Question

Two triangles can be formed with the given information. Use the Law of Sines to solve the triangles

A = 59°, a = 13, b = 14

Answer 1

angle A = 59o and sides a = 13, b = 14.
Lets find remaining angles are B and C and the remaining side c.
Law of sines : a/sinA = b/sinB ⇒ sinB = (b*sinA)/a
sinB = (14*sin59o)/13 = (14*0.857167)/13 ≅ 0.923
B ≅ sin-1(0.923) ≅ 67.384.
There are two triangles, B₁ = 67.4 and B₂ = 180 - 67.4 = 112.6
find the angle c for B₁ = 67.4
A+ B +C = 180.
Angle C = 180 - (59 + 67.4) = 53.6.
Law of sines : b/sinB = c/sinC ⇒ c = b*sinC/sinB
c = (14)*sin(53.6)/sin(67.4) = (14)*(0.805)/(0.923) = 12.21.

find the angle c for B₂ = 112.6
A+ B +C = 180.
Angle C = 180 - (59 + 112.6) = 8.4.
Law of sines : b/sinB = c/sinC ⇒ c = b*sinC/sinB
c = (14)*sin(8.4)/sin(112.6) = (14)*(0.146)/(0.9232) = 2.27.

Answer 2

67.4°, C = 53.6°, c = 12.2; B = 112.6°, C = 8.4°, c = 2.2

a/sinA=b/sinB=c/sinC

a/sinA=13/0.8572=15.1662

sinB=b/a *sinA

sinB=14/13 *sin59=0.9231

B=67.4

B1=90+22.6=112.6

because sin67.4=sin112.6=0.9231

C=180-59-67.4=53.6

C1=180-59-112.6=8.4

c=a* sinC/sinA=13* sin53.6/sin59=12.2

c1=a* sinC1/sin59=13*sin8.4/sin59=2.2