Tan function and isosceles triangles

I have a non-right-angled isosceles triangle with two longer sides, X, and a short base Y.

I know the length of the long sides, X.

I also know the acute, vertex angle opposite the base Y, let's call it angle 'a'

I have been told I can calculate the length of the base Y by:

Y = tan(a) x X

I've sketched this out with a few hand drawn triangles and it does seem to work...... But why?

I can't derive that formula from any of the trigonometry I know. What am I missing?

• Do you know the sine/cosine rules for triangles in general? (Derivation and statement) – астон вілла олоф мэллбэрг Mar 24 '17 at 23:13
• Y = tan(a) x X At best, that's an approximation only valid for very small angles $a\,$. Draw the altitude on the base, and you should figure out what the correct formula is, instead. – dxiv Mar 24 '17 at 23:15
• Thanks all - so it's an approximation!.... :-) To give the background - this was part of a lecture on planning wedge-shaped bone cuts in orthopaedic surgery. The tan formula I quoted was just brushed over as if it was Day 1 geometry..... a given fact.... and it threw me completely as I just couldn't make head nor tail of it. For the surgery in question - the angle is only ever going to be very acute (5 < a < 45) - and so I guess this tan approximation of the opposite side is valid for this angle range, given the degree of (in)accuracy that is possible in this sort of surgery. – Robert Mar 26 '17 at 19:15

bisect angle $a$

$\sin \frac a2 = \frac {y}{2x}\\ \cos \frac a2 = \frac {\sqrt{4x^2-y^2}}{2x}\\ \tan \frac a2 = \frac {y}{\sqrt{4x^2-y^2}}\\ \tan a = \frac {2\tan\frac a2}{1-\tan^2 \frac a2}\\ \tan a = \frac {2y}{\sqrt{4x^2-y^2}(1-\frac {y^2}{4x^2-y^2})}\\ \tan a = \frac {y\sqrt{4x^2-y^2}}{2x^2-y^2}$

$\frac yx$ is small, $\frac{\sqrt{4x^2-y^2}}{2x^2-y^2} \approx \frac {1}{x}$

and $\tan a \approx \frac yx$

Cut the iscoles triangle in half to get a two right triangles with opposite side $\frac 12 y$ and hypotenuse $x$.

$\frac 12 Y = \sin (\frac 12 a) x$

So apparently this is claiming $2\sin(\frac 12 a) = \tan a$ which isn't true but is apparently an approximation. $\tan a = \frac{\sin (\frac 12 a + \frac 12 a)}{\cos(\frac 12 a + \frac 12 a)} = \frac {2\sin \frac 12 a\cos \frac 12 a}{\cos^2 \frac 12 a - \sin^2 \frac 12 a}=2\sin\frac 12 a*\frac {\cos \frac 12a}{\cos^2 \frac 12 a - \sin^2 \frac 12 a}$

And $\frac {\cos \frac 12a}{\cos^2 \frac 12 a - \sin^2 \frac 12 a}=\frac {\cos \frac 12a}{1 - 2\sin^2 \frac 12 a}$ which, I guess for small values of $a$ is close to 1. (you said $x > y$ so $a < 60$ and $\frac 12 a < 30$ So for $a = 60$ then term is $\frac{4\sqrt{3}}6=1.1547$ and as $a$ decreases it gets closer to $1$... I guess.)