Anyone care to help me out with this easy problem?
"Find sin, cos and tan of a triangle with area :0,465 m^2
side 1: 0,74 cm
side 2: 0,0936 cm"
Any ideas? How should I go about figuring out the answer? Thanks in advance!

- anonymous

- chestercat

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- inkyvoyd

1/2 sin(included angle)* (AB)=area of triangle

- anonymous

Care to explain? :)

- inkyvoyd

1/2 sin(angle between 1 and 2)*0,74*0,0936=0,465
solve for sin(angle between 1 and 2)

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## More answers

- inkyvoyd

btw, there are 3 angles in a triangle, and you have not specfied which angle you want the sin cos and tangent of. I'm assuming it is the angle made by sides 1 and 2.

- inkyvoyd

the identity I have used is just the area formula for a triangle, which is 1/2 sin(included angle)*(ab)=area of triangle, where a and b are side lengths, and the included angle is the angle between a and b.
To find cos x simply use sqrt(1-sin x) (pythagorean identity), and to find tan x simply use sin x/cos x

- anonymous

Hmm,.. Think you could walk me through each step and explain more thoroughly? :/ Not sure I understand your reasoning.

- inkyvoyd

I am using the following formula as an identity: http://www.analyzemath.com/Geometry_calculators/area_triangle_sine.html

- inkyvoyd

look at the bottom of this site for a proof
http://www.mathopenref.com/triangleareasas.html

- anonymous

Ah, I think I understand! I'll get back to you if any additional questions arise. You rock.

- inkyvoyd

good luck! you might want to show a quick proof of the formula if you need to show work for the problem. The area formula with SAS is common but I don't think it's required as standard curriculum (it's very useful though)

- anonymous

1.) shouldn't you start off by converting the sides from centimeters to meters?
2.) I understand the proof and everything, but what do I do when I have the value for sin(angle between a and b)?

- inkyvoyd

Well, that sine was what you were looking for, right? you were asked to find the sin of the angle, and I'm assuming they wanted the sine of the included angle.

- inkyvoyd

And yes, I did not catch that - you would need to convert units (which I didnot do correctly)

- anonymous

But what the cos and tan? How do I go about finding them?

- anonymous

What about*

- anonymous

Tried everything, can't figure it out. :/

- inkyvoyd

sin^2 x+cos^2 x=1
sin x/cos x=tan x
use the folowing identities. First use the first one to find cos x, the second after you found cos x to find tan x.

- anonymous

Think you could explain more in depth? :/

- inkyvoyd

have you heard of the identity sin^2 x+cos ^2 x=1?

- anonymous

No, never, I'm afraid.

- inkyvoyd

I might be being a bit too advanced then o.o

- anonymous

Yes, bring it down to my level. :)

- anonymous

10th grade precalc.

- inkyvoyd

are you just getting introduced to trig fnctions? is this some online course in which you may be expected to know prereqs in trig that you have a knowledge gap in? Please answer these questions - the identities I'm using while valid might make your teacher angry etc

- anonymous

I'm kust starting to with trigonometric ratios and the like. Just need to know how to attain tan and cos from here.

- inkyvoyd

Okay. Is this a right triangle?

- anonymous

Just started with trigonometric ratios*

- anonymous

I don't know. Question doesn't specify.

- inkyvoyd

have you only been working with right triangles as of late?

- anonymous

Yes.

- inkyvoyd

if there is an included picture, please draw it. If there isn't one, this is the most ambiguous problem ever.

- anonymous

There isn't. But since they're asking us tp find tank, cos and sin I guess we'll have to assume it's a right triangle?

- anonymous

Think you could just explain the identity you suggested earlier?

- inkyvoyd

Yeah. I was assuming this was a general review question so I pulled a lot of formulas and identities out of my head. Throw those out for now - focuing on the ratios is what is important.
"Find sin, cos and tan of a triangle with area :0,465 m^2 side 1: 0,74 cm side 2: 0,0936 cm"

- inkyvoyd

assuming that it is a right triangle, you'll have to do some trial and error. Remember that a^2+b^2=c^2, so try around with values. IN addition, if a and b are legs of the right triangle, you should get ab=2*area (you've been given the area). if you find out which side is which, you can easily progress.

- anonymous

Assuming it isn't, could you explIn tp me the identity you suggested earlier?

- inkyvoyd

does the illustration in the second link (and provided explanation) make sense?
http://www.mathopenref.com/triangleareasas.html
make sure you go down to "how it works"

- anonymous

Or link to something that does

- anonymous

Yes. Males perfect sense.

- anonymous

Makes*

- inkyvoyd

Okay. So when I saw your problem I instantly figured that was what they were asking about. You can see this even specifies which angle - the included angle. There are two more trig identities that you apply to find the cos and tan of the included angle. I'll link you to the first one.

- inkyvoyd

http://www.regentsprep.org/Regents/math/algtrig/ATT9/pythagoreanid.htm
read the first paragraph and tell me if the first identity makes sense. ignore the others.

- inkyvoyd

Since you know the value of sin x you can just bring it into that identity which holds true for all real numbers x, and you can find cos x.
THe next step is to find tan x, which is really sin x/cos x
here's a simple proof:
\(\sin x=\frac{O}{H}\)
\(\cos x=\frac{A}{H}\)
\(\tan x=\frac{O}{A}\)
\(\sin x/\cos x=\frac{O}{H}\frac{H}{A}=\frac{O}{A}=\tan x\)
thus for all real x sin x/cos x=tan x

- inkyvoyd

in fact, since you mgiht be curious, I will bring the last thing up - if one wishes to find the sin cos and tangents of the other angles, you will need to use the law of sines and cosines (there is a law of tangents but it's not really used much and I don't know it)
THIS STUFF IS EXTRA.
Also, you can view the proofs, which are fairly advanced (you will most likely cover this at the end of "regular" trigonometric functions.
http://www.regentsprep.org/Regents/math/algtrig/ATT12/derivelawofsines.htm
I have to go to bed.
Good luck, and happy mathing!

- inkyvoyd

And, if anyone thinks you are too nerdy or it's too early to learn, that's complete bullcrap. I'm a 10th grader as well, and I'm happily enjoying my maths course. Don't let anyone keep you from your potential :P

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