Question

已知三角形的三边长为a=2根号2,b=2,c=根号6+根号2,求角A,B,C

a=2倍根号2

Answer 1

由余弦定理cosA=(b²+c²-a²)/(2bc)=(4+8+4√3-8)/[2*2*(√6+√2)]
=4(√3+1)/[4√2(√3+1)]
=√2/2
所以A=45°
cosB=(a²+c²-b²)/(2ac)=(8+8+4√3-4)/[2*2√2*(√6+√2)]
=4√3(√3+1)/[8(√3+1)]
=√3/2
所以B=30°
故C=180°-A-B=105°

Answer 2

a=2根号2,b=2,c=根号6+根号2,
a²=8,b²=4,c²=8+4√3
cosA=(c²+b²-a²)/2cb=[4+(8+4√3)-8]/2×2×(√6+√2)=(4+4√3)/4×(√6+√2)=√2/2
A=45º
cosB=(a²+c²-b²)/2ac=[8+(8+4√3)-4]/2×2√2×(√6+√2)=(4+4√3)/8×(√3+1)=√3/2
B=30º
C=180º-30º-45º=105º
或cosC=(a²+b²-c²)/2ab=[8+4-(8+4√3)]/2×2×2√2=(1-√3)/2√2=(√2-√6)/4
C=105º