<PRE><B>Question 979889</B>


Resultant is the sum of given vectors, &amp;nbsp;correct? 

The sum of vectors is the diagonal of a parallelogram built on these vectors. &amp;nbsp;Do you know it? &amp;nbsp;See the &amp;nbsp;&lt;B&gt;Figure&lt;/B&gt;&amp;nbsp; below.

&lt;Table&gt;
  &lt;TR&gt;
  &lt;TD&gt;
&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;{{{drawing( 242, 200, -6.0, 6.1, -4.0, 5.0,
            line(-5.0, -3.0, -2.0, 1.0),
            locate (-5.2,  -3.0, A), 
            locate (-2.4,   1.7, B),
            line( -2.0, 1.0, -2.5, 0.7),
            line( -2.0, 1.0, -2.1, 0.5),


            line(-2.0, 1.0, 4.0, 2.0),
            locate (4.1,   2.5, C), 


        red(line(-5.0, -3.0, 4.0, 2.0)),
            line( 4.0, 2.0,  3.5, 1.57), 
            line( 4.0, 2.0,  3.3, 1.80), 

            line(-5.0, -3.0, 1.0, -2.0),
            locate (1.1,  -1.9, D),
            line( 1.0, -2.0,  0.5, -1.85),
            line( 1.0, -2.0,  0.5, -2.25),

            line(1.0, -2.0, 4.0, 2.0)
)}}}

&amp;nbsp;&amp;nbsp;&lt;B&gt;Figure&lt;/B&gt;. &amp;nbsp;The sum of the vectors&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;&amp;nbsp;
&lt;B&gt;AB&lt;/B&gt; + &lt;B&gt;AD&lt;/B&gt; = &lt;B&gt;AC&lt;/B&gt; &amp;nbsp;(the &amp;nbsp;&lt;B&gt;parallelogram rule&lt;/B&gt;)
  &lt;/TD&gt;
  &lt;/TR&gt;
&lt;/Table&gt; 

So, &amp;nbsp;we are given the vectors &amp;nbsp;&lt;B&gt;AB&lt;/B&gt;&amp;nbsp; and &amp;nbsp;&lt;B&gt;AD&lt;/B&gt;&amp;nbsp; of the length &amp;nbsp;3&amp;nbsp; and &amp;nbsp;4&amp;nbsp; and the angle &amp;nbsp;&lt;I&gt;L&lt;/I&gt;&lt;B&gt;A&lt;/B&gt; = 80°&amp;nbsp; between them. 


The magnitude of the resultant is the length of the vector of sum, &amp;nbsp;which is the vector &amp;nbsp;&lt;B&gt;AC&lt;/B&gt;.


Apply the &amp;nbsp;&lt;B&gt;Law of Cosines&lt;/B&gt;&amp;nbsp; (see the lesson &amp;nbsp;&lt;A HREF=http://www.algebra.com/algebra/homework/Trigonometry-basics/Proof-of-the-Law-of-Cosines-revisited.lesson&gt;Proof of the Law of Cosines revisited&lt;/A&gt;&amp;nbsp; in this site)&amp;nbsp; to the triangle &amp;nbsp;{{{DELTA}}}&lt;/B&gt;ADC&lt;/B&gt;&amp;nbsp; to find the measure of its side &amp;nbsp;&lt;B&gt;AC&lt;/B&gt;.


{{{abs(AC)^2}}} = {{{abs(AD)^2}}} + {{{abs(DC)^2}}} - {{{2*abs(AD)*abs(DC)*cos(D)}}} = 3^2 + 4^2 - 2*3*4*cos(100°) = 9 + 16 - 24*(-0.17365) = 29.16752.


Hence, &amp;nbsp;|&lt;B&gt;AC&lt;/B&gt;| = {{{sqrt(29.16752)}}} = 5.40. 


&lt;B&gt;Answer&lt;/B&gt;. &amp;nbsp;The magnitude of the resultant is &amp;nbsp;5.40. 


</PRE>