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An Acute Problem

Prove that for an acute angled triangle ABC sinA + sin B + sin C > 2.

This is a problem that was posed to me at a meeting of the Further Maths Support Programme (@furthermaths) in Leeds a few months ago that stumped me for quite a while, but once I realised how to the answer it seemed fairly obvious.

I believe the question originally came from a step paper (ie Oxbridge entrance exam) so it is aimed at an able A Level student.

Without meaning to patronise, I want to clarify the wording incase anyone was unclear. An acute angled triangle is a triangle where all the angles are acute. The vertices are labelled A, B and C. The problem is looking for a proof that the sum of the sines of the angles is always greater than 2.

When I first encountered the problem I set off on all sorts of dead ends. Dropping perpendiculars, equating this, equating that, all to know avail. Then I drew the sine curve. At first I was just staring blankly at it, then I considered just the portion of the curve which relates to acute angles. Then I saw this:

image

A straight line from the origin to pi/2 (or 90… I don’t know why I did it in degrees on this picture!) would prove that for all acute angles sin x > Pi/2 (or sin x > 90 in degrees).

From there it was pretty straightforward:

image

So there you have it, for any acute angle, x, it’s sine, sin x, is greater than x/90. (I’m just going to work in degrees now, as that’s what’s on the photo!).

So:

Sin A > A/90

Sin B > B/90

Sin C > C/90

Hence:

Sin A + Sin B + Sin C > A/90 + B/90 + C /90

> (A+B+C)/90

> 180/90 (as A B C are the angles on a triangle)

>2

A lovely simple solution when you see it! But hard to find at first.

And in radians:

image

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