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?

  • $\begingroup$ Do you know the sine/cosine rules for triangles in general? (Derivation and statement) $\endgroup$ – астон вілла олоф мэллбэрг Mar 24 '17 at 23:13
  • $\begingroup$ 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. $\endgroup$ – dxiv Mar 24 '17 at 23:15
  • $\begingroup$ 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. $\endgroup$ – 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.)


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.