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lilsis76

  • 2 years ago

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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  1. lilsis76
    • 2 years ago
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    |dw:1348604530701:dw|

  2. lilsis76
    • 2 years ago
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    So far I figured: first triangle: AD________ <D=90 deg <A=30 deg <E= im going to say 60 deg. middle triangle: AE_________ AB_________ BE__________ <A_________ <B=90 deg <E=45 deg bottom triangle: BC____________ CE____________ <B____________ <C=60 deg. <E____________

  3. gezimbasha
    • 2 years ago
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    The sum of angles within each triangle is always 180 deg. So you were right when you assumed that <E is 60 deg on the first triangle. Try computing the rest of the angles and then I will help you with the lengths.

  4. lilsis76
    • 2 years ago
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    YAY! thank you for dble checking my 60 deg. okay thank you so far i have....

  5. lilsis76
    • 2 years ago
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    first triangle: <D=90 <A=30 <E=60 2nd triangle: <B=90 <E=45 <A=45 right? last Triangle: <B=90 <C=60 <E=75 right?

  6. gezimbasha
    • 2 years ago
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    Ok, your first two triangles are perfect. Your last triangle has a small mistake. Double check it. Its a small error on your calculation.

  7. lilsis76
    • 2 years ago
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    okay let me look

  8. lilsis76
    • 2 years ago
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    okay its <B=45

  9. gezimbasha
    • 2 years ago
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    No no, you were right at <B=90. If <B=90 and your other angle is <C=60. <E=180-(B+C) , hence your <E=30. Heh, small error. Dont worry. Let's continue with lengths now. Have you studied trigonometry?

  10. lilsis76
    • 2 years ago
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    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

  11. lilsis76
    • 2 years ago
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    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

  12. gezimbasha
    • 2 years ago
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    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.

  13. lilsis76
    • 2 years ago
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    okay thank you.

  14. lilsis76
    • 2 years ago
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    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

  15. gezimbasha
    • 2 years ago
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    |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,\[b=\frac{1}{\sqrt{3}}\]

  16. lilsis76
    • 2 years ago
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    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?

  17. gezimbasha
    • 2 years ago
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    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.

  18. lilsis76
    • 2 years ago
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    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...\[C \sqrt{3} / 2 = 2, then (C/1 )\sqrt{3}/2 = 1/C (C/1) \to get: C \sqrt{3} = 2/\sqrt{3} then i mult. both \right side by \sqrt{3} the \top and bottom \to get 2\sqrt{3} / \sqrt{9} giving me the answer 2\sqrt{3}/3\]

  19. lilsis76
    • 2 years ago
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    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

  20. gezimbasha
    • 2 years ago
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    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.

  21. lilsis76
    • 2 years ago
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    okay thank you i will.

  22. lilsis76
    • 2 years ago
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    okay im back.

  23. lilsis76
    • 2 years ago
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    okay let me review the work u placed again give me a min. please.

  24. gezimbasha
    • 2 years ago
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    Hey welcome back! Let's get to solving the probs then.

  25. lilsis76
    • 2 years ago
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    haha okay, hey where are you that the time is night?

  26. lilsis76
    • 2 years ago
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    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√

  27. gezimbasha
    • 2 years ago
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    I live in Kosovo, which is in Europe. I need some clarification. Which 2 are you asking about?

  28. lilsis76
    • 2 years ago
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    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

  29. lilsis76
    • 2 years ago
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    okay i dont know if u could see my last typing. but ill draw what i have gotten as an answer.

  30. gezimbasha
    • 2 years ago
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    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|

  31. lilsis76
    • 2 years ago
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    |dw:1348610123789:dw|

  32. lilsis76
    • 2 years ago
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    |dw:1348610422667:dw|

  33. lilsis76
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    okay im done, is that right tho?

  34. gezimbasha
    • 2 years ago
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    Hmm hang on a minute, ill solve the whole problem and check yoru answers one by one.

  35. lilsis76
    • 2 years ago
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    okay. haha oh man, i thought you had fallen asleep for a sec. thank u for waitin for me to finish

  36. gezimbasha
    • 2 years ago
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    lol, no dont worry, i'm awake. i will help you solve this prob completely.

  37. lilsis76
    • 2 years ago
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    thank you

  38. gezimbasha
    • 2 years ago
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    Have you tried solving the third triangle or the middle triangle?

  39. lilsis76
    • 2 years ago
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    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

  40. lilsis76
    • 2 years ago
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    But, i think Im going to u

  41. lilsis76
    • 2 years ago
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    haha but i think im going to use a=1 for AC=1 because i need something to use as an opposite

  42. gezimbasha
    • 2 years ago
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    Ok, here's what we have then..|dw:1348611838196:dw|

  43. lilsis76
    • 2 years ago
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    YA, thats what i have so far, haha but i forgot to place the 1 for x in the far right for length

  44. gezimbasha
    • 2 years ago
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    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\[AB=\frac{2}{\sqrt{3}}\]

  45. lilsis76
    • 2 years ago
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    |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?

