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10 questions from this quiz
Any triangle
Opposite vertex AAA
sinAa=sinBb\frac{\sin A}{a} = \frac{\sin B}{b}asinA=bsinB
asinA=bsinB\frac{a}{\sin A} = \frac{b}{\sin B}sinAa=sinBb
AAS or ASA
SOH CAH TOA and Pythagoras' theorem
Two possible angle solutions exist
115°115°115°
Check angles sum to 180°180°180°
Only round the final answer
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