Question 83147
 Two forces of 40 pounds and 20 pounds, respectively, act simultaneously on an object. The angle between the two forces is 40(degrees). Find the magnitude of the resultant, to the nearest tenth of a pound. Find the measure of the angle, between the resultant and the larger force.

<pre><font size = 3><b>
{{{drawing(400,142.935,-2,60,-2,20, arrow(0,0,15.301,12.856), locate(2.5,2.5,"40°"),
   
  arrow(0,0,40,0), locate(22,-1,"40#"), locate(2,7,"20#") 

)}}} 

Complete the parallelogram by drawing a line parallel to each vector
through the tip of the OTHER vector.

{{{drawing(400,142.935,-2,60,-2,20, arrow(0,0,15.301,12.856), locate(2.5,2.5,"40°"),
   
  arrow(0,0,40,0), locate(42,1,"40#"), locate(15,16.2,"20#") 
  line(40,0,55.321,12.856), line(15.321,12.356,55.321,12.856)
)}}} 

Since two adjacent angles of a parallelogram are supplementary, the
angle at the lower right corner of the parallelogram is 180°-40² or 140°

{{{drawing(400,142.935,-2,60,-2,20, arrow(0,0,15.301,12.856), locate(2.5,2.5,"40°"),
   
  arrow(0,0,40,0), locate(42,1,"40#"), locate(15,16.2,"20#") 
  line(40,0,55.321,12.856), line(15.321,12.356,55.321,12.856),
  locate(37,3,"140°")
)}}}

Now we draw in the resultant vector R

{{{drawing(400,142.935,-2,60,-2,20, arrow(0,0,15.301,12.856), locate(2.5,2.5,"40°"), locate(24,8,R),
   
  arrow(0,0,40,0), locate(42,1,"40#"), locate(15,16.2,"20#") 
  line(40,0,55.321,12.856), line(15.321,12.356,55.321,12.856),
  locate(37,3,"140°"), arrow(0,0,55.321,12.856)
)}}}

Isolate the lower triangle. (You could isolate the top one
just as well).  Label it triangle ABC

{{{drawing(400,142.935,-2,60,-2,20,
   
  triangle(0,0,40,0,55.321,12.356), locate(24,8,b),

  locate(-.5,-1,A), locate(41,-.5,B), locate(57,13,C),
  locate(48,6,"a=20"), locate(37,3,"140°"), locate(20,-.6,"c=40")  

  
)}}}

Solve for b (which is the magnitude of vector R, using the law 
of cosines:

b² = a² + c² - 2ac·cos(B)

b² = 20² + 40² - 2(20)(40)cos(140°)

b² = 400 + 1600 - 1600)(-.7660444431)

b² = 3225.671109

Use the principle of square roots:

b = 56.79499194

Since b is the magnitude of vector R, then
the resultant force R to the nearest tenth
of a pound is 56.8 lb.

Now we only need to find angle A for that
is the angle between the larger force and
the resultant.

The law of sines tells us that

{{{a/(sin(A)) = b/(sin(B))}}}

cross-multiplying:

{{{b*sin(A) = a*sin(B)}}}  

{{{sin(A) = (a*sin(B))/b}}} 

{{{sin(A) = 20*sin(140)/56.79499194}}}

{{{sin(A) = .226353623}}}

Use the inverse sine

       A = 13.08248883°

To the nearest degree, this is 13°

Edwin</pre>