In the triangle ABC, a = 3, B = root 3, a = 60 degree solution triangle

In the triangle ABC, a = 3, B = root 3, a = 60 degree solution triangle

In rtcbd, BD = √ (BC ^ 2-CD ^ 2) = 3 √ 3 / 2. AB = AD + BD = 2 √ 3. Sin angle ABC = CD / BC = 1 / 2 = sin30, angle ABC = 30 degrees, angle BCD = 90-30 = 60 degrees, angle ACB = 30 + 60 = 90 degrees

In △ ABC, a = √ 2, B = √ 3, B = 60 ° to solve triangle

Let C be CD ⊥ AB over D,
In RT Δ BCD, B = 60 ° a = √ 2,
∴CD=a*cosB=√6/2,BD=a*cosB=√2/2,
∴AD=√(b^2-CD^2)=√6/2,
∴A=45°,c=(√2+√6)/2,
∴C=180°-(A+B)=75°.
sinA=a*sinB/b=√2/2,
∵b>a,∴∠A=45°,∴∠C=75°,
sin75°=(√2+√6)/4,
∴ c=b*sinC/sinB=(√2+√6)/2.

In the triangle ABC, B = 8, C = 3, a = 60 ° solves the triangle

2bccosA=b^2+c^2-a^2,
Substituting B = 8, C = 3, a = 60 degree
The solution is a = 7