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In the isosceles triangle ABC, if
,and a point E divides AB internally in ratio
.Then cosine of the angle between
and isWhere
Last edited by juantheron (2011-10-19 21:11:36)
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hi juantheron
I cannot get a diagram here. You got BC twice and not AC. Please would you edit your post.
Does this have to be done by vectors? I think straight trig using the cosine rule will do it more easily.
Bob
Last edited by Bob (2011-10-19 19:57:19)
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Hi juantheron;
There is a nice piece of software called geogebra. It is just what the doctor ordered for geometry problems.
Try it out.
In mathematics, you don't understand things. You just get used to them.
If it ain't broke, fix it until it is.
Always satisfy the Prime Directive of getting the right answer above all else.
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hi juantheron,
Thanks for sorting that distance out. My diagram is much the same as bobbym's.
I would split triangle ABC down it's line of symmetry to get angle B and then use the cosine rule to find EC.
Then I'd use the cosine rule again in triangle ACE to get the angle you want.
Post again if you need help with the trig. or to tell us you've done it.
Bob
Children are not defined by school ...........The Fonz
You cannot teach a man anything; you can only help him find it within himself..........Galileo Galilei
Sometimes I deliberately make mistakes, just to test you! …………….Bob
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Thanks bobbundy. got it
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