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Using the inverse sine function 2000-05-31
Nelson Rothermel pose la question :
This has me completely baffled. I have to use the laws of sine or cosine to find the angles of a triangle when I have 3 sides, so I can't go 180-x-y when I have 2 angles. Now, I have a triangle with values of 3, 7, and 9. Here are the steps I used (A,B,C are angles; a,b,c are opposite sides):

angle A (16.1951 degrees): cos-1*((b2+c2-a2)/(2*b*c))
angle B (40.6011 degrees): sin-1*(b*sin(A)/a)
angle C (56.7962 degrees): sin-1*(c*sin(A)/a)

If you notice, A+B+C does not equal 180. According to the book, A and B are correct, but C is supposed to be 123.2038 degrees. Why doesn't it work???


Harley Weston lui répond.
 
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