三角形數學問題(10點)

The lengths of the sides of triangle ABC are AB=6, BC=4 and CA=5.
Then cosA=??


cosA=[AB^2+AC^2-BC^2]/2(AB)(AC)

=(36+25-16)/2(6)(5)

=45/60

=3/4 //

2008-07-05 13:52:56 補充：
A=41-24-34.64
sin A=0.6614 (4 sig. fig.)
tan A=0.8819 (4 sig. fig.)

If you have not learn the cosine rule, following is the alternative:
ABC is an isos. triangle, let AX be the perpendicular line from A to BC, therefore, BX = 2. Therefore angle BAX = arcsin(2/5) = 23.578 degree.
Therefore angle A = 2 x angle BAX = 47.156 degree.
Therefore, cosA = 0.68.

cosA=[AB2+AC2-BC2]/2(AB)(AC)
=(25+25-16)/2(5)(5)
=17/25//


2008-07-05 18:33:17 補充：
cosA=[AB^2+AC^2-BC^2]/2(AB)(AC)
=(36+25-16)/2(6)(5)
=45/60
=3/4

餘弦定律 cosine formula

a^2 + b^2 - c^2 = 2*a*b* cos(C)

5^2+5^2-4^2 = 2*5*5*cosA

34 = 50 * cos A

cos A = 0.68

A~ 47.15 deg

2008-07-04 17:38:55 補充：
係喎，題目問緊 cos A

咁都俾你執到雞 o_O ??

2008-07-05 11:19:47 補充：
HAHA, 又一次被 eelyw 技術性擊倒 tim~