Triganomenty..HELP PLEEASE

[cheerioHobo]

can anyone do triganometry, i know how to do it, but dont know which formulae to do for different things like finding lenth or angle. please help, would be much appreciated xxxxxxxxxxx

[fo3nix]

ok, so you have the nice thing to remember,

soh/cah/toa

this stands for sine (opposite/hypotenuse) / cosine (adjacent/hypotenuse) / tan(opposite/adjacent)

sine(angle) = opposite / hypotenuse
cosine(angle) = adjacent / hypotenuse
tan(angle) = opposite / adjacent

each one of these three things, soh, cah or toa, has theee parts to it -- each has an angle, and two lengths of the triangle. regardless of whether you're trying to work out an angle or a length, you must always have two of these things to be able to work out the third. so, all this really boils down to it figuring out what two things you have, and what third thing you are trying to find!

for example, if you are trying to calculate the angle in a triangle and have the opposite and hypotenuse lengths of that triangle, then you'd use sine. why? because the sine equation i've written above has three parts: an angle, the opposite side length, and the hypotenuse side length. so, you have two of these, and are trying to find the third. now, to actually do the calculation, put in the numbers you have. for example, you may have the opposite side being of length 1, and the hypotenuse side being of length 2. so, the equation would be

sine(angle) = 1/2 = 0.5

so, you know that the sine of the angle is 0.5. but, you're wanting the angle, obviously! to do this, you actually divide by sine on both sides of the equation, to get

angle = sine^-1 0.5

the ^-1 means sine to the power -1, which is the same as meaning divide by sine. usually you have to press shift sine on your calculator, and it'll show sin-1 or similar. just type sin-1 0.5, and press equals, and it'll tell you what the angle is.

[fo3nix]

next, imagine you have an angle as well as the length of the hypotenuse side, and want to know what the length of the adjacent side is. first, we see what equation to use. from looking at what we have and knowing what we want, we need to use

cosine(angle) = adjacent / hypotenuse

so, first we can move 'hypotenuse' in this equation to the left hand side, by multiplying both sides by 'hypotenuse':

hypotenuse cosine(angle) = adjacent

and here we have an equation that has the two things we know on the left, and the thing we want on the right. this is exactly what we want! so, let us pretend that the hypotenuse has a length of 3, and the angle is 60 degrees. on your calculator type in '3 cos 60', and press the equals button. this will tell you the length of the adjacent side.

[fo3nix]

to clear some things up:

the hypotenuse side is the side that is opposite the right angle of the triangle. the opposite side is the side opposite the other angle you've either been given in the question or are wanting to find. and that means that the other side must be the adjacent!

it basically just requires you to see what type things you have been given in the question, and what third thing you need and are being asked to find.

then once you know what formula to use, rearrange it to get it to show [thing you want] = [stuff you have]. this just requires you to know how to move stuff about in equations.

now, answers!

for the first example i gave, you'll get an angle of 30 degrees.

for the second problem you'll find that the length of the adjacent size is 1.5

does that make sense?

[Buenchico]

(I prepared this while Fo3nix was sending his/her excellent post. I don't know whether it adds anything but I'll post it anyway!)

Trigonometry is tricky without diagrams but I'll have a go!

Firstly, I'm going to assume that you're doing 'basic' trigonometry, where there's always a right angle somewhere in the triangle. As well as the right angle, there will be one other 'labelled angle'. (That label could be something like 36 degrees or it could simply be a letter, such as x).

Before proceeding any further, I'd like to check that you fully understand the labels 'hypotenuse', 'adjacent' and 'opposite', for the three sides of the a right-angled triangle. (When I was teaching maths, the majority of problems that people had with trigonometry were due to misunderstanding these terms):

'Hypotenuse', of course, is the easy one. It's the longest side (and the one which 'faces', rather than touches, the right angle).

The 'Adjacent' side is the one which is adjacent to (i.e. 'next door to' or 'touching') the labelled angle.

The 'Opposite' side is the one opposite to (i.e. 'looking across at' and 'not touching') the labelled angle.

[Buenchico]

As well as a 'labelled angle', there will also be two 'labelled sides' in the question. (These labels might be in, say, centimetres or one of them might just be a letter, such as z).

So, every question will have three labelled bits of information. You can now use these (or, more particularly, the two labelled sides) to decide which of three formulas you should be using:

You've got three formulas. In shortened form, these are

Sine = Opposite / Hypotenuse

Cosine = Adjacent / Hypotenuse

Tangent = Opposite / Adjacent

You choose which formula to use, based upon which two sides are 'labelled' in the question. For example, if there are labels on the Adjacent and Hypotenuse side, it must be a Cosine question because this is the only formula with those two sides in it.

Now write out the required formula, putting the numbers or letters into it. For example, if the angle is labelled x, the adjacent side is labelled 5cm and the hypotenuse is labelled 8cm, you'll write this:

cos x = 5 / 8

<=> cos x = 0.625

<=> x = 51.3� (to 3 s.f.)

[Buenchico]

In that example, we needed to find the angle. I'll now give an example where the unknown length is on the 'top' of the fraction:

Let's suppose that looking at the triangle shows that the labelled angle is 35�, the opposite side is labelled p and the adjacent side is marked as 4 cm.

The first thing to do is to decide which formula to use. The only one which involves both the opposite and adjacent sides is the tangent formula, so we write it out, putting in the labelled information:

tan 35� = p / 4

We've now got ' p divided by 4' but we want just p, 'on it's own'. So we multiply both sides by 4:

<=> p = 4 tan 35�

<=> p = 2.80 cm (to 3 s.f.)

Lastly, I'll give an example where the unknown side is on the 'bottom' of the fraction:

Let's suppose that looking at the triangle has revealed the information to be as follows: The labelled ange is 44�, the side opposite it is marked as 5cm and we're required to find the length of the hypotenuse, which is marked as z.

The first step is to find the formula which has the opposite and hypotenuse in it. The only one which suits our purpose is the sine formula, so we write it out, filling in the information from the labels:

sin 44� = 5 / z

The probem we've got is that z is 'on the bottom' whereas we need it 'on top'. So we multiply both sides by z:

<=> z sin 44� = 5

Now z is 'on top' but we want it on it's own, not multiplied by sin 44�. So we divide both sides by sin 44�:

<=> z = 5 / Sin 44�

<=> z = 7.20 cm (to 3 s.f.)

There's about 6 weeks of maths lessons there - all in one post!

Chris

[fo3nix]

Buenchico: male, and also known by some as Chris.

I assumed you'd be preparing an answer, knowing that you used to be a maths teacher.

Let's hope all is clear for cheerioHobo!

[Buenchico]

(With apologies to cheerioHobo, for 'stealing the thread')

Fo3nix:
When I clicked on the 'preview' button (to check that I'd got the bold text properly formatted) it showed that you had posted just before me. I read your excellent post and hesitated before posting myself because I thought that you might think that I was trying to imply that my explanation was somehow better than yours. (I was definitely not. As I've said, your post is excellent).

Then I remembered that you're far too intelligent and mature to take offence at something like that, so I decided to post anyway.

As you say, let's just hope that cheerioHobo can make some sense of at least one of our answers ;-)

Chris

[jake-the-peg]

Would a picture help?

http://revisioncentre.co.uk/gcse/maths/sin_cos _tan.html

[cheerioHobo]

thank you so so so so much, this has helped me ALOT.
rock on peace out
xxxxx J
cheerio Hobo

[postmanthe]

I used to remember them so:
Should old Harry
Catch a herring
Trawling off America

[TeeGee]

Or
Some Old Hag
Chased After Him
'Till Ordered Away