lilsis76
  • lilsis76
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
Mathematics
  • Stacey Warren - Expert brainly.com
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SOLVED
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katieb
  • katieb
I got my questions answered at brainly.com in under 10 minutes. Go to brainly.com now for free help!
lilsis76
  • lilsis76
|dw:1348604530701:dw|
lilsis76
  • lilsis76
So far I figured: first triangle: AD________
anonymous
  • anonymous
The sum of angles within each triangle is always 180 deg. So you were right when you assumed that

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lilsis76
  • lilsis76
YAY! thank you for dble checking my 60 deg. okay thank you so far i have....
lilsis76
  • lilsis76
first triangle:
anonymous
  • 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
  • lilsis76
okay let me look
lilsis76
  • lilsis76
okay its
anonymous
  • anonymous
No no, you were right at
lilsis76
  • 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
  • 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
  • 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
  • lilsis76
okay thank you.
lilsis76
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • 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
  • lilsis76
okay thank you i will.
lilsis76
  • lilsis76
okay im back.
lilsis76
  • lilsis76
okay let me review the work u placed again give me a min. please.
anonymous
  • anonymous
Hey welcome back! Let's get to solving the probs then.
lilsis76
  • lilsis76
haha okay, hey where are you that the time is night?
lilsis76
  • 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=b3√2 Multiply both sides by two, and you get 1=b3√ Divide by sqrt(3) and you get, b=13√
anonymous
  • anonymous
I live in Kosovo, which is in Europe. I need some clarification. Which 2 are you asking about?
lilsis76
  • 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=b3√2
lilsis76
  • lilsis76
okay i dont know if u could see my last typing. but ill draw what i have gotten as an answer.
anonymous
  • 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
  • lilsis76
|dw:1348610123789:dw|
lilsis76
  • lilsis76
|dw:1348610422667:dw|
lilsis76
  • lilsis76
okay im done, is that right tho?
anonymous
  • anonymous
Hmm hang on a minute, ill solve the whole problem and check yoru answers one by one.
lilsis76
  • lilsis76
okay. haha oh man, i thought you had fallen asleep for a sec. thank u for waitin for me to finish
anonymous
  • anonymous
lol, no dont worry, i'm awake. i will help you solve this prob completely.
lilsis76
  • lilsis76
thank you
anonymous
  • anonymous
Have you tried solving the third triangle or the middle triangle?
lilsis76
  • 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
  • lilsis76
But, i think Im going to u
lilsis76
  • lilsis76
haha but i think im going to use a=1 for AC=1 because i need something to use as an opposite
anonymous
  • anonymous
Ok, here's what we have then..|dw:1348611838196:dw|
lilsis76
  • 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
  • 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
  • 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
  • anonymous
Yes, you have the following values,\[AD=1\]\[AE=\frac{2}{\sqrt{3}}\]\[DE=\frac{1}{\sqrt{3}}\]
lilsis76
  • lilsis76
yes
lilsis76
  • 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
  • lilsis76
AH! sorry for saying sorry
anonymous
  • 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
  • lilsis76
okay, well right now im trying to figure out which is best, so im going to stick with |dw:1348612826468:dw|
anonymous
  • anonymous
Ok, very good. Calculate that and draw the triangle and tell me where x is.
lilsis76
  • lilsis76
how would u multiply |dw:1348612939771:dw|
lilsis76
  • lilsis76
|dw:1348613038024:dw|
lilsis76
  • lilsis76
|dw:1348613098124:dw|
anonymous
  • anonymous
Ok, hang on. You do know SOH CAH TOA right? That will help you alot in trigonometry.
lilsis76
  • 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
  • anonymous
|dw:1348613426139:dw|
lilsis76
  • lilsis76
okay, thank you, i thought it would be a 4 then squ rt 6, thank you for clarifying
lilsis76
  • lilsis76
okay but is that the answer for the side .....A-B? did i do it ?
anonymous
  • anonymous
Hmm, how did you get to that multiplication? You're close tho...
lilsis76
  • lilsis76
dang im always close. ill draw it.
