Solve triangle ABC. (If an answer does not exist, enter DNE. Round your answers to one decimal place. Below, enter your answers so that ∠A1 is smaller than ∠A2.)
b = 129, c = 168, ∠B = 48°
Angle A1Angle A2
Angle C1Angle C2
side a1side a2

∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9

Step-by-step explanation:

From Sine rule

∴ b / sinB = c / sinC

From the question,

b = 129, c = 168 and ∠B = 48°

∴ 129 / sin48° = 168 / sinC

Then, sinC = (168×sin48)/129

sinC = 0.9678

C = sin⁻¹(0.9678)

C = 75.42

∠C2=75.4°

and

∴∠C1 = 180° - 75.4°

∠C1 =104.6°

For ∠A

∠A1 = 180° - (104.6°+48°) [sum of angles in a triangle]

∠A1 = 27.4°

and

∠A2 = 180° - (75.4° + 48°)

∠A2 = 180° - (123.4°)

∠A2 = 56.6°

For side a

a1/sinA1 = b/sinB

a1/ sin27.4° = 129/sin48

a1 = (129×sin27.4°)/sin48

a1 = 79.8845

a1 = 79.9

and

a2/sinA2 = b / sinB

a2/ sin56.6° = 129/sin48

a2 = (129×sin56.6°)/sin48

a2 = 144.9184

a2 = 144.9

Hence,

∠A1 = 27.4°, ∠A2 = 56.6°, ∠C1 =104.6°, ∠C2=75.4°, a1 = 79.9 and a2 = 144.9

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step-by-step explanation: