frac{1.56m_{}+6.16m_{}}{2.67_{}m+1.51m_{}})=\tan ^{-1}(\frac{7.72_{}}{4.18_{}})=\tan ^{-1}(1.85) \\ \theta_S=\tan ^{-1}(1.85) \end{gathered}[/tex]