题目内容
设△ABC的内角A、B、C所对的边分别为a、b、c,已知a=1,b=2,cosC=
(I) 求△ABC的周长;
(II)求cos(A-C)的值.
| 1 |
| 4 |
(I) 求△ABC的周长;
(II)求cos(A-C)的值.
(I)∵c2=a2+b2-2abcosC=1+4-4×
=4,
∴c=2,
∴△ABC的周长为a+b+c=1+2+2=5.
(II)∵cosC=
,∴sinC=
=
=
.
∴sinA=
=
=
.
∵a<c,∴A<C,故A为锐角.则cosA=
=
,
∴cos(A-C)=cosAcosC+sinAsinC=
×
+
×
=
.
| 1 |
| 4 |
∴c=2,
∴△ABC的周长为a+b+c=1+2+2=5.
(II)∵cosC=
| 1 |
| 4 |
| 1-cos2C |
1-(
|
| ||
| 4 |
∴sinA=
| asinC |
| c |
| ||||
| 2 |
| ||
| 8 |
∵a<c,∴A<C,故A为锐角.则cosA=
1-(
|
| 7 |
| 8 |
∴cos(A-C)=cosAcosC+sinAsinC=
| 7 |
| 8 |
| 1 |
| 4 |
| ||
| 8 |
| ||
| 4 |
| 11 |
| 16 |
练习册系列答案
相关题目