In the triangle ABC, a = 18, B = 16, a = 150, solve the triangle

In the triangle ABC, a = 18, B = 16, a = 150, solve the triangle

∵SinA/a=SinB/b,
∴SinB=4/9,
∴B≈26°,
∴C=180-A-B=4°
∵c²=a²+b²-2ab*CosC,
C = root sign (18 & # 178; + 16 & # 178; - 2 * 18 * 16 * Cos4 °) ≈ 2.32
(C can also be obtained from sinc / C = Sina / a)

In △ ABC, a = 22, a = 30 ° and B = 45 ° are known to solve triangles

∵ a = 30 °, B = 45 °, C = 180 ° - 30 ° - 45 ° = 105 ° from Asina = bsinb, Asina = csinc, B = sinbsina · a = sin45 ° sin30 °× 22 = 2212 × 22 = 4, C = sincsina · a = sin105 ° sin30 °× 22 = 6 + 2412 × 22 = 2 (3 + 1)

In the triangle ABC, a = 45 ° B = 30 ° a = 2 solves the triangle

C=180-A-B=105°
b=sinB*a/sinA
=1/2*2/(√2/2)
=√2
sinC=sin(A+B)=sinAcosB+sinBcosA
=√2/2*√3/2+1/2*√2/2
=(√6+√2)/4
c=sinC*a/sinA
=(√6+√2)/4*2/(√2/2)
=(√3+1)/2

In the triangle ABC, a = 2 √ 2, B = 2 √ 3, C = 15 ° to solve the triangle

cosC=cos15°
=cos(45°-30°)
=cos45°cos30°+sin45°sin30°
=√2/2×√3/2+√2/2×1/2
=(√6+√2)/4
c²=a²+b²-2abcosC
=(2√2)²+(2√3)²-2×2√2×2√3×(√6+√2)/4
=8+12-2√6×(√6+√2)
=20-12-4√3
=8-4√3
c=√(8-4√3)
=2√(2-√3)
=2√[(4-2√3)/2]
=2√[(√3-1)²/2]
=2(√3-1)×√2/2
=√6-√2
cosA=(b²+c²-a²)/(2bc)
=[(2√3)²+(√6-√2)²-(2√2)²]/[2×2√3×(√6-√2)]
=(12+8-4√3-8)/(12√2-4√6)
=(12-4√3)/(12√2-4√6)
=(3-√3)/(3√2-√6)
=(3-√3)(3√2+√6)/[(3√2+√6)(3√2-√6)]
=(9√2-3√2)/(18-6)
=6√2/12
=√2/2
A=45°
B=180°-(A+C)=180°-(45°+15°)=120°