ATAR Notes: Forum

VCE Stuff => VCE Mathematics => VCE Mathematics/Science/Technology => VCE Subjects + Help => VCE Specialist Mathematics => Topic started by: Martoman on October 31, 2010, 12:22:17 am

Title: 08 exam
Post by: Martoman on October 31, 2010, 12:22:17 am: Just wondering whats up with the last question on exam 2?

My logic is circle area - segment area.

$\pi - \frac{1}{2}(1)^2(\frac{\pi}{3}-sin(\frac{\pi}{3})$ = 3.05

The pi/ 3 gotten from recognising that if you make up a triangle from the point (0,0),(0,1),( $\frac{-\sqrt{3}}{2}, \frac{1}{2})$ which is equilateral... so angle which the centre of the circle subtends the chord is 60 degrees.

Solutions say 2.53 wtf
Title: Re: 08 exam
Post by: 3Xamz on October 31, 2010, 02:19:11 am: Angle is 2pi/3 not pi/3 ?
Title: Re: 08 exam
Post by: Linkage1992 on October 31, 2010, 11:02:36 am: I think you have gotten the point of intersection wrong. It is (-root3/2 , 3/2)
Title: Re: 08 exam
Post by: Martoman on October 31, 2010, 11:44:02 am: Quote from: Linkage1992 on October 31, 2010, 11:02:36 am
I think you have gotten the point of intersection wrong. It is (-root3/2 , 3/2)

THANK YOUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUUU

On the previous page i wrote 3/2 on my drawing i decided to write 1/2 wtf