ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Martoman on July 16, 2010, 11:19:30 pm

Title: question on geometry
Post by: Martoman on July 16, 2010, 11:19:30 pm: Now I get to ask a question :D

Q8 in short answer from Ch:1 toolbox.

For the lazy ones I understand your pain so all you have to do is click this link to see the question:

http://img130.imageshack.us/f/capturefj.png/

I can get the answer. I want a more elegant way rather than the hacky crap i've been doing.

Thanks y'all ;D
Title: Re: question on geometry
Post by: theuncle on July 18, 2010, 10:57:13 pm: Not sure if you would call this hacky, but I'll give it a go!

Part a uses the alternate segment rule on page 21, as well as straight lines and triangles adding to 180.
Given the answer is <BCD=x, ACD is an isosceles triangle.

ABD is also an isosceles triangle given that AB=BD
an application of the sin rule gives two equations:
$\frac{a}{sin(x)} = \frac{y}{sin(180-2x)}$
$\frac{y}{sin(x)} = \frac{a+b}{sin(180-2x)}$
The rest is algebra, equating both to $\frac{sin(180-2x)}{sin(x)}$ and solving the remaining equation for y!
Title: Re: question on geometry
Post by: Martoman on July 18, 2010, 11:15:59 pm: yeah that was the *hacky* maths I was referring to, that was the route I took.

Thanks a lot for answering though, no one else seemed to love me enough :'(
Title: Re: question on geometry
Post by: 98.40_for_sure on July 18, 2010, 11:24:54 pm: Wouldn't have a clue! geometry is the topic i can't do :(
Title: Re: question on geometry
Post by: Martoman on July 18, 2010, 11:31:24 pm: Quote from: 99.95_for_sure on July 18, 2010, 11:24:54 pm
Wouldn't have a clue! geometry is the topic i can't do :(

He rhymes without meaning to! (lol c what i did thar?) ::)

I love geometry lots. its pretttttyyyy :smitten:
Title: Re: question on geometry
Post by: Ahmad on July 19, 2010, 11:41:31 am: This follows directly from ABD and ACD being similar triangles, AB/AD = AD/AC.

To show the similarity it suffices to show ABD = ADC since they share the common angle at A. Well, COD = 2 * CBD = 2 * (180 - ABD) and OCD is isosceles giving OCD = 1/2 (180 - 2(180 - ABD)) = ABD - 90. This gives ADC = 90 + ODC = ABD as required.
Title: Re: question on geometry
Post by: 98.40_for_sure on July 19, 2010, 04:59:59 pm: Quote from: Martoman on July 18, 2010, 11:31:24 pm
Quote from: 99.95_for_sure on July 18, 2010, 11:24:54 pm
Wouldn't have a clue! geometry is the topic i can't do :(

He rhymes without meaning to! (lol c what i did thar?) ::)

I love geometry lots. its pretttttyyyy :smitten:

Hahah Martoman, you are too good at picking up the nitty gritty... boo!