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BecomeMyFan=D

  • 3 years ago

One corner of a triangle has a 60° angle and the length of the two adjacent sides are in ratio 1 : 3. Calculate the angles of the other triangle corners (0,1°:s precision, 1 point / correct angle).

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  1. BecomeMyFan=D
    • 3 years ago
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    :)

  2. BecomeMyFan=D
    • 3 years ago
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    I think I have to first find the size of the opposite side to 60 degree angle right? i used cosine rule to do so and found 2.65, however I dont know what to do now

  3. BecomeMyFan=D
    • 3 years ago
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    2.645, not 2.65

  4. lokisan
    • 3 years ago
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    Yes, you're on the right track. Now use sine rule to find your angles...but, note you only have to use it once because the sum of the angles of a plane triangle is 180 degrees (and by the time you apply the sine rule, you'll have two of them).

  5. BecomeMyFan=D
    • 3 years ago
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    ok, so then I just sub the 2 angs form 180 and.... right?

  6. lokisan
    • 3 years ago
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    First find one of the other angles.

  7. BecomeMyFan=D
    • 3 years ago
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    ok I just did

  8. BecomeMyFan=D
    • 3 years ago
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    79.2

  9. BecomeMyFan=D
    • 3 years ago
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    something is not right

  10. lokisan
    • 3 years ago
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    I think you found the sum of 60 and one of the other ones, 19.1.

  11. BecomeMyFan=D
    • 3 years ago
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    is 2.645 right for the third side?

  12. lokisan
    • 3 years ago
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    \[\frac{a}{\sin \theta}=\frac{\sqrt{7} a}{\sin 60^o}\]

  13. BecomeMyFan=D
    • 3 years ago
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    i found one angle to be 19.1

  14. lokisan
    • 3 years ago
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    Your third side is right, assuming, in your ratio of 1:3, the side with the '1' has unit length. A more general assumption is that this side has length 'a', so that the other side has length 3a.

  15. lokisan
    • 3 years ago
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    The side opposite 60 degrees would then have length,\[\sqrt{7}a\]

  16. lokisan
    • 3 years ago
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    Yes, you're right.

  17. BecomeMyFan=D
    • 3 years ago
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    how?

  18. BecomeMyFan=D
    • 3 years ago
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    oh, yeah, i get it

  19. lokisan
    • 3 years ago
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    How? Cosine rule, like you used before (if you're asking how I get sqrt(7)a?).

  20. lokisan
    • 3 years ago
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    Once the other angle's found, you're done.

  21. BecomeMyFan=D
    • 3 years ago
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    ok, subing 19.1+60 from 180 gives 100.9

  22. BecomeMyFan=D
    • 3 years ago
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    but

  23. BecomeMyFan=D
    • 3 years ago
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    when i try to find the last angle using sine rule, it gives me 79.2

  24. BecomeMyFan=D
    • 3 years ago
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    which is right?

  25. lokisan
    • 3 years ago
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    Yeah, you're right...

  26. BecomeMyFan=D
    • 3 years ago
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    i did something wrong then

  27. lokisan
    • 3 years ago
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    Bizarre

  28. BecomeMyFan=D
    • 3 years ago
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    :)

  29. lokisan
    • 3 years ago
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    In mathematics, if you end up with a contradiction, it's because one of your assumptions is wrong...so what assumption(s) were made?

  30. BecomeMyFan=D
    • 3 years ago
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    the ratio of sides 1 to 3, the sine rule, the cosine rule and the 60 angle

  31. lokisan
    • 3 years ago
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    Does your question say "the length of *the* two adjacent sides" or "the length of two adjacent sides"?

  32. BecomeMyFan=D
    • 3 years ago
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    the

  33. BecomeMyFan=D
    • 3 years ago
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    whats the difference?

  34. lokisan
    • 3 years ago
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    'the' restricts your choice

  35. BecomeMyFan=D
    • 3 years ago
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    it says adjacent

  36. BecomeMyFan=D
    • 3 years ago
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    but i still dont see what I did wrong :(

  37. lokisan
    • 3 years ago
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    Omg, it just dawned on me - there are two possible solutions!

  38. lokisan
    • 3 years ago
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    One set will have (60, 9.1 and the other) and (60, 79.2, other)

  39. lokisan
    • 3 years ago
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    One assumption was missed - that there is only one solution.

  40. lokisan
    • 3 years ago
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    I constructed a triangle on the description in GeoGebra and have (60,9.1,100.9) as one solution.

  41. BecomeMyFan=D
    • 3 years ago
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    :) so there is just one solution?

  42. lokisan
    • 3 years ago
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    No, it's coming about because the arc of sine (in on rotation) has TWO angles whose sine is positive and the same value.

  43. lokisan
    • 3 years ago
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    in *one* rotation

  44. BecomeMyFan=D
    • 3 years ago
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    so, is it then imposible to find both of the angles correctly?

  45. lokisan
    • 3 years ago
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    No, it's just that you collect both possible solutions from the arc of sine on each angle, and then put them in the appropriate combinations (i.e. so they add to 180).

