Suppose AD=1. Find the length of each of the other line segments in the figure. When radicals appear in an answer, leave the answer in that form rather than using a calculator. No CALCULATOR.
I will post a drawing. and this is under RIGHT TRIANGLES in PRECAL @hartnn @timo86m

- lilsis76

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- lilsis76

|dw:1348604530701:dw|

- lilsis76

So far I figured:
first triangle:
AD________

- anonymous

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

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## More answers

- lilsis76

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

- lilsis76

first triangle:

- anonymous

Ok, your first two triangles are perfect. Your last triangle has a small mistake. Double check it. Its a small error on your calculation.

- lilsis76

okay let me look

- lilsis76

okay its

- anonymous

No no, you were right at

- lilsis76

okay thank you. No i haven't studied trignometry before, today my instructor guided us further into trig. functions, and triangles, but she goes so fast I am trying to keep up in our homework

- lilsis76

i thank you for taking the time to assit me with this, its a pain to me cuz i dont really understand it as much as i want to for some reason

- anonymous

Hey dont worry. It's difficult at first. Glad to help, and its good to see that people want to learn not just get their homework solved by someone else. Now let me solve one of the triangles of you, and then I will watch and you solve the other two. I will solve the first triangle. Hang on.

- lilsis76

okay thank you.

- lilsis76

OH! and we got this chart from the book i dont know if it will help:
heita or theta(circle with a line going horizontal)
Heita / Sin / Cos / Tan
30 deg 1/2 squ.rt.3 /2 squ rt 3 / 3
45 deg squ rt2/2 squrt. 2/2 1
60 deg squrt3/2 1/2 squrt 3

- anonymous

|dw:1348606638092:dw|
We know that a =1, because AD=1. Now we use the trigonometry equations, and the chart you provided,\[\sin (60)=\frac{ a }{ c } =\frac{ 1 }{ c }\]we know from the table that \[\sin(60)=\frac{ \sqrt{3}}{ 2 }\]when we make that equal to 1/c, we get;
\[\frac{ \sqrt{3} }{ 2 } = \frac{ 1 }{ c}\]and from there c is equal to the inverse of the result in the left,\[c=\frac{2}{\sqrt{3}}\]We repeat the same calculation for b, but this time instead of sin(60) we use sin(30). And we get,\[\sin(30)=\frac{b}{c}\]Now we know c from the result above so we substitute.\[\frac{ 1 }{ 2 } = \frac{b \sqrt{3}}{2}\]Multiply both sides by two, and you get\[1=b\sqrt{3}\]Divide by sqrt(3) and you get,\[b=\frac{1}{\sqrt{3}}\]

- lilsis76

WOOW. lol okay sorry okay i am barely on the first part Sin60=Opp/Hyp
so sin 60= 1/c okay sorry im taking such a step at a time. and Just letting you know ahead of time. Im going to take off at 220 to get sister from school and Ill be back by 245. Is it okay If i try and read thru your notes and get back to you around 245?

- anonymous

Ok hang on. What timezone you at?
Time over here is 11:13PM. I will probably be awake tho. I can help you. Dont worry.

- lilsis76

for the first one when we did the C, i multiplied up and down of the...i think its called a rational the squ 3...
well i did...\[C \sqrt{3} / 2 = 2, then (C/1 )\sqrt{3}/2 = 1/C (C/1) \to get: C \sqrt{3} = 2/\sqrt{3} then i mult. both \right side by \sqrt{3} the \top and bottom \to get 2\sqrt{3} / \sqrt{9} giving me the answer 2\sqrt{3}/3\]

- lilsis76

AH! okay my thingy didnt work with everything i typed. its 2:15pm in the US.
Okay if u have to go I understand. Well have a goodnite, i have to start taking off. thank you for everything

- anonymous

Lol, no i don't have to go. I just wanted to know what time it is because i didnt understand 220. But now its ok. I'll be around. Just reply when you're back.

- lilsis76

okay thank you i will.

- lilsis76

okay im back.

