## BecomeMyFan=D Group Title 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). 3 years ago 3 years ago

1. BecomeMyFan=D

:)

2. BecomeMyFan=D

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

2.645, not 2.65

4. lokisan

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

ok, so then I just sub the 2 angs form 180 and.... right?

6. lokisan

First find one of the other angles.

7. BecomeMyFan=D

ok I just did

8. BecomeMyFan=D

79.2

9. BecomeMyFan=D

something is not right

10. lokisan

I think you found the sum of 60 and one of the other ones, 19.1.

11. BecomeMyFan=D

is 2.645 right for the third side?

12. lokisan

$\frac{a}{\sin \theta}=\frac{\sqrt{7} a}{\sin 60^o}$

13. BecomeMyFan=D

i found one angle to be 19.1

14. lokisan

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

The side opposite 60 degrees would then have length,$\sqrt{7}a$

16. lokisan

Yes, you're right.

17. BecomeMyFan=D

how?

18. BecomeMyFan=D

oh, yeah, i get it

19. lokisan

How? Cosine rule, like you used before (if you're asking how I get sqrt(7)a?).

20. lokisan

Once the other angle's found, you're done.

21. BecomeMyFan=D

ok, subing 19.1+60 from 180 gives 100.9

22. BecomeMyFan=D

but

23. BecomeMyFan=D

when i try to find the last angle using sine rule, it gives me 79.2

24. BecomeMyFan=D

which is right?

25. lokisan

Yeah, you're right...

26. BecomeMyFan=D

i did something wrong then

27. lokisan

Bizarre

28. BecomeMyFan=D

:)

29. lokisan

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

the ratio of sides 1 to 3, the sine rule, the cosine rule and the 60 angle

31. lokisan

Does your question say "the length of *the* two adjacent sides" or "the length of two adjacent sides"?

32. BecomeMyFan=D

the

33. BecomeMyFan=D

whats the difference?

34. lokisan

35. BecomeMyFan=D

36. BecomeMyFan=D

but i still dont see what I did wrong :(

37. lokisan

Omg, it just dawned on me - there are two possible solutions!

38. lokisan

One set will have (60, 9.1 and the other) and (60, 79.2, other)

39. lokisan

One assumption was missed - that there is only one solution.

40. lokisan

I constructed a triangle on the description in GeoGebra and have (60,9.1,100.9) as one solution.

41. BecomeMyFan=D

:) so there is just one solution?

42. lokisan

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

in *one* rotation

44. BecomeMyFan=D

so, is it then imposible to find both of the angles correctly?

45. lokisan

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

so, how do I answer the quesstion in a work book? which set of angles do I choose?

47. lokisan

ok..I'm trying to figure out a way to do it so that it doesn't take a millennium to type.

48. sstarica

wouldn't the other angle be 30 and the other 90?

49. sstarica

since the other is 60?

50. lokisan

$\frac{\sin \alpha}{3a}=\frac{\sin 60}{\sqrt{7}a}\rightarrow \sin \alpha = \frac{3\sqrt{3}}{2\sqrt{7}}$

51. lokisan

Now

52. lokisan

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

where $\sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\approx 79.1^o$

54. lokisan

So the possible angles you get as solutions when considering this combination of sides is$79.1^o, 100.9^o$

55. lokisan

Try taking the sine of both of them in your calculator.

56. lokisan

So you have NO CHOICE but to accept two solutions for this first combination of sides.

57. BecomeMyFan=D

ok I get it

58. BecomeMyFan=D

THANKS ALOT

59. lokisan

hang on...

60. BecomeMyFan=D

ok

61. sstarica

just one question, is it a right triangle? ^_^

62. BecomeMyFan=D

the question does not specify

63. sstarica

because if so, then the other 2 angles are 90 and 30, otherwise , loki's answer is true :)

64. lokisan

Wait wait wait...

65. BecomeMyFan=D

:)

66. lokisan

When you take the arc of sine here you'll get two solutions for each angle, which are algebraically correct.

67. sstarica

lol, alright

68. lokisan

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

BUT

70. lokisan

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

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

then alpha is 100.9 and beta is 19.1 ?

73. lokisan

There are four possible combinations, but only ONE combination works

74. lokisan

Yes

75. sstarica

and he strikes again~ lol

76. lokisan

and i;m drunk - came back from a dinner

77. sstarica

but you've answered it , weirdly in such a state ._. did you get it andy?

78. lokisan

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

yes

80. BecomeMyFan=D

and, lokisan, you dont sound like drunk

81. sstarica

maybe half drunk ~

82. lokisan

$\left\{ \alpha, \beta|\alpha + \beta +60^o=180^o , \alpha \in \left\{ 79.1,100.9 \right\},\beta \in \left\{ 19.1,160.9 \right\} \right\}$

83. sstarica

discrete mathematics ^^" ...

84. lokisan

yes...so the above set is 19.1 and 100.9.

85. lokisan

Phew

86. lokisan

Good question.

87. sstarica

LOL

88. lokisan

Happy with that BMFan?

89. BecomeMyFan=D

yeah, it is hard, but i am damn happy

90. lokisan

awesome