Lever $AB$ is $2m$ long and initially horizontal as shown in the Figure below. Determine the angle $$ that lever $CD$ must rotate to cause lever $AB$ to rotate $30$ upward.

A$.$

$=si{n}^{-1}(\frac{2sin50}{1\mathrm{.}8})-\frac{2sin80}{\sqrt[\phantom{2}]{2^2-A{C}^2-4ACcos80}}$

where

$AC=\frac{1\mathrm{.}8}{sin50}sin[130-si{n}^{-1}(\frac{2sin50}{1\mathrm{.}8})]$

B$.$

$=si{n}^{-1}(\frac{2sin50}{1\mathrm{.}8})-\frac{2sin80}{\sqrt[\phantom{2}]{2^2-A{C}^2-4ACcos80}}$

where

$AC=\frac{4cos50+-\sqrt[\phantom{2}]{(4cos50{)}^2-4(2^2-{1\mathrm{.}8}^2)}}{2}$

C$.$

$=si{n}^{-1}(\frac{2sin80}{1\mathrm{.}8})-\frac{2sin80}{\sqrt[\phantom{2}]{2^2-A{C}^2-4ACcos80}}$

where

$AC=\frac{4+-\sqrt[\phantom{2}]{(4cos50{)}^2-4(2^2-{1\mathrm{.}8}^2)}}{2}$

D$.$

$=si{n}^{-1}(\frac{2sin80}{1\mathrm{.}8})-\frac{2sin50}{\sqrt[\phantom{2}]{2^2-A{C}^2-4ACcos80}}$

where

$AC=\frac{1\mathrm{.}8}{sin80}sin[130-si{n}^{-1}(\frac{2sin50}{1\mathrm{.}8})]$

E$.$

$=si{n}^{-1}(\frac{2sin80}{1\mathrm{.}8})$