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

  47. lilsis76
    • 2 years ago
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    yes

  48. lilsis76
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    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

  49. lilsis76
    • 2 years ago
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    AH! sorry for saying sorry

  50. gezimbasha
    • 2 years ago
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    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.

  51. lilsis76
    • 2 years ago
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    okay, well right now im trying to figure out which is best, so im going to stick with |dw:1348612826468:dw|

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

  53. lilsis76
    • 2 years ago
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    how would u multiply |dw:1348612939771:dw|

  54. lilsis76
    • 2 years ago
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    |dw:1348613038024:dw|

  55. lilsis76
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    |dw:1348613098124:dw|

  56. gezimbasha
    • 2 years ago
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    Ok, hang on. You do know SOH CAH TOA right? That will help you alot in trigonometry.

  57. lilsis76
    • 2 years ago
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    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

  58. gezimbasha
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    |dw:1348613426139:dw|

  59. lilsis76
    • 2 years ago
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    okay, thank you, i thought it would be a 4 then squ rt 6, thank you for clarifying

  60. lilsis76
    • 2 years ago
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    okay but is that the answer for the side .....A-B? did i do it ?

  61. gezimbasha
    • 2 years ago
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    Hmm, how did you get to that multiplication? You're close tho...

  62. lilsis76
    • 2 years ago
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    dang im always close. ill draw it.

  63. lilsis76
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    |dw:1348613700910:dw|

  64. lilsis76
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    |dw:1348613888683:dw|

  65. gezimbasha
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    Yes but you dont have a multiplication|dw:1348613976437:dw|

  66. lilsis76
    • 2 years ago
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    why is the X over the AE length?

  67. gezimbasha
    • 2 years ago
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    Because you are using sin which means opposite over hypotenuse.

  68. lilsis76
    • 2 years ago
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    |dw:1348614388796:dw| I rotated it, i think its confusing me when its laying down.

  69. lilsis76
    • 2 years ago
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    |dw:1348614492175:dw|

  70. lilsis76
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    Okay i see it Sin= Opp/hyp.

  71. gezimbasha
    • 2 years ago
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    Yes, but you have two 45, so you need to use 90-45-45.

  72. lilsis76
    • 2 years ago
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    AH haha sorry hold on

  73. lilsis76
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    |dw:1348614636423:dw|

  74. gezimbasha
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    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.

  75. lilsis76
    • 2 years ago
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    I got 4 ----- sq.rt. 6

  76. lilsis76
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    |dw:1348614899720:dw|

  77. lilsis76
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    |dw:1348614944183:dw|

  78. lilsis76
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    this is what im getting, im so confused

  79. gezimbasha
    • 2 years ago
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    |dw:1348615010146:dw| Close again haha :))

  80. gezimbasha
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    Divide the other way around.

  81. lilsis76
    • 2 years ago
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    oh... okay i got that. Now i see how you told me I was close. I had to make it a fraction

  82. lilsis76
    • 2 years ago
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    so that 2 / sq.rt 6 is lengths Ab & Ae?

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

  84. lilsis76
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    |dw:1348615274270:dw|

  85. lilsis76
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    |dw:1348615336270:dw| okay let me see the next part

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

  87. lilsis76
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    okay

  88. lilsis76
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    |dw:1348615597032:dw|

  89. gezimbasha
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    Perfect! Calculate that and show result..

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

  91. lilsis76
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    |dw:1348615813336:dw|

  92. lilsis76
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    then im stuck

  93. lilsis76
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    do i multiply the numerator and denomenator by sq. rt. 6? if I do I get: 2sqrt 6 / 3sqrt 6

  94. gezimbasha
    • 2 years ago
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    \[\tan(60)=\sqrt{3}\]And use this,\[x\sqrt{3}=\frac{2}{\sqrt{6}}\]

  95. lilsis76
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    okay let me look at it again

  96. lilsis76
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    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

  97. gezimbasha
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    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 :))

  98. lilsis76
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    so dang............I put it wrong ROAR! sorry

  99. lilsis76
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    okay so where did the x come from when u put|dw:1348616685558:dw|

  100. gezimbasha
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    |dw:1348616752298:dw|

  101. lilsis76
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    wait, is the X there because we're trying to find X right?

  102. lilsis76
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    okay |dw:1348617331171:dw|

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

  104. lilsis76
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    |dw:1348617643327:dw|

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

  106. lilsis76
    • 2 years ago