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Studying for my Math final. Could use some expert help!

Started by tjanuranus, August 19, 2011, 09:50:55 PM

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tjanuranus

Well i made this thread because i have a question but i figure i probably will have more by next thursday. I did really well on the first half of the course which was about Logs but the trig section i've struggled with. Here is a perfect example...

Find R and X.

A right triangle has an an angle of 45 degrees. The Opposite side is 2 * sqrt of 5. The hypotenuse is R and the adjacent side is X. Find both.

Any help with this problem is greatly appreciated.

Jamesman42

Since a right triangle obviously has a right angle of 90 degrees and it's given that another angle is 45 degrees, the third angle is 180-90-45 = 45 degrees as well. From geometry, those two angles corresponding sides are the same length since their angle measures are the same, so x = 2sqrt(5). Using the Pythagorean Theorem, You have:

[2sqrt(5)]2 + [2sqrt(5)]2 = R2 [Pythagorean Theorem]
4*5 + 4*5 = R2 [Square both terms]
20 + 20 = R2 [Simplify...]
40 = R2 [Simplify...]
R = sqrt(40) [Square root of both sides, keep the positive root only because we are dealing with distances, which can only be positive)
R = 2sqrt(10) [Simplifying radicals]

Or, another way is to know that a 45-45-90 triangle has the two legs equal to each other, so x = 2sqrt(5), and the hypotenuse is sqrt(2) times one of the legs. So R = 2sqrt(5)*sqrt(2) = 2sqrt(10), just like I got above.
\o\ lol /o/

tjanuranus

you sir are super awesome. I'm sure there will be more! lol.

Jamesman42

Just FYI, by next Monday i start teaching so you may see a lot less of me :) But you have Fiery Winds, Orbert, kari, rumborak, slycordinator and surely others to help!
\o\ lol /o/

tjanuranus

Ok this is what i was referring to in the facebook thread. I'm not seeing it! I can do it on the calculator but the book wants non decimals.

Right triangle with 30 degree angle. Adjacent side of 105. Find the other two sides...

Fiery Winds

Here's a helpful trick for quickly drawing up the basic trig values without having to memorize them.

https://www.youtube.com/watch?v=HFGUgfGnt8U

tjanuranus

wow that is sick! WTF? why doesn't my teacher do this? Any help with the last problem though? If i can get that one i can probably get them all.

Adami

You know, I have recently decided to take the GREs and this thread is scaring the hell out of me.
www. fanticide.bandcamp . com

Fiery Winds

Ok, you know one adjacent side, so you can use either Tan (Opposite over Adjacent) or Cos (Adjacent over Hypotenuse).  Let's go with Cos.

Cos 30 degrees = 105 / Hypotenuse

From that table we created, we know Cos 30 degrees = sqrt(3)/2

sqrt(3)/2 = 105 / Hypotenuse

Hypotenuse = 105 / (sqrt(3)/2)

Hypotenuse = 210 / sqrt(3)  (Rationalize if you need/want to)

Follow the same pattern for solving the other adjacent side.

tjanuranus

Quote from: Adami on August 19, 2011, 11:32:47 PM
You know, I have recently decided to take the GREs and this thread is scaring the hell out of me.

:rollin

tjanuranus

Quote from: Fiery Winds on August 19, 2011, 11:38:12 PM
Ok, you know one adjacent side, so you can use either Tan (Opposite over Adjacent) or Cos (Adjacent over Hypotenuse).  Let's go with Cos.

Cos 30 degrees = 105 / Hypotenuse

From that table we created, we know Cos 30 degrees = sqrt(3)/2

sqrt(3)/2 = 105 / Hypotenuse

Hypotenuse = 105 / (sqrt(3)/2)

Hypotenuse = 210 / sqrt(3)  (Rationalize if you need/want to)

Follow the same pattern for solving the other adjacent side.

Thank you, you are awesome as well. Going to try it now.

tjanuranus

Ok how do you do this? I know how to do it when it's just sin arc sin etc.. but ...

arcsin[cos(-pi/6/)]


Thanks!

Fiery Winds

First,

(-pi/6) = -30 degrees (Conversion from radians to degrees)

Now you have this,

arcsin(-30 degrees)

which simplifies to,

-arcsin(30 degrees)

Using a table of common arcsin values tells you that,

arcsin(30 degrees) = 0.5

Therefore,

-arcsin(30 degrees) = -0.5




tjanuranus

Thank you. The answer they have in the book though is pi/3. can you show me how to do it that way?

Fiery Winds

 :facepalm: I made a stupid mistake.

First,

(-pi/6) = -30 degrees (Conversion from radians to degrees)

cos(-30 degrees)

cos(30 degrees)  (same x-value on unit circle)

cos(30 degrees) = sqrt(3)/2

arcsin[sqrt(3)/2]

which means,

sin(theta) = sqrt(3)/2

Using the table of trig values, you'll see the corresponding theta is 60 degrees, or pi/3.


tjanuranus


Ħ

1) Draw a right triangle with θ=(-pi/6) on the unit circle.  You should get o=-1/2, a=(rad3)/2, and h=1.

