# Angle between the median and altitude to one side of an isosceles triangle

Can this be solved without trigonometry?

$$AB$$ is the base of an isosceles $$\triangle ABC$$. Vertex angle $$C$$ is $$50^\circ$$. Find the angle between the altitude and the median drawn from vertex $$A$$ to the opposite side.

I think I know how to do this using the Law of Cosines.

Call the side of the isosceles triangle $$x$$, $$CA=CB=x$$, and let $$E$$ be the midpoint of $$BC$$. Then in the triangle formed by the median, $$\triangle CAE$$, using the Law of Cosines:

$$AE = \sqrt{x^2 + \frac{x^2}4-2\frac{x^2}{2}\cos50^\circ}$$

From this you can find $$AE$$ in terms of $$x$$.

Then apply the Law of Cosines again to find $$\angle CAE$$ using $$CE=\frac{x}{2}= \sqrt{x^2 + AE^2-2xAE\cos\angle CAE}$$ and from $$\cos\angle CAE$$ you find $$\angle CAE$$, and the angle we want is $$40^\circ-\angle CAE$$.

But is it possible w/o trig?

• Is the goal to find a numerical answer? – quasi May 9 at 1:25
• Yes, an actual solution. – Ido Sarig May 9 at 1:32

As @quasi's answer suggests, the target angle almost-certainly isn't rational, so avoiding trig is unlikely.

That said, there's a pretty quick trigonometric approach to the target:

Let $$s$$ be the triangle's half-leg, $$M$$ the midpoint of $$\overline{BC}$$, and $$N$$ the foot of the altitude from $$A$$. Then, in right $$\triangle ACN$$ we have $$|AN|=2s\sin C \qquad |NC|=2s\cos C$$ Thus,

$$\tan\theta =\frac{|MN|}{|AN|}=\frac{|CN|-|CM|}{|AN|} = \frac{2s\cos C-s}{2s\sin C} = \frac{2\cos C-1}{2\sin C}$$

For $$C=50^\circ$$, this gives $$\theta=10.558\ldots^\circ$$.

Using trigonometry, the angle in question is approximately equal, in degrees, to $$10.558536057412143196227467316938626443256567512439$$ which, given the lack of an apparent repeating block, is almost certainly not a rational number.

Hence you should not expect a solution vis synthetic geometry.