Question 639217
<pre>
{{{drawing(400,350,-1,3,-1,2.5,
locate(-.8,.2,"60°"), locate(1.6,.2,"40°"),locate(.05,.81,12),
triangle(-1,0,2.064177772,0,0,sqrt(3)), green(line(0,0,0,sqrt(3))) )}}}

Since the formula for the area of a triangle is

            A = {{{1/2}}}b·h

and we have the height = h = 12, we must find the base. Let the left 
part of the base be x and the right part of the base be y.

{{{drawing(400,350,-1,3,-1,2.5,locate(-.5,0,x), locate(1,0,y),
locate(-.8,.2,"60°"), locate(1.6,.2,"40°"),locate(.05,.81,12),
triangle(-1,0,2.064177772,0,0,sqrt(3)), green(line(0,0,0,sqrt(3))) )}}}

To find x, we use {{{TANGENT=OPPOSITE/ADJACENT}}}

tan(60°) = {{{12/x}}}

Multiply both sides by x:

x·tan(60°) = 12

Divide both sides by tan(60°)

         x = {{{12/tan("60°")}}}
         x = 6.9282


To find y, we also use {{{TANGENT=OPPOSITE/ADJACENT}}}

tan(40°) = {{{12/y}}}

Multiply both sides by y:

y·tan(40°) = 12

Divide both sides by tan(40°)

         y = {{{12/tan("40°")}}}
         y = 14.3010

the base = b = x + y = 6.9282 + 14.3010 = 21.2292

            A = {{{1/2}}}b·h
            A = {{{1/2}}}(21.2292)(12)
            A = 127.3752  

Edwin</pre>