2) cos(-pi/6) = a/h = (rad3)/2

3) arcsin[(rad3)/2] = pi/3

EDIT: Ninja'd by Windu.

tjanuranus

Thank you. i think i need to take a break because now i can't seem to figure out simple things! Ok can someone help me with this?

A right triangle, you don't know either of the other two angles. You are given the Adjacent side which is 12 ft. the opposite side is 6 feet. and the hyp is 13.42 ft. Solve for the angles. I keep getting the wrong answers!

Ħ

a = 12
o = 6
h = 13.42

Let's call the unknown angles A and B.  Let's say A is the angle that is opposite of o.

There's multiple ways to do it.  Here's one.

sin(A) = o/h
sin(A) = 6/13.42
A = arcsin(6/13.42) = use a calculator

90 - A = B

Fiery Winds

You know the opposite and adjacent, so use tangent.

tan(theta) = 6 / 12

tan(theta) = 1 / 2

arctan(1/2) = theta

You'll need to use a calculator as it's not a nice value you'd find in a table.  Other angle is 90 - theta.

EDIT: Ninja'd by H

tjanuranus

Thanks you guys for the speedy help. Going to try these out in a bit. Taking a break, getting burned out.

kári

Also, but maybe you haven't learned this yet, but when it comes to solving triangles I find it way easier to just use the sine and cosine rules. I know that what you are using are derived from those but I find it much simpler to just use the "general" rules, even in right triangles.
So.. in any triangle (so not just a right triangle)


and the cosine rule which is a little more complicated states that a^2 = b^2 + c^2 - 2*b*c*cos(A). Just think of it as a^2 = (b-c)^2 but also multiply 2bc by cos(A). Or think of it as the general version of Pythagoras's rule, but now you also have to subtract 2*b*c*cos(A), which, when A is a right angle, is zero, so you get a^2 = b^2 + c^2.

Of course you can also say that b^2 = a^2 + c^2 - 2ac*cos(B) or c^2 = a^2 + b^2 - 2ab*cos(C).

Sorry if this confuses you but using those two formulas you can calculate everything there is to calculate about any triangle, I find using those way easier than remembering all the tan = opp/adj etc. which only work for right triangles anyway.

Ħ

Based on the types of questions he's asking, it doesn't sound like he's learned that yet.

kári

I know, but I just find it way easier. And I'm sure he can handle it.

Jamesman42

\o\ lol /o/

kári

James could you clean up your inbox? I've got something I want to PM you. :)

EDIT: nvm I'll just email you.

tjanuranus

Quote from: kári on August 23, 2011, 01:08:11 AM
Also, but maybe you haven't learned this yet, but when it comes to solving triangles I find it way easier to just use the sine and cosine rules. I know that what you are using are derived from those but I find it much simpler to just use the "general" rules, even in right triangles.
So.. in any triangle (so not just a right triangle)


and the cosine rule which is a little more complicated states that a^2 = b^2 + c^2 - 2*b*c*cos(A). Just think of it as a^2 = (b-c)^2 but also multiply 2bc by cos(A). Or think of it as the general version of Pythagoras's rule, but now you also have to subtract 2*b*c*cos(A), which, when A is a right angle, is zero, so you get a^2 = b^2 + c^2.

Of course you can also say that b^2 = a^2 + c^2 - 2ac*cos(B) or c^2 = a^2 + b^2 - 2ab*cos(C).

Sorry if this confuses you but using those two formulas you can calculate everything there is to calculate about any triangle, I find using those way easier than remembering all the tan = opp/adj etc. which only work for right triangles anyway.


I got the law of Cosines and Sines down today pretty good. Thank you.

tjanuranus

Ok i could use some advice on these problems. I have a list of identities in front of me and I have no idea which ones to use when. Clueless on some of these! If you could tell me how would tackle these i would appreciate it. What properties to use and why, etc...

Simplify the expression...

1. sinx(cscx - sinx)

2. cscx/cotx

3. secx * sinx/tanx

4. sin(pi/2 - x)cscx

and lastly..

5. cos^2y/1 - siny

Thank you in advance. I REALLY appreciate it!

Aefenwelg

Oh, these are fun.

Most of the time, you just want to put everything in terms of sin and cos and everything will cancel out.

For instance:

1. cscx = 1/sinx so,
sinx(cscx - sinx) = 1 - (sinx)^2  = (cosx)^2

2 and 3 are similar.

4. sin(pi/2 - x) = cosx
cosx*cscx = cotx

5. cos^2y = 1 - sin^2y = (1 + siny)(1 - siny)
cos^2y/1 - siny = 1 + siny

tjanuranus

Ok so I got pretty good at those but of course then one that was on the final I choked on! It was...


Sin^2 / sec - 1 = cos + cos^2

Quadrochosis


zxlkho


tjanuranus

thanks for that. I would like to know how to do it though. Probably just overlooked something.

Aefenwelg

Sin^2 / sec - 1 = cos + cos^2

sin^2/(1/cos)-1 = cos(1 + cos)

sin^2/((1-cos)/cos) = cos(1 + cos)

(sin^2 * cos)/(1 - cos) = cos(1 + cos)

sin^2 * cos = cos(1+cos)(1-cos)

sin^2 * cos = cos(1 - cos^2)

sin^2 * cos = cos * sin^2

Done.