Question 399602
One angle of a parallelogram is 120 degrees,and two consecutive sides have lengths of 8 inches and 15 inches. What is the area of the parallelogram? 
<pre><font face = "batangche" color = "indigo" size = 4><b>

{{{drawing(400,1200/7,-1,20,-1,8,line(0,0,15,0),line(15,0,19,4sqrt(3)),
line(19,4sqrt(3),4,4sqrt(3)),line(4,4sqrt(3),0,0), locate(4,6.6,"120°"),
red(arc(4,4sqrt(3),4,-4,240,360)), locate(1.2,4,8), locate(7,0,15)


   )}}}

Since two adjacent angles of a parallelogram are supplementary, we find
that the lower left angle is 180°-120° = 60°:

{{{drawing(400,1200/7,-1,20,-1,8,line(0,0,15,0),line(15,0,19,4sqrt(3)),
line(19,4sqrt(3),4,4sqrt(3)),line(4,4sqrt(3),0,0), locate(4,6.6,"120°"),
red(arc(4,4sqrt(3),4,-4,240,360)), locate(1.2,4,8), locate(7,0,15),
locate(.9,1.2,"60°"), red(arc(0,0,5,-5,0,60))


   )}}}

Area = base × height

We have the base, which is 15 inches, but we need the height, the
ACTUAL height, not the length of the slanted sides.  So we draw a 
perpendicular from a vertex to the opposite side.  I'll draw it
in green and label it h:

{{{drawing(400,1200/7,-1,20,-1,8,line(0,0,15,0),line(15,0,19,4sqrt(3)),
line(19,4sqrt(3),4,4sqrt(3)),line(4,4sqrt(3),0,0), green(line(4,4sqrt(3),4,0)), locate(1.2,4,8), locate(7,0,15), locate(4.2,3.4,h),
locate(.9,1.2,"60°"), red(arc(0,0,5,-5,0,60)), locate(11,4sqrt(3),15)

   )}}}

The right triangle on the left has a 60° angle, therefore it is a
30°60°90° right triangle, and therefore its shorter side (the bottom
side) is {{{1/2}}} the length of the hypotenuse, and {{{1/2}}} of 8 is
4. so the bottom side of the quadrilateral has been split into
two parts, 4 inches and 11 inches:

{{{drawing(400,1200/7,-1,20,-1,8,line(0,0,15,0),line(15,0,19,4sqrt(3)),
line(19,4sqrt(3),4,4sqrt(3)),line(4,4sqrt(3),0,0), green(line(4,4sqrt(3),4,0)), locate(1.2,4,8), locate(9,0,11), locate(2,0,4), locate(4.2,3.4,h),
locate(.9,1.2,"60°"), red(arc(0,0,5,-5,0,60)),locate(11,4sqrt(3),15)
   )}}}

We now use the Pythagorean theorem to calculate the length of the
green altitude:

{{{a^2+b^2=c^2}}}
{{{4^2+h^2=8^2}}}
{{{16+h^2=64}}}
{{{h^2=48}}}
{{{h=sqrt(48)}}}
{{{h=sqrt(16*3)}}}
{{{h=sqrt(16)*sqrt(3)}}}
{{{h=4sqrt(3)}}}

{{{drawing(400,1200/7,-1,20,-1,8,line(0,0,15,0),line(15,0,19,4sqrt(3)),
line(19,4sqrt(3),4,4sqrt(3)),line(4,4sqrt(3),0,0), green(line(4,4sqrt(3),4,0)), locate(1.2,4,8), locate(9,0,11), locate(2,0,4), locate(4.2,3.4,h=4sqrt(3)),
locate(.9,1.2,"60°"), red(arc(0,0,5,-5,0,60)), locate(11,4sqrt(3),15)

   )}}}

                                              __
Now we have the height of the parallelogram 4&#8730;3, and we know the base of
the parallelogram is 15.

Since 

Area = base × height,
               __
Area = 15 × 4&#8730;3
          __
Area = 60&#8730;3 square inches

Edwin</pre>