Question 85735
<pre>If the given values in a right triangle are Cos A = 1/3 and a=2,
find the hypotenuse c.

{{{drawing(400,380,-1,3,-1,3, triangle(0,0,1,0,1, sqrt(8)), locate(-.15,.1,A), locate(.5,-.05,b), locate(1.12,1.4,"a=2"), locate(.3,1.4,c), locate(1.1,.05,C),    
rectangle(.8,0,1,.2)  )}}}

We have the system of equations:

{{{a^2 + b^2 = c^2}}}
{{{cos(A) = 1/3}}}

or

{{{2^2 + b^2 = c^2}}}
{{{b/c = 1/3}}}

or

{{{4 + b^2 = c^2}}}
{{{c = 3b}}}

Now we can use substitution to solve:

{{{4 + b^2 = (3b)^2}}}
{{{4 + b^2 = 9b^2}}}
{{{4 = 8b^2}}}
{{{1 = 2b^2}}}
{{{1/2 = (2b^2)/2}}}
{{{1/2 = b^2}}}
{{{sqrt(1/2) = sqrt(b^2)}}}
{{{sqrt(2/4) = b}}}
{{{sqrt(2)/2 = b}}}
Substitute that in {{{c=3b}}}
{{{c=(3sqrt(2))/2}}}

Edwin</pre>