11/10/06: Ambiguous case at the sine law
In todays lesson we learnt about the sine law, and how it can be used to determine angles in particular triangles. The basic diagram is shown below,
and it basically shows that the outside angle of a triangle is equal to 180 - the inside angle. So therefore, we can say that [tex]cos(angle)=\frac{x}{1}[/tex]
Example: Suppose Triangle ABC: Angle A = 60 Line AB = 30m and Line BC = 35
[tex]\frac{sin(60)}{35}=\frac{sinC}{30}[/tex]
[tex]c= {sin^{-1}}} \times \frac{30sin60}{35}[/tex] And now the unknown angle can either be 132.07 degrees, (which is impossible as the other angle is 60, adding up to more than the permitted 180 degrees), or the more plausible 47.93.
and it basically shows that the outside angle of a triangle is equal to 180 - the inside angle. So therefore, we can say that [tex]cos(angle)=\frac{x}{1}[/tex]
Example: Suppose Triangle ABC: Angle A = 60 Line AB = 30m and Line BC = 35
[tex]\frac{sin(60)}{35}=\frac{sinC}{30}[/tex]
[tex]c= {sin^{-1}}} \times \frac{30sin60}{35}[/tex] And now the unknown angle can either be 132.07 degrees, (which is impossible as the other angle is 60, adding up to more than the permitted 180 degrees), or the more plausible 47.93.
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