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 Group Title

:)

2. BecomeMyFan=D Group Title

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 Group Title

2.645, not 2.65

4. lokisan Group Title

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 Group Title

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

6. lokisan Group Title

First find one of the other angles.

7. BecomeMyFan=D Group Title

ok I just did

8. BecomeMyFan=D Group Title

79.2

9. BecomeMyFan=D Group Title

something is not right

10. lokisan Group Title

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

11. BecomeMyFan=D Group Title

is 2.645 right for the third side?

12. lokisan Group Title

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

13. BecomeMyFan=D Group Title

i found one angle to be 19.1

14. lokisan Group Title

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 Group Title

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

16. lokisan Group Title

Yes, you're right.

17. BecomeMyFan=D Group Title

how?

18. BecomeMyFan=D Group Title

oh, yeah, i get it

19. lokisan Group Title

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

20. lokisan Group Title

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

21. BecomeMyFan=D Group Title

ok, subing 19.1+60 from 180 gives 100.9

22. BecomeMyFan=D Group Title

but

23. BecomeMyFan=D Group Title

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

24. BecomeMyFan=D Group Title

which is right?

25. lokisan Group Title

Yeah, you're right...

26. BecomeMyFan=D Group Title

i did something wrong then

27. lokisan Group Title

Bizarre

28. BecomeMyFan=D Group Title

:)

29. lokisan Group Title

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 Group Title

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

31. lokisan Group Title

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

32. BecomeMyFan=D Group Title

the

33. BecomeMyFan=D Group Title

whats the difference?

34. lokisan Group Title

35. BecomeMyFan=D Group Title

36. BecomeMyFan=D Group Title

but i still dont see what I did wrong :(

37. lokisan Group Title

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

38. lokisan Group Title

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

39. lokisan Group Title

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

40. lokisan Group Title

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

41. BecomeMyFan=D Group Title

:) so there is just one solution?

42. lokisan Group Title

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 Group Title

in *one* rotation

44. BecomeMyFan=D Group Title

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

45. lokisan Group Title

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 Group Title

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

47. lokisan Group Title

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

48. sstarica Group Title

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

49. sstarica Group Title

since the other is 60?

50. lokisan Group Title

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

51. lokisan Group Title

Now

52. lokisan Group Title

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 Group Title

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

54. lokisan Group Title

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

55. lokisan Group Title

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

56. lokisan Group Title

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

57. BecomeMyFan=D Group Title

ok I get it

58. BecomeMyFan=D Group Title

THANKS ALOT

59. lokisan Group Title

hang on...

60. BecomeMyFan=D Group Title

ok

61. sstarica Group Title

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

62. BecomeMyFan=D Group Title

the question does not specify

63. sstarica Group Title

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

64. lokisan Group Title

Wait wait wait...

65. BecomeMyFan=D Group Title

:)

66. lokisan Group Title

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

67. sstarica Group Title

lol, alright

68. lokisan Group Title

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 Group Title

BUT

70. lokisan Group Title

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 Group Title

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 Group Title

then alpha is 100.9 and beta is 19.1 ?

73. lokisan Group Title

There are four possible combinations, but only ONE combination works

74. lokisan Group Title

Yes

75. sstarica Group Title

and he strikes again~ lol

76. lokisan Group Title

and i;m drunk - came back from a dinner

77. sstarica Group Title

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

78. lokisan Group Title

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 Group Title

yes

80. BecomeMyFan=D Group Title

and, lokisan, you dont sound like drunk

81. sstarica Group Title

maybe half drunk ~

82. lokisan Group Title

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

83. sstarica Group Title

discrete mathematics ^^" ...

84. lokisan Group Title

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

85. lokisan Group Title

Phew

86. lokisan Group Title

Good question.

87. sstarica Group Title

LOL

88. lokisan Group Title

Happy with that BMFan?

89. BecomeMyFan=D Group Title

yeah, it is hard, but i am damn happy

90. lokisan Group Title

awesome