vvvva·b35vv∴a·。 b=﹣3,∴向量b在a方向上的投影为v??a5故答案为:D.
uuuruuuruuuruuuruuur2.已知?ABC的外接圆半径为1,圆心为O,且满足OA?2OB?4OC?0,则AB?OC?( )
A. ?1577 B. ? C. 161616 D.
15 16【答案】C
uuuruuuruuuruuuruuuruuuruuur2uuur2uuuruuuruuur2【解析】由OA?2OB?4OC?0,得2OB?4OC??OA,两边平方,得4OB?16OC?16OB?OC=OA.因uuuruuuruuuruuuruuuruuuruuur19为?ABC的外接圆半径为1,所以|OA|?|OB|?|OC|?1,所以4?16?16OB?OC?1,所以OB?OC??.同
16uuuruuuruuuruuuruuuruuuruuuruuuruuuruuuruuur1319137理可求得OA?OC??,所以AB?OC?(OB?OA)?OC?OB?OC?OA?OC=? ?(?)?,故选C.
816816rrrrrr2?3.已知向量a,b的夹角为,且a?(3,?4),|b|?2,则|2a?b|?( )
3A.23 B.2 C.221 D.84 【答案】C
rr2r2rr2?r2?1??b?4??32?42??4?32?42?2?????22?84,所以【解析】因为|2a?b|?4a?4agbcos3?2?rr|2a?b|?84?221,故选C.
4.先将函数y?2sinx的图像纵坐标不变,横坐标压缩为原来一半,再将得到的图像向左平移得图像的对称轴可以为( ) A.x???12个单位,则所
?12 B.x??11? C.x??
612 D.x??6
【答案】D
5. ?ABC中,角A,B,C的对边分别为a,b,c,A?103?,sinB?,D为BC边中点,AD?1.
104
?I?求
b的值; c?II?求?ABC的面积.
【答案】(Ⅰ)
2;(Ⅱ)2. 2
3.练原创
1.如图,从高为h的气球(A)上测量铁桥(BC)的长,如果测得桥头B的俯角是?,桥头C的俯角是?,则该桥的长可表示为( )
A.
sin(???)sin(???)sin(???)cos(???)?h D.?h ?h B.?h C.
cos?cos?cos?cos?sin?sin?cos?sin?【答案】A
【解析】过A作垂线AD交CB于D,则在Rt?ADB中,?ABD??,AB?h ,由正弦定理,得sin?BCABhsin(???)AB?sin(???)h?sin(???)?? ∴BC?,即桥梁BC的长度为 ,故选A. sin(???)sin?sin??sin?sin?sin??sin?2.设?ABC的内角A,B,C所对边分别为a,b,c,若?a?b?c??a?b?c??ab,则角C?_________. 【答案】2? 3
sin(A?B)a2?b23.在△ABC中,若?22,则△ABC的形状一定是
sin(A?B)a?b【答案】等腰或直角三角形
sin(A?B)sin2A?sin2B2??sinA(sin(A?B)?sin(A?B))? 【解析】原式可化为
sin(A?B)sin2A?sin2B?sin2B(sin(A?B)?sin(A?B))?0??sin2AcosAsinB?sin2BsinAcosB?0??sin2A?sin2B?0?sin2A?sin2B?A?B,或A?B?3. 2?,故该三角形是等腰或直角三角形. 24. 已知f(x)?cosxsinx?3cosx?(1)求f(x)的单调增区间;
2uruuuruuur3uu(2)在?ABC中,A为锐角且f(A)?,AB?AC?3AD,AB?3,AD?2,求sin?BAD.
2【答案】(1)[k???12,k??5?35?1],k?Z.(2) 128
因此sin?BAD?sin(?3??AEB)?3151135?1. (12分) ????24248r5.设向量a??r???3sinx,sinx,b??cosx,sinx?,x??0,?.
?2??rr(1)若a?b,求x的值;
rr(2)设函数f(x)?a?b,求f(x)的最大值.
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