Refreshing my memory again, but with triangles?
Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles. A = 55°, a = 12, b = 14
How do I find the second one again? @phi
@satellite73
@surjithayer ?
|dw:1401740149666:dw|
I can figure the first one, but I can never remember how to fins the second one!
<C+55+ <B=180 if you know <B, then you can find <C
That's the second triangle???
where is the data for second triangle?
finding the two ambiguous triangles can be confusing. notice when you solve for angle B, you set up this ratio \[ \frac{\sin 55}{12}= \frac{\sin B}{14} \] multiply both sides by 14 to get \[ 14\cdot \frac{\sin 55}{12}= \sin B \\ \sin B = \frac{7}{6} \sin 55 \\ \sin B= 0.95568 \] You should always keep in mind the graph of the sin, between 0 and 180º (these are the angles that might be in a triangle). It looks like this: |dw:1401802865640:dw|
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