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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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BecomeMyFan=D
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:)
BecomeMyFan=D
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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
BecomeMyFan=D
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2.645, not 2.65
lokisan
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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).
BecomeMyFan=D
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ok, so then I just sub the 2 angs form 180 and....
right?
lokisan
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First find one of the other angles.
BecomeMyFan=D
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ok I just did
BecomeMyFan=D
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79.2
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something is not right
lokisan
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I think you found the sum of 60 and one of the other ones, 19.1.
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is 2.645 right for the third side?
lokisan
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\[\frac{a}{\sin \theta}=\frac{\sqrt{7} a}{\sin 60^o}\]
BecomeMyFan=D
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i found one angle to be 19.1
lokisan
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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.
lokisan
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The side opposite 60 degrees would then have length,\[\sqrt{7}a\]
lokisan
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Yes, you're right.
BecomeMyFan=D
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how?
BecomeMyFan=D
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oh, yeah, i get it
lokisan
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How? Cosine rule, like you used before (if you're asking how I get sqrt(7)a?).
lokisan
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Once the other angle's found, you're done.
BecomeMyFan=D
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ok, subing 19.1+60 from 180 gives 100.9
BecomeMyFan=D
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but
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when i try to find the last angle using sine rule, it gives me 79.2
BecomeMyFan=D
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which is right?
lokisan
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Yeah, you're right...
BecomeMyFan=D
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i did something wrong then
lokisan
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Bizarre
BecomeMyFan=D
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:)
lokisan
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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?
BecomeMyFan=D
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the ratio of sides 1 to 3, the sine rule, the cosine rule and the 60 angle
lokisan
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Does your question say "the length of *the* two adjacent sides" or "the length of two adjacent sides"?
BecomeMyFan=D
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the
BecomeMyFan=D
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whats the difference?
lokisan
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'the' restricts your choice
BecomeMyFan=D
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it says adjacent
BecomeMyFan=D
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but i still dont see what I did wrong :(
lokisan
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Omg, it just dawned on me - there are two possible solutions!
lokisan
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One set will have (60, 9.1 and the other) and (60, 79.2, other)
lokisan
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One assumption was missed - that there is only one solution.
lokisan
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I constructed a triangle on the description in GeoGebra and have (60,9.1,100.9) as one solution.
BecomeMyFan=D
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:) so there is just one solution?
lokisan
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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.
lokisan
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in *one* rotation
BecomeMyFan=D
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so, is it then imposible to find both of the angles correctly?
lokisan
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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).
BecomeMyFan=D
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so, how do I answer the quesstion in a work book? which set of angles do I choose?
lokisan
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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.
sstarica
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wouldn't the other angle be 30 and the other 90?
sstarica
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since the other is 60?
lokisan
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\[\frac{\sin \alpha}{3a}=\frac{\sin 60}{\sqrt{7}a}\rightarrow \sin \alpha = \frac{3\sqrt{3}}{2\sqrt{7}}\]
lokisan
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Now
lokisan
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in the arc of 360 degrees, \[\alpha = \sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\] degrees AND\[\alpha = 180- \sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\]
lokisan
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where \[\sin^{-1}\frac{3\sqrt{3}}{2\sqrt{7}}\approx 79.1^o\]
lokisan
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So the possible angles you get as solutions when considering this combination of sides is\[79.1^o, 100.9^o\]
lokisan
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Try taking the sine of both of them in your calculator.
lokisan
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So you have NO CHOICE but to accept two solutions for this first combination of sides.
BecomeMyFan=D
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ok
I get it
BecomeMyFan=D
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THANKS ALOT
lokisan
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hang on...
BecomeMyFan=D
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ok
sstarica
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just one question, is it a right triangle? ^_^
BecomeMyFan=D
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the question does not specify
sstarica
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because if so, then the other 2 angles are 90 and 30, otherwise , loki's answer is true :)
lokisan
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Wait wait wait...
BecomeMyFan=D
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:)
lokisan
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When you take the arc of sine here you'll get two solutions for each angle, which are algebraically correct.
sstarica
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lol, alright
lokisan
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So, calling those angles alpha and beta, you have\[\alpha \in \left\{ 79.1,100.9 \right\}\]and\[\beta \in \left\{ 19.1, 160.9 \right\}\]
lokisan
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BUT
lokisan
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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.
lokisan
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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?
sstarica
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then alpha is 100.9 and beta is 19.1 ?
lokisan
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There are four possible combinations, but only ONE combination works
lokisan
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Yes
sstarica
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and he strikes again~ lol
lokisan
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and i;m drunk - came back from a dinner
sstarica
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but you've answered it , weirdly in such a state ._. did you get it andy?
lokisan
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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.
BecomeMyFan=D
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yes
BecomeMyFan=D
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and, lokisan, you dont sound like drunk
sstarica
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maybe half drunk ~
lokisan
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\[\left\{ \alpha, \beta|\alpha + \beta +60^o=180^o , \alpha \in \left\{ 79.1,100.9 \right\},\beta \in \left\{ 19.1,160.9 \right\} \right\}\]
sstarica
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discrete mathematics ^^" ...
lokisan
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yes...so the above set is 19.1 and 100.9.
lokisan
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Phew
lokisan
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Good question.
sstarica
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LOL
lokisan
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Happy with that BMFan?
BecomeMyFan=D
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yeah, it is hard, but i am damn happy
lokisan
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awesome