SOLUTION: Find the length of the legs of a 45�-45�-90� triangle with a hypotenuse of 4 times the square root of 2

Question 141962: Find the length of the legs of a 45�-45�-90� triangle with a hypotenuse of 4 times the square root of 2
Answer by jojo14344(1513) (Show Source): You can put this solution on YOUR website!
well, I guess since only the hypotenuse is given we have to use trigonometric function to solve the other 2 sides (opposite and adjacent side). And mind you they'll have both same measurements since it's 45deg-45deg to a right triangle. We'll find out;
First, we use the function "sine" to get one side, = opp/hyp= "a"/hyp
sin45=a/4*sqrt2
a=(sin45)(4*sq rt2), measurement of one leg
For the other leg we use "cosine" because we deal the adjacent side this time= adj/hyp= "b"/hyp
cos45=b/4*sq rt2
b=(cos45)(4*sq rt2), measurement of the other leg.
Why they're the same?
Because sin45=cos45, check it out.
why i use sine or cosine? review your trigo functions i guess in dealing with those legs- opposite and adjacent side.
Hope this help.
Thank you,
Jojo
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