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Triangle ABC is such that its $3$ angles $A,B$ and $C$ are in the ratio $3:4:5$. What is the ratio of its $3$ sides $BC:CA:AB$?

I tried using the sine law but it was of no use, please give me help

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    For simplicity, scale the triangle so that one side has length 1. Then use the sine law. – 2017-01-27
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    @quasi how can i scale it if theyrre variables.. how can i sclae BC to 1???? – 2017-01-27
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    this is what i did exactly:::::: Bc/Ca=3/4, BC/AB=3/5, CA/AB=4/5 however when i tried solving i got AB=AB ... etc – 2017-01-27
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    The angles hae the ratios 3:4:5, not the sides. Suppose the smallest angle is x. Then 3x + 4x + 5x = 180. – 2017-01-27
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    Since the given information only determines the angles, the triangle is determined only up to similarity, hence you can freely choose a value for one of the sides without changing the ratios of the sides. – 2017-01-27

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HINT: The angles are $45, 60, 75$ degrees. The sine values for these angles are well-known, so you can answer your question using the Sine's Law.

We have $BC:CA:AB = \frac{BC}{AB} : \frac{CA}{AB} : 1 = \frac{\sin(A)}{\sin(C)} : \frac{\sin(B)}{\sin(C)} : 1$

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    wait.. how do u know the values of the angles??????????? – 2017-01-27
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    @exchangehelpforuni: angles are $3k,4k,5k$ and the sum is $180°$ – 2017-01-27
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    @exchangehelpforuni Let $A = x$. Then we have $B = \frac 43 A = \frac 43 x$. Similarly $C = \frac 53 x$. Then what is the sum of all angles in a triangle? – 2017-01-27
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    @stefan4024 thanks makes sense – 2017-01-27
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    @stefan4024 i used these angles and using the sine law i had 3 different equations with 3 different varibales, however the variables tend to cancel out leaving me with no answer, do u know what i should of – 2017-01-27
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    @exchangehelpforuni I'll add some further explanation in the answer. – 2017-01-27
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    @stefan4024 waittttttttttttt – 2017-01-27
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    @stefan204 i figured it out. i picked BC to have side length 1 and then calcuated CA: 3/sqrt6, AB 1/sqrt3 -1... however now how can i determine the ratios, as im doing an online test which has mutliple choice asnwers – 2017-01-27
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    so do i just use the process of elimination – 2017-01-27
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    @exchangehelpforuni You can check the explanation I added. – 2017-01-27
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    @stefan4024 why is the thing i did wrong?? why should i insert he ratios from the beginig, why cant i just pick 1 to be a side – 2017-01-27
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    @exchangehelpforuni Actually you can do it. Then you will find an exact values for the rest two sides. In fact what I did was doing implicitly the very same thing. Pick $AB = 1$ and you will get the same. – 2017-01-27
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    @stefan4024 ok now what do i do next?? – 2017-01-27
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    @exchangehelpforuni Then you have $BC = \frac{\sin(A)}{\sin(C)} \cdot AB = \frac{\sin(A)}{\sin(C)}$. Similarly $AC = \frac{\sin(B)}{\sin(C)} \cdot AB = \frac{\sin(B)}{\sin(C)}$ – 2017-01-27
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    @stefan4024 these r the choices: A. 3^1/2 : 4^1/2 : 5^1/2 B. (1 + 2^1/2) : 6^1/2 : (1 + 3^1/2) C. 2 : 6^1/2 : (1 + 3^1/2) D. 3 : 4 : 5 E. 6^1/2 : (1 + 3^1/2) : (2^1/2 + 3^1/2) – 2017-01-27
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    @Stefan4024 iknow i calucated and got all the values, however how can i match it and make it a ratio, similar to what the question is askingn ? – 2017-01-27
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    Let us [continue this discussion in chat](http://chat.stackexchange.com/rooms/52624/discussion-between-exchangehelpforuni-and-stefan4024). – 2017-01-27