Use the Law of Sines to solve for all possible triangles tha

Use the Law of Sines to solve for all possible triangles that satisfy the given conditions. (If an answer does not exist, enter DNE Round your answers to one decimal place. Below, enter your answers so that A_1, is smaller than A_2 )

Solution

b = 46 , c = 43 , <C = 36^o

using law of cosines a/sinA = b/sinB = c/sinC

c/sinC = 43 / sin(36^o)

==> c/sinC = 73.156

==> b/ sinB = 73.156

==> sinB = b/ 73.156

==> sinB = 46/73.156

==> sinB = 0.629

==> B = sin^-1(0.629) = 38.961 , (180 - 38.961)      since sine function is positive in I and II quadrants

==> B = 38.961^o , 141.039^o

Sum of angles in a triangle = 180^o

==> A + B + C = 180^o

==> A = (180^o - B - C)

for B = 38.961^o

==> A = (180^o - 38.961 - 36)

==> A = 105.039^o

for B = 141.039^o

==> A = (180^o - 141.039 - 36)

==> A = 2.961^o

a/sinA = 73.156

==> a = 73.156(sinA)

for A = 105.039^o

==> a = 73.156(sin105.039^o)

==> a = 70.65

for A = 2.961^o

==> a = 73.156(sin2.961^o)

==> a = 3.779

Hence for smaller A :

<A = 2.961^o , <B = 141.039^o , a = 3.779

for larger A :

<A = 105.039^o , <B = 38.961^o , a = 70.65