# Solve the triangle using Law of Sines

by Zachary
(CA, USA)

How do i find all possible triangles that satisfy the given conditions?

### Comments for Solve the triangle using Law of Sines

 May 22, 2012 Solve the triangle using Law of Sines by: Staff Question: by Zachary (CA, USA) How do i find all possible triangles that satisfy the given conditions? Answer: Law of Sines a/(Sin A) = b/(Sin B) = c/(Sin C) a/(Sin A) = b/(Sin B) 72/(Sin 21°) = 103/(Sin B) Sin B = 103*(Sin 21°) / 72 Sin B = 103*(0.3583679495453) / 72 Sin B = 0.5126652611551 ∠B = Sin⁻¹(0.5126652611551) ∠B = 30.8415250916575° ∠C = 180° - ∠A - ∠B ∠C = 180° - 21° – 30.8415250916575° ∠C = 128.1584749083425° ∠C = 128.2° a/(Sin A) = c/(Sin C) 72/(Sin 21°) = c/(Sin 128.2°) c = (Sin 128.2°) *72/(Sin 21°) c = (0.7858568931754) *72/(0.3583679495453) c = 157.887155869882 ∠A₁ = 21° a₁ = 72 ∠B₁ = 30.8° b₁ = 103 ∠C₁ = 128.2° c₁ = 157.9 a₁/(Sin A₁) = b₁/(Sin B₁) = c₁/(Sin C₁) a₁/(Sin A₁) = 72/(Sin 21°) = 200.9108238930242 b₁/(Sin B₁) = 103/(Sin 30.8415250916575°) = 200.9108238930173 c₁/(Sin C₁) = 157.887155869882/(Sin 128.1584749083425°) = 200.7963581230925 ∠B₂ = 180° - 30.8415250916575° = 149.1584749083425° ∠C₂ = 180° - ∠A₂ - ∠B₂ ∠C₂ = 180° - 21° – 149.1584749083425° ∠C₂ = 9.8415250916575° ∠C₂ = 9.8° 72/(Sin 21°) = c₂/(Sin 9.8415250916575°) c₂ = (Sin 9.8415250916575°) *72/(Sin 21°) c₂ = (0.1709236283535) *72/(0.3583679495453) c₂ = 34.3404069952868 ∠A₂ = 21° a₂ = 72 ∠B₂ = 180° - 30.8415250916575° = 149.1584749083425° b₂ = 103 ∠C₂ = 9.8415250916575° c₂ = 34.3404069952868 ∠A₂ = 21° a₂ = 72 ∠B₂ = 149.2° b₂ = 103 ∠C₂ = 9.8° c₂ = 34.3 a₂/(Sin A₂) = b₂/(Sin B₂) = c₂/(Sin C₂) a₂/(Sin A₂) = 72/(Sin 21°) = 200.9108238930242 b₂/(Sin B₂) = 103/(Sin 149.1584749083425°) = 200.9108238930173 c₂/(Sin C₂) = 34.3404069952868/(Sin 9.8415250916575°) = 200.9108238930011 >>> the final answer is: **Triangle 1 ∠A₁ = 21° a₁ = 72 ∠B₁ = 30.8° b₁ = 103 ∠C₁ = 128.2° c₁ = 157.9 **Triangle 2 ∠A₂ = 21° a₂ = 72 ∠B₂ = 149.2° b₂ = 103 ∠C₂ = 9.8° c₂ = 34.3 Thanks for writing. Staff www.solving-math-problems.com