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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

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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

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YAY! thank you for dble checking my 60 deg. okay thank you so far i have....
first triangle:
Ok, your first two triangles are perfect. Your last triangle has a small mistake. Double check it. Its a small error on your calculation.
okay let me look
okay its
No no, you were right at
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
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
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.
okay thank you.
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
|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,\[b=\frac{1}{\sqrt{3}}\]
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?
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.
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...\[C \sqrt{3} / 2 = 2, then (C/1 )\sqrt{3}/2 = 1/C (C/1) \to get: C \sqrt{3} = 2/\sqrt{3} then i mult. both \right side by \sqrt{3} the \top and bottom \to get 2\sqrt{3} / \sqrt{9} giving me the answer 2\sqrt{3}/3\]
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
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.
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?
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=b3√2 Multiply both sides by two, and you get 1=b3√ Divide by sqrt(3) and you get, b=13√
I live in Kosovo, which is in Europe. I need some clarification. Which 2 are you asking about?
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=b3√2
okay i dont know if u could see my last typing. but ill draw what i have gotten as an answer.
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|
|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?
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
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
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\[AB=\frac{2}{\sqrt{3}}\]
|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?
Yes, you have the following values,\[AD=1\]\[AE=\frac{2}{\sqrt{3}}\]\[DE=\frac{1}{\sqrt{3}}\]
yes
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
AH! sorry for saying sorry
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.
okay, well right now im trying to figure out which is best, so im going to stick with |dw:1348612826468:dw|
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.
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
|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|
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.
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 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
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 :))
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
ugh, so frustration. its what about 2 o clock your time?? I think i just need to figure out one more length
Yeah lol, its 2:09, dont worry about me. Relax and concentrate on your problem.
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
WOW looks more complicated
Use sin(30) because now you now opposite of 30, but you don;t know the hypotenuse. So this will help you solve it.
okay so this is what i know
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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.
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
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