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June 15th, 2012, 09:09 PM   #1
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From: Maldives

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Vector Question

Hi, This is my first question.. I might be posting it in the wrong section, in that case please direct me. I took this question from IGCSE Mathematics book.
I got the answer to part a) and b)

Now I am struggling with the difficult part, question c) I could not get it.
Can you please lend a hand. (steps are important)

Code:
ANSWER: c) (1 + SQRT(2))b
Attached Images
 vector_reg_octagon.jpg (21.0 KB, 893 views)

June 15th, 2012, 09:40 PM   #2
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Joined: Jul 2010
From: St. Augustine, FL., U.S.A.'s oldest city

Posts: 12,211
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Math Focus: Calculus/ODEs
Re: Vector Question

We know $\vec{BG}$ is parallel to b and in the same direction, we just need the magnitude, which we may find from the Pythagorean theorem.

If we draw line segments from vertices A and H to $\vec{BG}$:

[attachment=0:32yo1y44]octagon.jpg[/attachment:32yo1y44]

The interior angles ? of an n-gon have the value, in radians, found by:

$\theta=\frac{\pi}{n}$$n-2$$$ hence for an octagon, we have:

$\theta=\frac{3\pi}{4}$ and so, referring to the sketch, we find:

$\alpha=\frac{3\pi}{4}-\frac{\pi}{2}=\frac{\pi}{4}$

Thus, we have:

$\alpha=\beta$ and consequently $c=h$ and so:

$b=\sqrt{2}c\:\therefore\:c=\frac{b}{\sqrt{2}}$

Now, the magnitude of $\vec{BG}$ is:

$b+2c=b+\sqrt{2}b=b$$1+\sqrt{2}$$$
Attached Images
 octagon.jpg (8.8 KB, 883 views)

 June 16th, 2012, 12:28 AM #3 Newbie   Joined: Jun 2012 From: Maldives Posts: 19 Thanks: 0 Re: Vector Question Thank you very much for the answer. The thing is that the radiant is not included in the IGCSE O' Level Syllabus. If it is possible to find the answer without using the radiant it will be great. and I could not understand the part how you got: b = sqrt(2c) I think it should be b = sqrt(2c^2) once again thanks for the help it is very much appreciated.
 June 16th, 2012, 12:45 AM #4 Senior Member     Joined: Jul 2010 From: St. Augustine, FL., U.S.A.'s oldest city Posts: 12,211 Thanks: 521 Math Focus: Calculus/ODEs Re: Vector Question $\pi$ radians is equal to $180^{\circ}$. So, using this, you may convert the angular measures from radians to degrees. And the square root, or radical, only has the two under it...it is $\sqrt{2}c$ not $\sqrt{2c}$.
 June 16th, 2012, 09:28 AM #5 Math Team   Joined: Dec 2006 From: Lexington, MA Posts: 3,267 Thanks: 408 Re: Vector Question Hello, suood! Here's a non-Trig approach to MarkFL's solution . . . Code:  H 1 A * - - - * /: :\ 1 / : : \ 1 / : : \ G * - * - - - * - * B P Q $\text{Suppose }HA= 1\;\text{ . . . that is, }\vec b\text{ is a unit vector.}$ $\Delta HPG\text{ and }\Delta AQB\text{ are both isosceles right triangles: }\:GP = HP \;\;\;GP^2\,+\,HP^2 \:=\:1^2 \;\;\;\Rightarrow\;\;\;GP^2\,+\,GP^2 \:=\:1\;\;\;\Rightarrow\;\;\;2\,\cdot\,GP^2 \:=\:1 \;\;\;GP^2 \:=\:\frac{1}{2} \;\;\;\Rightarrow\;\;\;GP \:=\:\frac{1}{\sqrt{2}} \:=\:\frac{\sqrt{2}}{2} \:=\:QB \;\;\;\text{Hence: }\:GB \:=\:\frac{\sqrt{2}}{2}\,+\,1\,+\,\frac{\sqrt{2}}{ 2} \;=\;1 + \sqrt{2}$ $\text{Therefore: }\:\vec{BG} \;=\;(1\,+\,\sqrt{2})\vec{AH} \;=\;(1\,+\,\sqrt{2})\vec b$
 June 22nd, 2012, 09:55 PM #6 Newbie   Joined: Jun 2012 From: Maldives Posts: 19 Thanks: 0 Re: Vector Question Thank you very much for the support..

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