ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Yitzi_K on April 11, 2010, 02:41:05 pm

Title: Circle theorem question
Post by: Yitzi_K on April 11, 2010, 02:41:05 pm

Find the magnitude of each of the angles of a cyclical quadrilateral ABCD, the length of whose sides are AB=4, BC=5, CD=6 and DA=7. (Assume the quadrilateral is named in anticlockwise cyclic order).

There's probably something very simple that I'm missing here but I can't work it out

Title: Re: Circle theorem question
Post by: /0 on April 11, 2010, 03:05:39 pm

Try the cosine rule on triangles ABC and CDA, as well as BCD and DAB.

Title: Re: Circle theorem question
Post by: Yitzi_K on April 11, 2010, 03:11:17 pm

I'd thought of that, but how do I know the lengths
of AC and BD?

Title: Re: Circle theorem question
Post by: Ahmad on April 11, 2010, 03:14:21 pm

You should be able to eliminate the unknown length, making use of the fact that opposite angles of a cyclic quadrilateral add up to 180 degrees.

Title: Re: Circle theorem question
Post by: Yitzi_K on April 11, 2010, 03:27:39 pm

This is what I get: if angle D=x

then 85-84cos(x) = 41-40cos(180-x)

which on the CAS gives me four different (positive) solutions

Title: Re: Circle theorem question
Post by: Ahmad on April 11, 2010, 04:14:34 pm

Of course you also have the identity, cos(180 - x) = -cos(x). So you can solve for cos(x) explicitly without a calculator. Are you restricting x between 0 and 180 degrees?

Title: Re: Circle theorem question
Post by: Yitzi_K on April 11, 2010, 08:52:02 pm

Quote from: Ahmad on April 11, 2010, 04:14:34 pm

Of course you also have the identity, cos(180 - x) = -cos(x). So you can solve for cos(x) explicitly without a calculator. Are you restricting x between 0 and 180 degrees?

yep with that identity works fine. Thanks