# trigonometry equilateral triangle

first of all, sorry for the lame question.

Having a starting point, A and a height (catet) of y, what's the formula to calculate x?

Thank you, i don't have any trig basis.

• An equilateral triangle can be cut into two 30-60-90 triangles. If $y$ is the altitude, then $x=\frac{y}{\sqrt 3}$. You can use the Pythagorean theorem for this: $y^2+x^2=4x^2$. – J. M. isn't a mathematician Jan 6 '12 at 16:31
• You don't have enough information here. You need either one angle or the base length. – John Alexiou Jan 6 '12 at 16:40
• @ja72,the triangle is equilateral. – leo Jan 6 '12 at 16:49
• ja72, it's an equilateral triangle so all the angles are 60 deg. J.M. would you like to put that as an answer? so i can accept it? and thanks – André Alçada Padez Jan 6 '12 at 16:49

Since it's an equilateral triangle, each angle has measure 60°. If you consider one of the smaller, right triangles formed by the altitude, the legs have length $y$ (the altitude) and $x$ (along the base) and the angle opposite the $y$ leg has measure 60°, so $$\tan 60°=\frac{y}{x}.$$ Since $\tan 60°=\sqrt{3}$, $$x=\frac{y}{\sqrt{3}}.$$
• One thing I find problematic about this is that it assumes $\tan 60^\circ=\sqrt{3}$ rather than taking the opportunity to show how that fact is derived from the Pythagorean theorem. – Michael Hardy Jan 6 '12 at 20:19
Use the Pythagorean theorem. The length of each side of the triangle is $2x$, so you have a right triangle (either the left half or the right half of the equilateral triangle) in which the hypotenuse has length $2x$ and one leg has length $x$. The height must therefore be $$y=\sqrt{(2x)^2-x^2} = \sqrt{4x^2-x^2}=\sqrt{3x^2}=x\sqrt{3}.$$