In triangleOAB,(→OA)=a and (→OB)=b
It is given that |a| =6,|b| =3 and angleAOB=2π/3.C and D are points on AB such that AD:DC:CB=3:2:1
(a)Express (→OC) and (→OD) in terms of a and b
(b)Find a‧b
Hence,find angleCOD,correst to the nearest degree
http://i214.photobucket.com/albums/cc72/yesterday0721/14.jpg
amaths
2008-01-13 6:17 pm
回答 (1)
2008-01-13 8:53 pm
✔ 最佳答案
a)BC:CA = 1:5
(→OC) = (a+5b) / 6
BD:DA = 3:3 = 1:1
(→OD) = (a+b) / 2
b)
a‧b
= |a| |b| cos(angleAOB)
= -9
(→OC)‧(→OA)
= [(a+5b) / 6]‧a
= [ |a| ^2 +5(a‧b)] / 6
= (36 - 45) / 6
= -3/2
|→OC| cos(angleCOA) = (-3/2) / 6 = -1/4
|→OC| sin(angleCOA) = OB sin(angleAOB) (5/6) = 5sqrt(3) / 4
thus, tan(angleCOA) = -5sqrt(3)
angleCOA = 96.59 degrees
Similar for angle DOA,
(→OD)‧(→OA)
= [(a+b) / 2]‧a
= [ |a| ^2 +(a‧b)] / 2
= (36 - 9) / 2
= 27/2
|→OD| cos(angleDCOA) = (27/2) / 6 = 9/4
|→OD| sin(angleDOA) = OB sin(angleAOB) (3/6) = 3sqrt(3) / 4
thus, tan(angleDOA) = sqrt(3) / 3
angleDOA = 30 degrees
Therefore,
angleCOD
= angleCOA - angleDOA
= 67 degrees (to the neareast degree)
收錄日期: 2021-04-29 19:45:19
原文連結 [永久失效]:
https://hk.answers.yahoo.com/question/index?qid=20080113000051KK00866