Member $AB$ rotates $150$ clockwise from the original position shown in the Figure below. Determine the length $C{B}^{\text{'}}.($Hint: Use length $AC$ as a common length to the two triangles formed on either side of it$.)$

A$.C{B}^{\text{'}}=\sqrt[\phantom{2}]{A{C}^2+{0\mathrm{.}2}^2-2(AC)(0\mathrm{.}2)c\frac{os}{{?}_{C}A{B}^{\text{'}}}}m$

where, $\frac{?}{{?}_{C}A{B}^{\text{'}}}=180-si{n}^{-1}(\frac{0\mathrm{.}5s\frac{in}{{?}_{C}BA}}{AC})$

and $AC=\sqrt[\phantom{2}]{{0\mathrm{.}5}^2+{0\mathrm{.}2}^2-2(0\mathrm{.}5)(0\mathrm{.}2)c\frac{os}{{?}_{C}BA}}m$

with $\frac{?}{{?}_{C}BA}=6$

B$.C{B}^{\text{'}}=\sqrt[\phantom{2}]{A{C}^2+{0\mathrm{.}2}^2-2(AC)(0\mathrm{.}2)c\frac{os}{{?}_{C}A{B}^{\text{'}}}}m$

where, $\frac{?}{{?}_{C}A{B}^{\text{'}}}=si{n}^{-1}(\frac{0\mathrm{.}5s\frac{in}{{?}_{C}BA}}{AC})$

and $AC=\sqrt[\phantom{2}]{{0\mathrm{.}5}^2+{0\mathrm{.}2}^2-2(0\mathrm{.}5)(0\mathrm{.}2)c\frac{os}{{?}_{C}BA}}m$

with $\frac{?}{{?}_{C}BA}=6$

C$.C{B}^{\text{'}}=\sqrt[\phantom{2}]{A{C}^2+{0\mathrm{.}2}^2-2(AC)(0\mathrm{.}2)c\frac{os}{{?}_{C}A{B}^{\text{'}}}}m$

where, $\frac{?}{{?}_{C}A{B}^{\text{'}}}=si{n}^{-1}(\frac{0\mathrm{.}5s\frac{in}{{?}_{C}BA}}{AC})$

and $AC=\sqrt[\phantom{2}]{{0\mathrm{.}5}^2+{0\mathrm{.}2}^2-2(0\mathrm{.}5)(0\mathrm{.}2)c\frac{os}{{?}_{C}BA}}m$

with $\frac{?}{{?}_{C}BA}=20$

D$.C{B}^{\text{'}}=\sqrt[\phantom{2}]{A{C}^2+{0\mathrm{.}5}^2-2(AC)(0\mathrm{.}5)c\frac{os}{{?}_{C}A{B}^{\text{'}}}}m$

where, $\frac{?}{{?}_{C}A{B}^{\text{'}}}=180+si{n}^{-1}(\frac{0\mathrm{.}5s\frac{in}{{?}_{C}BA}}{AC})$

and $AC=\sqrt[\phantom{2}]{{0\mathrm{.}5}^2+{0\mathrm{.}2}^2-2(0\mathrm{.}5)(0\mathrm{.}2)c\frac{os}{{?}_{C}BA}}m$

with $\frac{?}{{?}_{C}BA}=26$

E$.C{B}^{\text{'}}=\sqrt[\phantom{2}]{A{C}^2+{0\mathrm{.}2}^2-2(AC)(0\mathrm{.}2)cos150}m$

where, $AC=\sqrt[\phantom{2}]{{0\mathrm{.}5}^2+{0\mathrm{.}2}^2-2(0\mathrm{.}5)(0\mathrm{.}2)cos26}m$