# Using trig to find distance?

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 :)

Homework Helper
Gold Member
## \sin(\theta)=##length of opposite/hypotenuse . ## \tan(\theta)=## length of opposite/length of adjacent.

## \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
Mentor
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}}##

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)}##

• opus
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}}##

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
Mentor
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.