## anonymous 5 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).

1. anonymous

:)

2. anonymous

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. anonymous

2.645, not 2.65

4. anonymous

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. anonymous

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

6. anonymous

First find one of the other angles.

7. anonymous

ok I just did

8. anonymous

79.2

9. anonymous

something is not right

10. anonymous

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

11. anonymous

is 2.645 right for the third side?

12. anonymous

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

13. anonymous

i found one angle to be 19.1

14. anonymous

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. anonymous

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

16. anonymous

Yes, you're right.

17. anonymous

how?

18. anonymous

oh, yeah, i get it

19. anonymous

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

20. anonymous

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

21. anonymous

ok, subing 19.1+60 from 180 gives 100.9

22. anonymous

but

23. anonymous

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

24. anonymous

which is right?

25. anonymous

Yeah, you're right...

26. anonymous

i did something wrong then

27. anonymous

Bizarre

28. anonymous

:)

29. anonymous

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. anonymous

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

31. anonymous

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

32. anonymous

the

33. anonymous

whats the difference?

34. anonymous

35. anonymous

36. anonymous

but i still dont see what I did wrong :(

37. anonymous

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

38. anonymous

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

39. anonymous

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

40. anonymous

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

41. anonymous

:) so there is just one solution?

42. anonymous

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. anonymous

in *one* rotation

44. anonymous

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

45. anonymous

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. anonymous

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

47. anonymous

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

48. anonymous

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

49. anonymous

since the other is 60?

50. anonymous

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

51. anonymous

Now

52. anonymous

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. anonymous

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

54. anonymous

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

55. anonymous

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

56. anonymous

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

57. anonymous

ok I get it

58. anonymous

THANKS ALOT

59. anonymous

hang on...

60. anonymous

ok

61. anonymous

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

62. anonymous

the question does not specify

63. anonymous

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

64. anonymous

Wait wait wait...

65. anonymous

:)

66. anonymous

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

67. anonymous

lol, alright

68. anonymous

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. anonymous

BUT

70. anonymous

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. anonymous

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. anonymous

then alpha is 100.9 and beta is 19.1 ?

73. anonymous

There are four possible combinations, but only ONE combination works

74. anonymous

Yes

75. anonymous

and he strikes again~ lol

76. anonymous

and i;m drunk - came back from a dinner

77. anonymous

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

78. anonymous

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. anonymous

yes

80. anonymous

and, lokisan, you dont sound like drunk

81. anonymous

maybe half drunk ~

82. anonymous

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

83. anonymous

discrete mathematics ^^" ...

84. anonymous

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

85. anonymous

Phew

86. anonymous

Good question.

87. anonymous

LOL

88. anonymous

Happy with that BMFan?

89. anonymous

yeah, it is hard, but i am damn happy

90. anonymous

awesome