Answer to Question #142867 in Trigonometry for Raffy

Question #142867
Solve the following triangles. Identify the case # in each given
triangle.

5.) Given parts:
A= 78°48' b= 726 c= 938
1
Expert's answer
2020-11-15T18:16:58-0500

a=b2+c22bccosAa = \sqrt{b^2 +c^2 -2b*c*cos\angle{A}}

a=7262+93822726938cos78°48=1650.168a = \sqrt{726^2 +938^2 -2*726*938*cos78\degree 48'} = 1650.168

a/sinA=b/sinBa / sin\angle A = b/sin\angle B

sinB=bsinA/a=726sin78.8°/1650.168=0.432sin\angle B = b*sin\angle A / a = 726*sin78.8\degree / 1650.168 = 0.432

if B>90°,\angle B > 90\degree, when C<11°12\angle C < 11\degree 12' and side the smallest, but the smallest side is b,

therefore B<90°,\angle B < 90\degree,

B=28.59°=28°36\angle B = 28.59\degree = 28\degree36'

C=180°AC=180°28°3678°48\angle C = 180\degree -\angle A -\angle C = 180\degree - 28\degree36' - 78\degree 48'

C=72°36\angle C = 72\degree36'

answer :

a=1650.168C=72°36B=28°36a =1650.168\\\angle C = 72\degree36'\\\angle B = 28\degree36'




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