在三角形ABC中,a=4,b=2√2,面积为4,求角C跟边c
答案:5 悬赏:10
解决时间 2021-01-22 03:45
- 提问者网友:我们很暧昧
- 2021-01-21 15:38
在三角形ABC中,a=4,b=2√2,面积为4,求角C跟边c
最佳答案
- 二级知识专家网友:野慌
- 2021-01-21 17:04
解由S=1/2absinC=1/2×4×2√2sinC=4
即sinC=√2/2
知C=π/4或C=3π/4
则当C=π/4时,c^2=a^2+b^2-2abcosc=4^2+(2√2)^2-2×4×2√2×√2/2
=16+8-16=8
故c=2√2
当C=3π/4时,c^2=a^2+b^2-2abcosc=4^2+(2√2)^2-2×4×2√2×(-√2/2)
=16+8+16=40
故c=2√10
即sinC=√2/2
知C=π/4或C=3π/4
则当C=π/4时,c^2=a^2+b^2-2abcosc=4^2+(2√2)^2-2×4×2√2×√2/2
=16+8-16=8
故c=2√2
当C=3π/4时,c^2=a^2+b^2-2abcosc=4^2+(2√2)^2-2×4×2√2×(-√2/2)
=16+8+16=40
故c=2√10
全部回答
- 1楼网友:人间朝暮
- 2021-01-21 20:21
C角45度 根据余弦定理求c边
- 2楼网友:独行浪子会拥风
- 2021-01-21 19:04
面积S=1/2 a b sinC
4=1/2*4*2√2sinC
sinC=√2/2
所以角c=45°
4=1/2*4*2√2sinC
sinC=√2/2
所以角c=45°
- 3楼网友:蓝房子
- 2021-01-21 17:43
面积为4
(1/2)ab sinC=4
(1/2)(4)(2√2)sinC = 4
√2sinC=1
sinC= √2/2
=> C =π/4 or 3π/4
case 1:C =π/4
c^2 = a^2+b^2-2abcosC
= 16+8-2(4)(2√2)cos(π/4)
=16 +8 - 16
= 8
c= 2√2
case 2:C =3π/4
c^2 = a^2+b^2-2abcosC
= 16+8-2(4)(2√2)cos(3π/4)
=16 +8 + 16
= 40
c= 2√10
(1/2)ab sinC=4
(1/2)(4)(2√2)sinC = 4
√2sinC=1
sinC= √2/2
=> C =π/4 or 3π/4
case 1:C =π/4
c^2 = a^2+b^2-2abcosC
= 16+8-2(4)(2√2)cos(π/4)
=16 +8 - 16
= 8
c= 2√2
case 2:C =3π/4
c^2 = a^2+b^2-2abcosC
= 16+8-2(4)(2√2)cos(3π/4)
=16 +8 + 16
= 40
c= 2√10
- 4楼网友:怀裏藏嬌
- 2021-01-21 17:34
由S=1/2absinC
所以sinC=2S/ab=8/(4*2√2)=√2/2
所以C=π/4或3π/4
若C=π/4
则c²=a²+b²-2abcosC
=16+8-2*4*2√2*√2/2
=8
c=2√2
若C=3π/4
则c²=a²+b²-2abcosC
=16+8-2*4*2√2*(-√2/2)
=32
c=4√2
综上
C=π/4,c=2√2或C=3π/4,c=4√2
所以sinC=2S/ab=8/(4*2√2)=√2/2
所以C=π/4或3π/4
若C=π/4
则c²=a²+b²-2abcosC
=16+8-2*4*2√2*√2/2
=8
c=2√2
若C=3π/4
则c²=a²+b²-2abcosC
=16+8-2*4*2√2*(-√2/2)
=32
c=4√2
综上
C=π/4,c=2√2或C=3π/4,c=4√2
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