Question 934849
Let

{{{b}}} = base of the triangle
{{{h}}} = altitude of the triangle

Area of a triangle {{{A= (1/2)bh =40}}}

Therefore, {{{b = 2*40/h}}} or

{{{b = 80/h}}} ........eq. 1

Angle {{{C = 180 - 30 - 50 = 100}}}

Using the Law of Sines,

{{{sin (100)/(AB) = sin (50)/b}}}

where

{{{AB}}} = length of the side included between the angles {{{A = 30}}} and {{{B =50}}}

Solving for "{{{b}}}",

{{{b = (AB)(sin (50)/sin (100))}}} ............eq. 2

Since eq.1 = eq.2, then

{{{80/h = (AB)(sin (50)/sin (100))}}}

Solving for {{{AB}}},

{{{AB = 80(sin (100))/(h*sin (50)) .......eq.33

Look at your triangle set-up and note that

{{{sin (30) = h/AB}}} OR {{{h = AB(sin (30)) = AB/2}}}

since {{{h = AB/2}}}, substitute this in eq.3,

{{{AB = 80(sin (100))/((AB/2)*sin(50))}}}

{{{AB = 2*80(sin (100))/(AB*sin(50))}}}



Simplifying the above,

{{{(AB)^2 = 160(sin (100)/sin (50))}}}

{{{(AB)^2 = 160(0.9848/0.7660)}}}

{{{(AB)^2 = 205.70}}}

{{{AB = 14.34cm}}}