Question 86487
I have a right triangle. a=2 and b=4 what does Cos(A)=?
I am having trouble with the problem. I know Cos(A)= b/c so i have to figure out what c= i believe? Thanks for the help
{{{drawing(400,266.7,-1,5,-1,3, triangle(0,0,4,0,4,2),locate(-.1,0,"A"),
locate(4.1,0,"C"),locate(4.1,2,"B"), locate(4.1,1,"a = 2"), locate(2,0,"b=4"),
locate(2,1.5,"c=?"), rectangle(3.8,0,4,.2)
 
 
  )}}}

Yes, you have to calculate c using the Pythagorean theorem:

{{{c^2}}} = {{{a^2}}} + {{{b^2}}}

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

{{{c^2}}} = 4 + 16

{{{c^2}}} = 20

c = {{{sqrt(20)}}}

c = {{{sqrt(4*5)}}}

c = 2{{{sqrt(5)}}}


{{{drawing(400,266.7,-1,5,-1,3, triangle(0,0,4,0,4,2),locate(-.1,0,"A"),
locate(4.1,0,"C"),locate(4.1,2,"B"), locate(4.1,1,"a = 2"), locate(2,0,"b=4"),
 locate(2,1.6,"c=2*V5"), locate(2.6,1.83,"_"), rectangle(3.8,0,4,.2) 
 
  )}}}

so cos(A) = {{{b/c}}} = {{{4/(2sqrt(5))}}} = {{{2/sqrt(5)}}} = {{{(2*sqrt(5))/(sqrt(5)sqrt(5))}}} = {{{2sqrt(5)/5}}}