# Math Help - some more rt triangle trig

1. ## some more rt triangle trig

so there is a right triangle
90 degrees on the bottom left 39 on the bottom right
and 12.1 in length of the hypotenuse
im told to look for all side
i tried using cosine and sine but my answers were
larger than the hypotenuse
any ideas?

2. How did you set up your equations? Did you make sure your calculator was in degree mode?

Excellent checking, by the way! That's the right instinct.

3. Originally Posted by gabriel
so there is a right triangle
90 degrees on the bottom left 39 on the bottom right
and 12.1 in length of the hypotenuse
im told to look for all side
i tried using cosine and sine but my answers were
larger than the hypotenuse
any ideas?
so, how did you set up those equations?

4. Originally Posted by gabriel
so there is a right triangle
90 degrees on the bottom left 39 on the bottom right
and 12.1 in length of the hypotenuse
im told to look for all side
i tried using cosine and sine but my answers were
larger than the hypotenuse
any ideas?
then after

$sin39^o=\frac{opposite}{hypotenuse}=\frac{a}{12.1}$

you should get "a" by multiplying both sides by 12.1.

Did you end up with $a=\frac{12.1}{sin39^o}$ instead?

Also

$cos39^o=\frac{adjacent}{hypotenuse}=\frac{b}{12.1}$

so you get "b" by multiplying both sides by 12.1.

You will get a value larger than the hypotenuse if you look for $\frac{12.1}{cos39^o}$ instead.

Why is that and why multiply both sides by 12.1?

5. skeeter's diagram is exactly how the picture is formed
so i tried the following
cos(39)=b/12.1
but now that u mention it
i forgot to multiply "12.1" to the cos(39) and
i saw that i screwed up the problem there resulting in 9.4 instead of 15
so i think thats where i went wrong