  46. BecomeMyFan=D
    • 3 years ago
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    so, how do I answer the quesstion in a work book? which set of angles do I choose?

  47. lokisan
    • 3 years ago
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    ok..I'm trying to figure out a way to do it so that it doesn't take a millennium to type.

  48. sstarica
    • 3 years ago
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    wouldn't the other angle be 30 and the other 90?

  49. sstarica
    • 3 years ago
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    since the other is 60?

  50. lokisan
    • 3 years ago
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    \[\frac{\sin \alpha}{3a}=\frac{\sin 60}{\sqrt{7}a}\rightarrow \sin \alpha = \frac{3\sqrt{3}}{2\sqrt{7}}\]

  51. lokisan
    • 3 years ago
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    Now

  52. lokisan
    • 3 years ago
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    in the arc of 360 degrees, \[\alpha = \sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\] degrees AND\[\alpha = 180- \sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\]

  53. lokisan
    • 3 years ago
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    where \[\sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\approx 79.1^o\]

  54. lokisan
    • 3 years ago
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    So the possible angles you get as solutions when considering this combination of sides is\[79.1^o, 100.9^o\]

  55. lokisan
    • 3 years ago
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    Try taking the sine of both of them in your calculator.

  56. lokisan
    • 3 years ago
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    So you have NO CHOICE but to accept two solutions for this first combination of sides.

  57. BecomeMyFan=D
    • 3 years ago
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    ok I get it

  58. BecomeMyFan=D
    • 3 years ago
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    THANKS ALOT

  59. lokisan
    • 3 years ago
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    hang on...

  60. BecomeMyFan=D
    • 3 years ago
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    ok

  61. sstarica
    • 3 years ago
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    just one question, is it a right triangle? ^_^

  62. BecomeMyFan=D
    • 3 years ago
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    the question does not specify

  63. sstarica
    • 3 years ago
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    because if so, then the other 2 angles are 90 and 30, otherwise , loki's answer is true :)

  64. lokisan
    • 3 years ago
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    Wait wait wait...

  65. BecomeMyFan=D
    • 3 years ago
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    :)

  66. lokisan
    • 3 years ago
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    When you take the arc of sine here you'll get two solutions for each angle, which are algebraically correct.

  67. sstarica
    • 3 years ago
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    lol, alright

  68. lokisan
    • 3 years ago
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    So, calling those angles alpha and beta, you have\[\alpha \in \left\{ 79.1,100.9 \right\}\]and\[\beta \in \left\{ 19.1, 160.9 \right\}\]

  69. lokisan
    • 3 years ago
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    BUT

  70. lokisan
    • 3 years ago
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    only certain combinations of those angles will give you a true conclusion here, since you have an additional constraint: that the angles\[\alpha, \beta, 60^o\]must sum to 180 degrees.

  71. lokisan
    • 3 years ago
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    So you have to find those combinations elements from the set of alpha and beta that will allow you to get 180 (after you add 60 to them). You see?

  72. sstarica
    • 3 years ago
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    then alpha is 100.9 and beta is 19.1 ?

  73. lokisan
    • 3 years ago
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    There are four possible combinations, but only ONE combination works

  74. lokisan
    • 3 years ago
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    Yes

  75. sstarica
    • 3 years ago
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    and he strikes again~ lol

  76. lokisan
    • 3 years ago
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    and i;m drunk - came back from a dinner

  77. sstarica
    • 3 years ago
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    but you've answered it , weirdly in such a state ._. did you get it andy?

  78. lokisan
    • 3 years ago
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    It's similar to a situation when you have to solve the quadratic equation, which might have something to do with length, and you get two solutions - one positive, one negative. You apply an additional constraint (i.e. physical measurements aren't negative) and discard one of the solutions. Here, the constraint is that you can only take those angles whose sum will be 180.

  79. BecomeMyFan=D
    • 3 years ago
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    yes

  80. BecomeMyFan=D
    • 3 years ago
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    and, lokisan, you dont sound like drunk

  81. sstarica
    • 3 years ago
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    maybe half drunk ~

  82. lokisan
    • 3 years ago
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    \[\left\{ \alpha, \beta|\alpha + \beta +60^o=180^o , \alpha \in \left\{ 79.1,100.9 \right\},\beta \in \left\{ 19.1,160.9 \right\} \right\}\]

  83. sstarica
    • 3 years ago
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    discrete mathematics ^^" ...

  84. lokisan
    • 3 years ago
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    yes...so the above set is 19.1 and 100.9.

  85. lokisan
    • 3 years ago
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    Phew

  86. lokisan
    • 3 years ago
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    Good question.

  87. sstarica
    • 3 years ago
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    LOL

  88. lokisan
    • 3 years ago
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    Happy with that BMFan?

  89. BecomeMyFan=D
    • 3 years ago
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    yeah, it is hard, but i am damn happy

  90. lokisan
    • 3 years ago
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    awesome

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