Law of sines

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 = 59^o 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*sin59^o)/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.