Re: Can anyone help me solve this porblem?

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Posted by Edward on September 15, 1998 at 22:12:50:

In Reply to: Can anyone help me solve this porblem? posted by Ivan Li on September 13, 1998 at 22:32:51:

Hello Ivan,

The solution is as follows:
Suppose ABCDEFGH is in order by clockwise, putting it in complex plane and assume A is the origin. Since ABCDEFGH is regualr octagon, q = p(cos(-45*) + i sin(-45*))

and

AH = P(cos(-135*) + i sin(-135*))
= p(cos(-45*) + i sin(-45*))^3
= p(q/p)^3
= q^3/p^2.

Since cos(-45*)+ isin(-45*) = q/p.

Note: The "<" sign means angle. The "^" sign means "power of", "cos^2" is same as "cosine square". The "*" sign means "degree".

Hopefully this can help you a little bit.

Best regards,

Edward

: I have a problem that I cannot solve.Can anyone
: help?

: ABCDEFGH is a regular octagon. The vector AB=p,
: vector BC=q. Express vector AH in terms of p and q.

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Comments: : Hello Ivan, : The solution is as follows: : Suppose ABCDEFGH is in order by clockwise, putting it in complex plane and assume A is the origin. Since ABCDEFGH is regualr octagon, <ABC = <HAB = 135*. By the geometric meaning of multiplication of complex numbers, we have : q = p(cos(-45*) + i sin(-45*)) : and : AH = P(cos(-135*) + i sin(-135*)) : = p(cos(-45*) + i sin(-45*))^3 : = p(q/p)^3 : = q^3/p^2. : Since cos(-45*)+ isin(-45*) = q/p. : Note: The "<" sign means angle. The "^" sign means "power of", "cos^2" is same as "cosine square". The "*" sign means "degree". : Hopefully this can help you a little bit. : Best regards, : Edward : : I have a problem that I cannot solve.Can anyone : : help? : : ABCDEFGH is a regular octagon. The vector AB=p, : : vector BC=q. Express vector AH in terms of p and q.

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