Geometry II Homework help... >.>

[Lucid]

Alrighty, I never thought I'd ever have to ask for help with homework but...
It's Feb. vacation, and I've got a ton of this crap to do so...

Here it goes...
Excuse the crappy drawing.

 

[Ikerous]

tan(30) = 15/x

1/sqrt(3) = 15/x

x = 15sqrt(3)

 

[Lucid]

tan(30) = 15/x

1/sqrt(3) = 15/x

x = 15sqrt(3)

That was quick o.o'

Mind explaining what you did?

The thing I posted was just an example problem to give me a general idea on what to do with the other problems.

 

[Axyon]

tan30 = 15/x
x = 15/tan30
x = 25.98

y = root25.98^2 + 15^2
y = 29.9993

Beaten to some extent.

 

[Ikerous]

Tan = opposite/adjacent

So the tan(30) is going to be equal to 15/x

The tan(30) happens to be 1/sqrt(3), so you can solve from there


Although, iirc all you need to do at this point is memorize the fact that for a 30 degree triangle, the base will be height*sqrt(3)

 

[Nat Turner]

x = 15 * sqrt (3).

special triangles, good to memorize

 

[BabyHeadCrab]

thats a 30, 60, 90 triangle

the answer is 15 * square root of 3

x = 15 * sqrt (3).

special triangles, good to memorize

qft, the others in this thread are makeing this WAAAAAY harder than it used to be

special triangles 30/60/90 and 45/45/90

in a 45 45 90 you just use root 2 times the side instead for the side, sorry for typos on my paint drawing I did it in a jiff,

but since you know that theres a 90deg angle and a 30deg angle the other has to be 60

 

[JNightshade]

Yeah. I forgot how to do this, but I remembered that you don't need to deal with tangents.

 

[BabyHeadCrab]

Yeah. I forgot how to do this, but I remembered that you don't need to deal with tangents.

haha yeah I laughed when they made it so hard, then I cried

 

[Ikerous]

Thats why i added

"Although, iirc all you need to do at this point is memorize the fact that for a 30 degree triangle, the base will be height*sqrt(3)" :/

And the other way isn't hard XD I love trig
Id rather use trig than memorize things

 

[Lucid]

Much obliged.
Made my work alot easier to do.

 

[Dan]

It's as easy as one, two, (square root) three.

one is the shortest side, two is the longest side, and square root threee is the side that's somewhere in the middle

 

[Captain M4d]

Well, looks like I'll know where to go next time I need help on my math homework. ^_^

 

[OvA]

Geometry 2? I did this in Geometry 1.

 

[Teta_Bonita]

Geometry 2? I did this in Geometry 1.

Geometry 1? I did this in 6th grade math.
Homeschool ftw!

/geek boasting

 

[Qonfused]

Yeah. I got lost at "Alrighty."

 

[BabyHeadCrab]

doing this in basic geometry atm, lol i'm a math retard

 

[Lucid]

Likewise :|

I failed Geometry II last quarter with a 61...(65 needed to pass)

Anyways seeing as I'm without my Geometry book, I've got another problem for ya'll this time it's calculating the length of the hypotenuse ... and one about finding leg lengths of right triangles...

 

[Que-Ever]

dude, a squared plus b squared is c squared!
the first one is 25, I think (did it in my head) and the other is 6 (actually c squared minus a squared)

 

[Axyon]

dude, a squared plus b squared is c squared!
the first one is 25, I think (did it in my head) and the other is 6 (actually c squared minus a squared)

Nah, the first's 18.03, as it's the square root of 325.

 

[Hectic Glenn]

seriously pythagorus theorum whatever its called. I'm SO AWFUL at maths in general, and this bit was simple rearranging formulas. Learn the a squared + b squared = c squared, and you'll never go wrong again!

 

[Lucid]

Nevermind x.x

Thanks for all the help.
You guys are definitely useful when it comes to homework

So for the second one I subtract 10² and 8² and that should give me 36 which I then square root to get 6?

 

[Hectic Glenn]

First one, you do a squared which is 15 squared (225), then b squared which is 10 squard (100) add them together = 325. And thats your value for c squared. But you just want C, so the squared becomes square root, and thats 18.03 (according to axyon). I think thats you do it, been 2 years now/

edit: i can do maths! I did the 2nd one on paper Que is right its 6.

a squared + b squared = c squared

a squared (unknown side) + 8 squared = 10 squared

a squared + 64 = 100

a squared = 100 - 64 (you take it over to the other side and minus it)

a squared = 36 (need to get rid of squared, so you move it to the other side and becomes square root)

a = (square root of) 36

a = 6

Hope you understand that, by no means am i any good at maths, nor am i pretending to be, i'm just enjoying remembering this.

edit2: exactly lucid, you got it.

 

[Lucid]

Aye, Tis a cake-walk.

I just rarely ever pay attention in class, so when I see it now it's like "How the **** am I supposed to do this crap? *dies* "

But after actually doing it out, it's relatively easy... with or without a calculator.

Thanks again.

 

[Teta_Bonita]

The exact value for the first one is 5√13 BTW

 

[Lucid]

Alrighty, Another vacation... another crap load of homework...
Now I'm not quite sure what I'm supposed to do on either of these, my geometry teacher was in such a god damn hurry to gtfo of the school that he just printed random work sheets and gave them to us.

 

[kirovman]

http://regentsprep.org/regents/math/rtritrig/Ltrig.htm

 

[Lucid]

http://regentsprep.org/regents/math/rtritrig/Ltrig.htm

... this is Geometry... not trig. :X

 

[kirovman]

Trigonometry helps you solve Geometric problems

I'm sure that page contains everything you need.

If we do your homework for you, it doesn't really help you. The meaning is, "find the value of x"
where x is the length of a side.

 

[MiccyNarc]

First one, it's a 45/45/90 triangle. These are pretty easy.
The base = the adjacent leg (opposite the labeled angle).

So your triangle is 5/5/x
To solve for X
Find the square root of 50.
Always remember for right triangles (I love right triangles):
A^2 (one leg) + B^2 (2nd leg) = C^2 (hypotenuse)
There are 180 degrees in any triangle.

The other one is a 30/60 right triangle, those are harder, so I'll let someone smarter than I answer it. I know it involves the length multiplied by either the square root of 2 or square root of 3.

 

[Lucid]

First one, it's a 45/45/90 triangle. These are pretty easy.
The base = the adjacent leg (opposite the labeled angle).

So your triangle is 5/5/x
To solve for X
Find the square root of 50.
Always remember for right triangles (I love right triangles):
A^2 (one leg) + B^2 (2nd leg) = C^2 (hypotenuse)
There are 180 degrees in any triangle.

The other one is a 30/60 right triangle, those are harder, so I'll let someone smarter than I answer it. I know it involves the length multiplied by either the square root of 2 or square root of 3.

Square root of 50 is 7.07 >.>

So I take it to solve it I go...
5² + 5² + 7.07?

 

[MiccyNarc]

Nono, 5^2 + 5^2 = 7.07^2
So the answer is 7.07.

 

[Lucid]

Nono, 5^2 + 5^2 = 7.07^2
So the answer is 7.07.

Ah.

That certainly makes more sense

I believe for the second one it's a 30 60 90 triangle... so wouldn't it be a square root of 3?
And if that's the case...the answer would be 108, right?

 

[MiccyNarc]

Ah.

That certainly makes more sense

I believe for the second one it's a 30 60 90 triangle... so wouldn't it be a square root of 3?

Yeah, it's the square root of 3. I have a hard time remembering that for whatever reason o.0