# some more rt triangle trig

• Aug 4th 2010, 12:12 PM
gabriel
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?
• Aug 4th 2010, 12:25 PM
Ackbeet
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 PM
skeeter
Quote:

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?
• Aug 4th 2010, 01:24 PM
Quote:

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

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

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 PM
gabriel
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