Jrahs q9 (1 Viewer)

[kawaiipotato]

community.boredofstudies.org/attachments/13/mathematics-extension-1/30632d1404894953-ruse-2013-trial-solutions-2013-jrah-extn1-trial.pdf
Q9

Firstly, I use 2pi*r = circumference = 9y
Gives me r = 9y/2pi
Then Constructing a chord to complete the triangle, I then construct the radius from the centre to the two corners. Labelling the angle in the newly constructed triangle as theta,
I use r(theta) = 5y
Giving me theta = 10pi/9
Which is already greater than the angle sum in a triangle
Is this saying that I can't construct a triangle with these points?? Confused

 

[psyc1011]

Is your theta to the left or right of the centre? Is it facing 5y or x?

 

[kawaiipotato]

Theta is the one at the centre facing 5y

 

[calamebe]

60° I'll post a picture soon

 

[calamebe]

http://m.imgur.com/nZr8r5U

The picture didn't really have the best explanation. So as we have y+y+2y+5y=9y circumference, we can divide 360° by 9 to find the angle for each y. 360/9=40°. So we can calculate what angle each of the arcs form. The two y arcs form 40° and the 2y arc forms 80°, so altogether it forms 160°. From that, we know that the sides of all the triangles formed will be isosceles, so we can easily find out the base angles, which are 70°. Now it is a quadrilateral, so to find x we just subract all of these angles from 360, to get x=60°.

 

[kawaiipotato]

Thanks

But the answer is A. It didn't have to go through the centre.
And also, was just wondering why I got the result I did as stated in the OP.

edit: it was B so ur correct

 

[psyc1011]

Thanks

But the answer is A. It didn't have to go through the centre.
And also, was just wondering why I got the result I did as stated in the OP.

edit: it was B so ur correct

It can't go through the centre. Arc length on each side has to be equal for that construction to cross the centre

 

[psyc1011]

View attachment 32463
Theta is the one at the centre facing 5y

I think your diagram is incorrect. The chord completing the triangle should be to the right of the centre because of the arc length ratios

 

[kawaiipotato]

I think your diagram is incorrect. The chord completing the triangle should be to the right of the centre

Ahh, so that's why I got theta to be greater than pi radians?

 

[calamebe]

The only thing I can come up with is maybe your diagram should look like this: http://m.imgur.com/XRuutkk

 

[psyc1011]

Ahh, so that's why I got theta to be greater than pi radians?

Yep, if you look at your diagram and label the arc lengths, you can see that your diagram demonstrates 5y < 4y

 

[calamebe]

Thanks

But the answer is A. It didn't have to go through the centre.
And also, was just wondering why I got the result I did as stated in the OP.

edit: it was B so ur correct

Oh thanks so much for the edit I got so shit scared. I worked it out another 2 ways and got the same answer lol, I thought I forgot how to geometry.

 

[psyc1011]

Oh thanks so much for the edit I got so shit scared. I worked it out another 2 ways and got the same answer lol, I thought I forgot how to geometry.

My bad, the picture you posted looked like it had a diameter (chord to complete triangle) lol.

 

[calamebe]

My bad, the picture you posted looked like it had a diameter (chord to complete triangle) lol.

Nah that's my crappy drawing, sorry. It's all good, it's been solved.