Right triangles: help, please

**aikenfan** · October 22nd 2007, 07:24 PM

There are only eight problems...I think that I have the first two down...and possibly the last two..not sure if they are correct, if maybe someone can check my work so that I know if I am doing them correctly...The others, I don't really understand where to start...if maybe someone could help me...

The directions for 9 -11 got cut off a bit...should say determine the third side and then find the other five trig functions

**topsquark** · October 22nd 2007, 07:33 PM

Originally Posted by aikenfan

There are only eight problems...I think that I have the first two down...and possibly the last two..not sure if they are correct, if maybe someone can check my work so that I know if I am doing them correctly...The others, I don't really understand where to start...if maybe someone could help me...

The directions for 9 -11 got cut off a bit...should say determine the third side and then find the other five trig functions

In 4, you can simplify the $\sqrt{468}$ .

For 39:
$\frac{sin(\theta)}{cos(\theta)} + \frac{cos(\theta)}{sin(\theta)} = csc(\theta)sec(\theta)$

Just add the fractions:
$\frac{sin^2(\theta) + cos^2(\theta)}{sin(\theta)cos(\theta)} = csc(\theta)sec(\theta)$

In 46 is the argument in radians? That's the only way I can replicate your answer.

The rest look good!

-Dan

**earboth** · October 22nd 2007, 07:50 PM

Hello,

for #11:

$\sec(\theta)=4~\implies~\frac1{\cos(\theta)}=4 ~\implies~\cos(\theta)=\frac14$

for #33: change only the $\cot(\theta)$ :

$\tan(\theta) \cdot \frac1{\tan(\theta)}=1$

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