# Thread: Need lil help with An Acute Angle word problem.

1. ## Need lil help with An Acute Angle word problem.

It is actually a couple word problems i missed on my quiz. Wasn't sure exactly how to solve them. So any help would be awesome thx!
The answers are given i just am trying to figure out how they are solved.

http://i48.photobucket.com/albums/f2...99/prob1-1.jpg

http://i48.photobucket.com/albums/f247/ace799/Prob2.jpg

http://i48.photobucket.com/albums/f247/ace799/prob3.jpg

2. That looks a lot like a quiz for class credit. If so, we can't help you. It's forum policy not to help with questions that count towards a grade.

3. quiz ends in 20 mins and then i can't work on it. After it ends then i would greatly appreciate how to solve this. thx guys.

4. Ok, no problem. Let us know when you're done.

5. finished it.

6. For the first problem, there are two tricks. First is to get a trigonometric relation involving the $24^{\circ}18'$, the $4.55$ ft., and the $x$. Use the following:

$\frac{\text{opposite}}{\text{hypotenuse}}=\sin(\th eta)=\frac{\text{Oscar}}{\text{had}}$
$\frac{\text{adjacent}}{\text{hypotenuse}}=\cos(\th eta)=\frac{\text{a}}{\text{hold}}$
$\frac{\text{opposite}}{\text{adjacent}}=\tan(\thet a)=\frac{\text{on}}{\text{Arthur.}}$

Which of those do you think applies?

The second trick for the first problem is to know how to convert that $24^{\circ}18'$ into something you can plug into a calculator. Note: there are sixty minutes in a degree, and sixty seconds in a minute.

Note: I recommend solving for your target variable $x$, before you plug in any numbers. It's a better practice whenever you can do it. It saves time in the long run, especially if the teacher gives you lots of related problems, only tweaking the initial values a little.

For the second problem, again, I would solve for the target variable, in this case $\theta_{2}$. How would you do that? Once you have that equation, you can plug in what you know and just turn the crank.

This one is very similar to the first one. In fact, you'll use the same general knowledge and technique; your expression for $d$ will be different from what you had for $x$ in the first problem.

So, with these hints, see what you can do.