在三角形ABC中,a=2√2,b=2√3,c=15° 解三角形
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在三角形ABC中,a=2√2,b=2√3,c=15° 解三角形
在三角形ABC中,a=2√2,b=2√3,c=15° 解三角形
在三角形ABC中,a=2√2,b=2√3,c=15° 解三角形
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°
余弦定理：
c²=a²+b²-2ab×cos15°
=(2√2)²+(2√3)²-2×2√2×2√3×(√6+√2)/4
=8+12-12-4√3
=8-4√3
=2(√3-1)²
∴c=√2(√3-1)=√6-√2
cosA=（b²+c²-a²)...
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余弦定理：
c²=a²+b²-2ab×cos15°
=(2√2)²+(2√3)²-2×2√2×2√3×(√6+√2)/4
=8+12-12-4√3
=8-4√3
=2(√3-1)²
∴c=√2(√3-1)=√6-√2
cosA=（b²+c²-a²)/2bc
=(12+8-4√3-8)/[2×2√3×(√6-√2)]
=(3-√3)/[√3×(√6-√2)]
=(√3-1)/(√6-√2)
=(√3-1)(√6+√2)/4
=√2/2
∴A=45°
∴B=120°
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