1992 Sydney Grammar School Paper: Triangle geometry problem (1 Viewer)

[asianese]

8c) Last question
(c) The lengths of the sides of a triangle forman arithmetic progression and the largest angle of the triangle exceeds the smallest by 90◦. Find the ratio of the lengths of the sides.

What I've got so far:

a, b, c are in arithmetric series, so



And if the largest angle exceeds the smallest by 90, then



Using sine rule






Any help is appreciated. I don't know where I'm going with this lol. I generally hate geometry but this seemed like a fun problem.

 

[Fus Ro Dah]

Assuming your working is correct, a/b = tan alpha and a/c = tan alpha, so a/b = a/c, so b = c.

If b = c, then b/c = 1.

Sub b/c = 1 into your ratio for b/c then solve for the unique solution of alpha, then sub back into the expressions for a/b and a/c, then get your ratios.

 

[asianese]

weh

 

[bleakarcher]

a/b=sin(alpha)/cos(2*alpha) *

 

[bleakarcher]

check your final line, beta=pi/2-2*alpha not pi/2-alpha

 

[bleakarcher]

I just did part of the problem and got that the ratio of the largest side to the smallest is [4+sqrt(7)]/3.

 

[asianese]

Oh shoot..Will redo later

 

[seanieg89]

I just did part of the problem and got that the ratio of the largest side to the smallest is [4+sqrt(7)]/3.

This is correct. The question is just using the sine rule and then solving a quadratic.

 

[asianese]

I can't seem to get the quadratic..Is it a quadratic in one of the sides?