- lilsis76

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

- anonymous

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

- lilsis76

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

- lilsis76

okay from this: first part, where did the 2 come from?
We repeat the same calculation for b, but this time instead of sin(60) we use sin(30). And we get,
sin(30)=bc
Now we know c from the result above so we substitute.
12=b3âˆš2
Multiply both sides by two, and you get
1=b3âˆš
Divide by sqrt(3) and you get,
b=13âˆš

- anonymous

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

- lilsis76

wow, ive always wanted to go to europe. okay the 2 as in the number 2
----Now we know c from the result above so we substitute.
sin 30 12=b3âˆš2

- lilsis76

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

- anonymous

Hehe, its a great place really. Europe is nice. I visited a couple of countries myself this summer. Was pretty cool. The two comes from the fact that:|dw:1348610270623:dw|

- lilsis76

|dw:1348610123789:dw|

- lilsis76

|dw:1348610422667:dw|

- lilsis76

okay im done, is that right tho?

- anonymous

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

- lilsis76

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

- anonymous

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

- lilsis76

thank you

- anonymous

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

- lilsis76

im trying to work on them right now, but it looks as tho i need a length, but im still going by the degrees. right now im trying to do the middle one and im using the first triangle measurements as an example to look back onto for guidence

- lilsis76

But, i think Im going to u

- lilsis76

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

- anonymous

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

- lilsis76

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

- anonymous

You dont need to compute AC, lets compute AB first. We can use \[\tan (45)=\frac{ AB }{ \frac{2}{\sqrt{3}} }\]This gives you\[1=\frac{AB}{\frac{2}{\sqrt{3}}}\]And from that\[AB=\frac{2}{\sqrt{3}}\]

- lilsis76

|dw:1348612135479:dw|
the part where i put a dot it would be the same exact answer right for the length to solve the next triangle?

- anonymous

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

- lilsis76

yes

- lilsis76

now with filling out the middle triangle, i use either of the answers from the first one to get the answer for the middle one. okay let me see. Sorry im taking so long. i feel like im keeping u

- lilsis76

AH! sorry for saying sorry

- anonymous

lol relax don't worry. I am enjoying this. Otherwise I wouldn't be here. You try to solve that, and I will wait for your answer.

- lilsis76

okay, well right now im trying to figure out which is best, so im going to stick with |dw:1348612826468:dw|

- anonymous

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

- lilsis76

how would u multiply |dw:1348612939771:dw|

- lilsis76

|dw:1348613038024:dw|

- lilsis76

|dw:1348613098124:dw|

- anonymous

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

- lilsis76

OH SORRY. how would i go about mulitplying...|dw:1348613263475:dw|
yea I got the hang of SOH CAH TOA, but i cant seem to figure out a way to remember the other ones like cot or sec i think the other was scs? haha

- anonymous

|dw:1348613426139:dw|

- lilsis76

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

- lilsis76

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

- anonymous

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

- lilsis76

dang im always close. ill draw it.

- lilsis76

|dw:1348613700910:dw|

- lilsis76

|dw:1348613888683:dw|

- anonymous

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

- lilsis76

why is the X over the AE length?

- anonymous

Because you are using sin which means opposite over hypotenuse.

- lilsis76

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

- lilsis76

|dw:1348614492175:dw|

- lilsis76

Okay i see it Sin= Opp/hyp.

- anonymous

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

- lilsis76

AH haha sorry hold on

- lilsis76

|dw:1348614636423:dw|

- anonymous

Ok now just take AE and divide it by sqrt(2) and you will get x. Which means you effectively solve both AB and BE.

- lilsis76

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

- lilsis76

|dw:1348614899720:dw|

- lilsis76

|dw:1348614944183:dw|

- lilsis76

this is what im getting, im so confused

- anonymous

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

- anonymous

Divide the other way around.

- lilsis76

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

- lilsis76

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

- anonymous

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

- lilsis76

|dw:1348615274270:dw|

- lilsis76

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

- anonymous

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

- lilsis76

okay

- lilsis76

|dw:1348615597032:dw|

- anonymous

Perfect! Calculate that and show result..

