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 2 years ago
What are the two cases in which the Laws of Sines can be applied to solve a nonright triangle?
Case I. You know the measures of two angles and any side of the triangle.
Case II. You know the measures of two sides and an angle opposite one of the two known sides.
Case III. You know the measures of two sides and the included angle.
Case IV. You know the measures of all three sides.
A. Case I and Case III
B. Case II and Case IV
C. Case III and Case IV
D. Case I and Case II
 2 years ago
What are the two cases in which the Laws of Sines can be applied to solve a nonright triangle? Case I. You know the measures of two angles and any side of the triangle. Case II. You know the measures of two sides and an angle opposite one of the two known sides. Case III. You know the measures of two sides and the included angle. Case IV. You know the measures of all three sides. A. Case I and Case III B. Case II and Case IV C. Case III and Case IV D. Case I and Case II

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SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1The law of sines let's you set up a ratio a/Sin(A) = b/Sin(B) now, you can only set up that ratio if you have a side and its opposite angle. Go through those cases and see which ones describe a situation like that where you know an angle and the side opposite it.

Abbie23
 2 years ago
Best ResponseYou've already chosen the best response.1Oooo, okay thank you @SmoothMath :)

thushananth01
 2 years ago
Best ResponseYou've already chosen the best response.0Bro @SmoothMath for the case I) they say there are 2 angles u then obviously will know the other one...so case 1 is correct...and the case 111) they have mention the included angle...

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Case 1 is correct. Yes. Case 3 looks like this: dw:1335238727974:dw If I knew a and A, then I could write a/sin(A) if I knew b and B, then I could write b/sin(B) if I knew c and C, then I could write c/sin(C)

Abbie23
 2 years ago
Best ResponseYou've already chosen the best response.1I think you guys both helped me!

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1What a sweetheart. :3

SmoothMath
 2 years ago
Best ResponseYou've already chosen the best response.1Haha how could I be complaining about you being sweet?
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