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|dw:1348604530701:dw|

So far I figured:
first triangle:
AD________

The sum of angles within each triangle is always 180 deg. So you were right when you assumed that

YAY! thank you for dble checking my 60 deg.
okay thank you so far i have....

first triangle:

okay let me look

okay its **
**

No no, you were right at **
**

okay thank you.

okay thank you i will.

okay im back.

okay let me review the work u placed again give me a min. please.

Hey welcome back! Let's get to solving the probs then.

haha okay, hey where are you that the time is night?

I live in Kosovo, which is in Europe. I need some clarification. Which 2 are you asking about?

okay i dont know if u could see my last typing. but ill draw what i have gotten as an answer.

|dw:1348610123789:dw|

|dw:1348610422667:dw|

okay im done, is that right tho?

Hmm hang on a minute, ill solve the whole problem and check yoru answers one by one.

okay. haha oh man, i thought you had fallen asleep for a sec. thank u for waitin for me to finish

lol, no dont worry, i'm awake. i will help you solve this prob completely.

thank you

Have you tried solving the third triangle or the middle triangle?

But, i think Im going to u

haha but i think im going to use a=1 for AC=1 because i need something to use as an opposite

Ok, here's what we have then..|dw:1348611838196:dw|

YA, thats what i have so far, haha but i forgot to place the 1 for x in the far right for length

Yes, you have the following values,\[AD=1\]\[AE=\frac{2}{\sqrt{3}}\]\[DE=\frac{1}{\sqrt{3}}\]

yes

AH! sorry for saying sorry

Ok, very good. Calculate that and draw the triangle and tell me where x is.

how would u multiply |dw:1348612939771:dw|

|dw:1348613038024:dw|

|dw:1348613098124:dw|

Ok, hang on. You do know SOH CAH TOA right? That will help you alot in trigonometry.

|dw:1348613426139:dw|

okay, thank you, i thought it would be a 4 then squ rt 6, thank you for clarifying

okay but is that the answer for the side .....A-B? did i do it ?

Hmm, how did you get to that multiplication? You're close tho...

dang im always close. ill draw it.

|dw:1348613700910:dw|

|dw:1348613888683:dw|

Yes but you dont have a multiplication|dw:1348613976437:dw|

why is the X over the AE length?

Because you are using sin which means opposite over hypotenuse.

|dw:1348614388796:dw|
I rotated it, i think its confusing me when its laying down.

|dw:1348614492175:dw|

Okay i see it Sin= Opp/hyp.

Yes, but you have two 45, so you need to use 90-45-45.

AH haha sorry hold on

|dw:1348614636423:dw|

I got 4
-----
sq.rt. 6

|dw:1348614899720:dw|

|dw:1348614944183:dw|

this is what im getting, im so confused

|dw:1348615010146:dw| Close again haha :))

Divide the other way around.

oh... okay i got that.
Now i see how you told me I was close. I had to make it a fraction

so that 2 / sq.rt 6 is lengths Ab & Ae?

Yes.. you're catching quick!. Let's continue with the third triangle.

|dw:1348615274270:dw|

|dw:1348615336270:dw| okay let me see the next part

Ok, try to solve and send the solution to me.

okay

|dw:1348615597032:dw|

Perfect! Calculate that and show result..

Be careful with the nominator. You have division not multiplication there.

|dw:1348615813336:dw|

then im stuck

do i multiply the numerator and denomenator by sq. rt. 6? if I do I get:
2sqrt 6 / 3sqrt 6

\[\tan(60)=\sqrt{3}\]And use this,\[x\sqrt{3}=\frac{2}{\sqrt{6}}\]

okay let me look at it again

so dang............I put it wrong ROAR! sorry

okay so where did the x come from when u put|dw:1348616685558:dw|

|dw:1348616752298:dw|

wait, is the X there because we're trying to find X right?

okay |dw:1348617331171:dw|

nevermind that, it was my mistake. I just divided by sqrt(3)

|dw:1348617643327:dw|

Correct, but you can simplify \[\sqrt{18} = 3\sqrt{2}\]

|dw:1348617948304:dw|

okay so that would be the BC length

Yeah lol, its 2:09, dont worry about me. Relax and concentrate on your problem.

WOW looks more complicated

okay so this is what i know

|dw:1348618611092:dw|

wait, what do u mean? that x, 2x, and xsqrt 3 ?

yes.. the one against 30 is multiplied by two and you get the last unknown.

|dw:1348618995244:dw|

You did forget 3 though.|dw:1348619146057:dw|

No wait.. wrong. It's the other one.. Haha it seems im tired lol

|dw:1348619217214:dw|