- anonymous

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

- lilsis76

|dw:1348615813336:dw|

- lilsis76

then im stuck

- lilsis76

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

- anonymous

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

- lilsis76

okay let me look at it again

- lilsis76

so i wasn't close then. :/
hmm. okay on the = squrt 3, where did that come from??
on the tan60 = 3 from that chart thingy i first drew

- anonymous

Heita / Sin / Cos / Tan
30 deg 1/2 squ.rt.3 /2 squ rt 3 / 3
45 deg squ rt2/2 squrt. 2/2 1
60 deg squrt3/2 1/2 squrt 3
It's sqrt(3) not 3 :))

- lilsis76

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

- lilsis76

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

- anonymous

|dw:1348616752298:dw|

- lilsis76

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

- lilsis76

okay |dw:1348617331171:dw|

- anonymous

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

- lilsis76

|dw:1348617643327:dw|

- anonymous

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

- lilsis76

|dw:1348617948304:dw|

- lilsis76

okay so that would be the BC length

- lilsis76

ugh, so frustration. its what about 2 o clock your time?? I think i just need to figure out one more length

- anonymous

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

- lilsis76

okay then. Ugh im a worry wort. i think thats how u spell it.
OKAY!
the last length would be CE. so!, again we would use Tangent and we get
Tan 30 = 2 / 3sqrt2 (ALL OVER) 2/sqrt 6

- lilsis76

WOW looks more complicated

- anonymous

Use sin(30) because now you now opposite of 30, but you don;t know the hypotenuse. So this will help you solve it.

- lilsis76

okay so this is what i know

- lilsis76

|dw:1348618611092:dw|

- anonymous

Now you can use the properties of 90-30-60 triangle, and multiply the length opposite to 30 by 2 and get the last unkown length.

- lilsis76

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

- anonymous

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

- lilsis76

|dw:1348618995244:dw|

- anonymous

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

- anonymous

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

- anonymous

|dw:1348619217214:dw|

- lilsis76

wait the 3 came from....? haha

- lilsis76

okay i see the length we got from BC= 2 / 3sqrt2

- anonymous

Yep!

- lilsis76

|dw:1348619345557:dw|
so we got all the sides!
ugh....no wonder im not good at math, it makes my brain hurt

- lilsis76

This means were done now right? nothing else to solve because I dont have any more instruction to this problem

- anonymous

Haha you're good at math. Are you a high school student or uni?

- lilsis76

not a university, Im in a community college at the moment I hope to transfer to a university in about 3 years. what about y

- lilsis76

you?

- anonymous

I am a university student. Engineering. What is the difference between the community college and uni?

- lilsis76

Community College, CC; is like if you dont have good grades or want to get the easiest classes out of the way like General Eduation you can take it at a Comminity college then transfer to a university. but Ive been in the CC a few years now. Im estimated to transfer in about 2-3 years since im taking about 2-3 classes a semester

- lilsis76

right now Im part of a STEM club, Science Technology Engineering and Mathematics major
I want to major in the sciences.
what do you do in engineering?

- anonymous

Oh nice. I like STEM alot. I wanted to get a STEM club going in my high school. I was the president for a year then I graduated.
I am studying mechanical engineering. It's really good to hear that you want to do the sciences. What is your interest?

- lilsis76

Medical field. I have always wanted to become a Pediatric since I was in elemetry And so im going to stick to it. But since i was in a car accident it put me behind 3 years in school

- anonymous

Oh :(, that is terrible.
Its good that you are going to continue studying what you've always wanted. I admire that.
I have followed you in openstudy so whenever you need help just message me. I will be glad to help you.

- lilsis76

Thank you very much. I'm better now, I just have to get back into the act of getting used to math again which was never my strongest subject.
Thank you for helping me.
I'll let you sleep now. :)
Goodnite, And if need help later I will be sure to ask you. thank you
ill close this post now.

- anonymous

Heh, good night to you too. Don't hesitate to ask me. I am surely enjoying this whole open study experience. It's great to be part of this. Makes me happy to know people are so keen on sharing knowledge.

- lilsis76

not too many people us Keen I like it. But ya, I wont hesitate don't worry. Ill write a message if and when I need help. I know ill need help this weekend thats for sure. I havent even started the rest of todays homework since we practically worked on this most of the day. its already going to be 6pm over here in California

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