Question

12. Solve the following triangles using Law of Sines or Law of Cosines (round to nearest tenth when necessary and find all solutions) *Must show path/process/work for full credit": a. A-58 a. B- b=12

Answer 1

sin(B)/b = sin(A)/a

sin(B)/12 = sin(58)/a

a = 12(sin(58)/sin(B))

Now we can use the Law of Cosines to find the remaining sides of the triangle:

a^2 = b^2 + c^2 - 2bc*cos(A)

a^2 = 12^2 + c^2 - 2(12)(c)*cos(58)

c^2 - 24c*cos(58) + 144 - a^2 = 0

Using the quadratic formula, we get:

c = (24*cos(58) ± sqrt((24*cos(58))^2 - 4(1)(144 - a^2)))/2(1)

c = 12*cos(58) ± sqrt(144*cos(58)^2 - 4(144 - a^2))

c = 12*cos(58) ± sqrt(576*cos(58)^2 - 4a^2)

c = 12*cos(58) ± sqrt(576*(1 - sin(58)^2) - 4a^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 4a^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 4(12(sin(58)/sin(B)))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/sin(B))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(180 - A - B)))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(180 - 58 - B)))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - B)))^2)

Now we can substitute the value we found for a into the equation for c to get:

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - B)))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(a/b))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(12/a))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(12/(12(sin(58)/sin(B)))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(sin(58)/sin(B))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(sin(58)/(12*sin(58)/a))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(a/12))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - arcsin(1/12)*a))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(122 - 4.98)*a))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(sin(117.02)*a))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(sin(58)/(0.97*a))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(1.03*a/sin(58))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(1.03*(12*sin(58)/sin(B))/sin(58))))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(1.03*(12/sin(B)))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(1.03*(12/sin(180 - A - B)))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(1.03*(12/sin(180 - 58 - B)))))^2)

c = 12*cos(58) ± sqrt(576 - 576*sin(58)^2 - 576(1.03*(12/sin(122 - B)))))^2)

Now we can solve for c using the two possible values of B:

B = arcsin(b*sin(A)/a)

B = arcsin(12*sin(58)/a)

B = arcsin(12*sin(58)/(12*sin(58)/sin(B)))

B = arcsin(sin(B))

B = 58

or

B = 180 - arcsin(b*sin(A)/a)

B = 180 - arcsin(12*sin