Using trig to find distance?

[OldWorldBlues]

Hi there! I haven't yet taken a trigonometry course (I'm in High-school), but I have an amateur interest in surveying. Recently I began thinking about how I could calculate the height of a point relative to me, or the distance of the object from me. Naturally, I immediately thought of the Pythagorean Theorem. However, the formula's need for 2 known lengths of a triangle proved unwieldy for my purposes. I did some research, and came across a formula from Clark University:
sinθ = length of opposite / length of adjacent
Doing some algebra, I got:
(sinθ)length of adjacent = length of opposite
Where θ is the acute angle directly adjacent to the triangle's right angle. I worked out a few problems on paper, which seemed to fit together well, but like I said: I'm no expert. Is there anything I need to know about trig functions (special rules, etc.)? Thanks in advance for any help you guys can give me :)

 

[Charles Link]

$\sin(\theta)=$length of opposite/hypotenuse . $\tan(\theta)=$ length of opposite/length of adjacent.

 

[OldWorldBlues]

$\sin(\theta)=$length of opposite/hypotenuse . $\tan(\theta)=$ length of opposite/length of adjacent.

Oh, darn! Thanks for the correction. I think I meant to write tan() but got mixed up.

 

[Mark44]

There's a simple mnemonic device that's helpful to learn the relationships: SOH-CAH-TOA
It represents these relationships:
$\sin(\theta) = \frac {\text{opposite}}{\text{hypotenuse}}$
$\cos(\theta) = \frac {\text{adjacent}}{\text{hypotenuse}}$
$\tan(\theta) = \frac {\text{opposite}}{\text{adjacent}}$

If you have these memorized, the other three trig functions are pretty straightforward.
$\csc(\theta) = \frac 1 {\sin(\theta)}$
$\sec(\theta) = \frac 1 {\cos(\theta)}$
$\cot(\theta) = \frac 1 {\tan(\theta)}$

 

[OldWorldBlues]

There's a simple mnemonic device that's helpful to learn the relationships: SOH-CAH-TOA
It represents these relationships:
$\sin(\theta) = \frac {\text{opposite}}{\text{hypotenuse}}$
$\cos(\theta) = \frac {\text{adjacent}}{\text{hypotenuse}}$
$\tan(\theta) = \frac {\text{opposite}}{\text{adjacent}}$

If you have these memorized, the other three trig functions are pretty straightforward.
$\csc(\theta) = \frac 1 {\sin(\theta)}$
$\sec(\theta) = \frac 1 {\cos(\theta)}$
$\cot(\theta) = \frac 1 {\tan(\theta)}$

Thanks, that's pretty useful :)
Is there a specific place that θ HAS to be, or is it just the angle between the adjacent & hypotenuse?

 

[Mark44]

Thanks, that's pretty useful :)
Is there a specific place that θ HAS to be, or is it just the angle between the adjacent & hypotenuse?

Yes, the formulas I wrote assume that θ is the angle between the adjacent & hypotenuse.