lilsis76
  • lilsis76
|dw:1348613700910:dw|
lilsis76
  • lilsis76
|dw:1348613888683:dw|
anonymous
  • anonymous
Yes but you dont have a multiplication|dw:1348613976437:dw|
lilsis76
  • lilsis76
why is the X over the AE length?
anonymous
  • anonymous
Because you are using sin which means opposite over hypotenuse.
lilsis76
  • lilsis76
|dw:1348614388796:dw| I rotated it, i think its confusing me when its laying down.
lilsis76
  • lilsis76
|dw:1348614492175:dw|
lilsis76
  • lilsis76
Okay i see it Sin= Opp/hyp.
anonymous
  • anonymous
Yes, but you have two 45, so you need to use 90-45-45.
lilsis76
  • lilsis76
AH haha sorry hold on
lilsis76
  • lilsis76
|dw:1348614636423:dw|
anonymous
  • 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
  • lilsis76
I got 4 ----- sq.rt. 6
lilsis76
  • lilsis76
|dw:1348614899720:dw|
lilsis76
  • lilsis76
|dw:1348614944183:dw|
lilsis76
  • lilsis76
this is what im getting, im so confused
anonymous
  • anonymous
|dw:1348615010146:dw| Close again haha :))
anonymous
  • anonymous
Divide the other way around.
lilsis76
  • lilsis76
oh... okay i got that. Now i see how you told me I was close. I had to make it a fraction
lilsis76
  • lilsis76
so that 2 / sq.rt 6 is lengths Ab & Ae?
anonymous
  • anonymous
Yes.. you're catching quick!. Let's continue with the third triangle.
lilsis76
  • lilsis76
|dw:1348615274270:dw|
lilsis76
  • lilsis76
|dw:1348615336270:dw| okay let me see the next part
anonymous
  • anonymous
Ok, try to solve and send the solution to me.
lilsis76
  • lilsis76
okay
lilsis76
  • lilsis76
|dw:1348615597032:dw|
anonymous
  • anonymous
Perfect! Calculate that and show result..
anonymous
  • anonymous
Be careful with the nominator. You have division not multiplication there.
lilsis76
  • lilsis76
|dw:1348615813336:dw|
lilsis76
  • lilsis76
then im stuck
lilsis76
  • lilsis76
do i multiply the numerator and denomenator by sq. rt. 6? if I do I get: 2sqrt 6 / 3sqrt 6
anonymous
  • anonymous
\[\tan(60)=\sqrt{3}\]And use this,\[x\sqrt{3}=\frac{2}{\sqrt{6}}\]
lilsis76
  • lilsis76
okay let me look at it again
lilsis76
  • 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
  • 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
  • lilsis76
so dang............I put it wrong ROAR! sorry
lilsis76
  • lilsis76
okay so where did the x come from when u put|dw:1348616685558:dw|
anonymous
  • anonymous
|dw:1348616752298:dw|
lilsis76
  • lilsis76
wait, is the X there because we're trying to find X right?
lilsis76
  • lilsis76
okay |dw:1348617331171:dw|
anonymous
  • anonymous
nevermind that, it was my mistake. I just divided by sqrt(3)
lilsis76
  • lilsis76
|dw:1348617643327:dw|
anonymous
  • anonymous
Correct, but you can simplify \[\sqrt{18} = 3\sqrt{2}\]
lilsis76
  • lilsis76
|dw:1348617948304:dw|
lilsis76
  • lilsis76
okay so that would be the BC length
lilsis76
  • lilsis76
ugh, so frustration. its what about 2 o clock your time?? I think i just need to figure out one more length
anonymous
  • anonymous
Yeah lol, its 2:09, dont worry about me. Relax and concentrate on your problem.
lilsis76
  • 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
  • lilsis76
WOW looks more complicated
anonymous
  • 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
  • lilsis76
okay so this is what i know
lilsis76
  • lilsis76
|dw:1348618611092:dw|
anonymous
  • 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
  • lilsis76
wait, what do u mean? that x, 2x, and xsqrt 3 ?
anonymous
  • anonymous