This week's book giveaway is in the OO, Patterns, UML and Refactoring forum. We're giving away four copies of Refactoring for Software Design Smells: Managing Technical Debt and have Girish Suryanarayana, Ganesh Samarthyam & Tushar Sharma on-line! See this thread for details.

Cosine rule for two triangles:ABD and ABC cos B = a**2+c**2-b**2/2*a*c gives BC = 2. [ June 05, 2004: Message edited by: Ram Abdullah D'Souza ]

MH

Nick George
Ranch Hand

Joined: Apr 04, 2004
Posts: 815

posted

0

Well, that looks like the right answer, howerver, i am a bit lost as to how it is obtained. Using law of cosines, I was able to get the following three equations from the three triangles:

s=length of BD (half of BC) beta1+beta2=angle BAC

4s�=52-48cos(beta1+beta2)

s�=41-40cos(beta1)

s�=61-60cos(beta1)

after solving all three together, i found s to be something like 1.223.

can i have a little more explanation as to how you came at your answers?

Arjun Shastry
Ranch Hand

Joined: Mar 13, 2003
Posts: 1888

posted

0

We can apply cosine rule only to two triangles.namely,ABD and ABC. Let 2x be length of BC.then BD=DC = x. For triangle,ABD, cosB = (4*4+x*x-5*5)/2*4*x-----------(1) For triangle ,ABC cosB = (4*4+(2x*2x)-6*6)/2*4*2x--------(2) Equating 1 &2 ,will give x=1-->BC=2

Nick George
Ranch Hand

Joined: Apr 04, 2004
Posts: 815

posted

0

quite understood... Thanks!

I’ve looked at a lot of different solutions, and in my humble opinion Aspose is the way to go. Here’s the link: http://aspose.com