a solve the triangle A34 deg C101 deg a23 b a6 b11 C123 deg

a) solve the triangle. A=34 deg C=101 deg, a=23

b) a=6, b=11, C=123 deg

c) Two sides of an angle (SSA) of a triangle are given.Determine whether the given measurements produce 2 results. a=16, b=17 , A=53 deg

a) A = 34 ; C = 101 ; a =23

Angle B = 180 - 34 -101 = 45 deg

Use Sine rule:

b/sinB = a/sinA ----> b = sin45*23/sin34 = 29.1

c/sinC = a/sinA ----> c= sin101*23/sin34 = 40.38

b) a=6, b=11, C=123 deg

Use cosine rule:

c^2 = a^2 +b^2 -2abcos123 = 6^2 +11^2 -2*6*11cos123 = 228.89

c =15.13

Use Sine rule:

a/sinA = c/sinC ---> sinA = 6*sin123/15.13 =0.332

A1 = 19.40 deg ; A2 = 160.6 deg

B1 = 180 -123- 19.40 = 37.6 deg ; B2 = is not possible

So, we only one A = 19.40 deg and B = 37.6 deg

c) a=16, b=17 , A=53 deg

use sine rule:

a/sinA = b/sinB

sinB = 17*sin53/16 =0.848

B1 = 58 deg ; B2 = 180 -58 = 122 deg

C1 = 69 deg ; C2 = 5 deg

So, there are two triangles with A = 53 , B1 = 58 , C1 = 69

A = 58 , B2 = 122 , C2 = 5 deg