Question

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

5.) Given parts:
A= 78°48' b= 726 c= 938

Answer 1

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

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

$a / sin\angle A = b/sin\angle B$

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

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

therefore $\angle B < 90\degree,$

$\angle B = 28.59\degree = 28\degree36'$

$\angle C = 180\degree -\angle A -\angle C = 180\degree - 28\degree36' - 78\degree 48'$

$\angle C = 72\degree36'$

answer :

$a =1650.168\\\angle C = 72\degree36'\\\angle B = 28\degree36'$