1
$\begingroup$

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.

enter image description here

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?

$\endgroup$
2
  • $\begingroup$ Is the goal to find a numerical answer? $\endgroup$ Commented May 9, 2021 at 1:25
  • $\begingroup$ Yes, an actual solution. $\endgroup$ Commented May 9, 2021 at 1:32

2 Answers 2

Reset to default
3
$\begingroup$

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.

$\endgroup$
1
$\begingroup$

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:

enter image description here

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 $\theta=10.558\ldots^\circ$.

$\endgroup$

You must log in to answer this question.

Start asking to get answers

Find the answer to your question by asking.

Ask question

Explore related questions

See similar questions with these tags.