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

[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.

[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!

[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.

[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]

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

[tjanuranus]

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]

  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]

Thank you very much. I get it now!

[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]

Laws of Sines and Cosines up in hurr

[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]

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]

just use your ipad and you will get your degree

[zxlkho]

just use your ipad and you will get your degree

iDegree

[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.