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?

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- Aug 4th 2010, 12:12 PMgabrielsome 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? - Aug 4th 2010, 12:25 PMAckbeet
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. - Aug 4th 2010, 12:58 PMskeeter
- Aug 4th 2010, 01:24 PMArchie Meade
If your answers were larger than the hypotenuse,

then after

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

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

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

Also

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

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

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

Why is that and why multiply both sides by 12.1? - Aug 4th 2010, 04:11 PMgabriel
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