试题与答案

△ABC的三内角A,B,C所对边长分别是a,b,c,设向量m=(a+b,sinC

题型:填空题

题目:

△ABC的三内角A,B,C所对边长分别是a,b,c,设向量
m
=(a+b,sinC)
n
=(
3
a+c,sinB-sinA)
,若
m
n
,则角B的大小为 ______.

答案:

m
n

∴(a+b)(sinB-sinA)=sinC(

3
a+c)

由正弦定理知

(a+b)(b-a)=c(

3
a+c)

a2+c2-b2=-

3
ac

由余弦定理知

2accosB=-

3
ac

∴cosB=-

3
2

B∈(0,π)

∴B=

6

故答案为

6

试题推荐
题型:阅读理解

读短文,选答案。

      Usually I go to school on foot but this morning I went to school by bus, because I got up late. My best

friend Mike didn't come to school today. He went swimming yesterday and had a cold. The doctor asked him

to stay in bed and take some medicine. After class, John and I went to visit Mike. He is much better. I hope he

can go to school tomorrow.

1. I usually go to school ______________.[ ]

A. by bus

B. on foot

2. I went to school by bus this morning because ______________.[ ]

A. I got up late

B. I had a cold

3. Mike ______________ yesterday.[ ]

A. went shopping

B. went swimming

4. John and I went to ______________..[ ]

A. the hospital

B. visit Mike

5. How does Mike feel now?  ______________.[ ]

A. He is angry.

B. He is better.

查看答案
题型:选择题

2011年11月3日凌晨1时29分,经历近43小时飞行和五次变轨的“神舟八号”飞船飞抵距地面343公里的近圆轨道,与在此轨道上等待已久的“天宫一号”成功对接;11月16日18时30分,“神舟八号”飞船与“天宫一号”成功分离,返回舱于11月17日19时许返回地面.下列有关“天宫一号”和“神舟八号”说法正确的是(  )

A.对接前“天宫一号”的运行速率约为11.2km/s

B.若还知道“天宫一号”运动的周期,再利用万有引力常量,就可算出地球的质量

C.在对接前,应让“天宫一号”与“神舟八号”在同一轨道上绕地球做圆周运动,然后让“神舟八号”加速追上“天宫一号”并与之对接

D.“神舟八号”返回地面时应先减速

查看答案
微信公众账号搜索答案