AB,AC,BC and h are known and its a isosceles triangle
how to find angle BAC?
[Math] Find the Angle BAC
trianglestrigonometry
Related Solutions
I saw the following solution may years ago:
On side $AD$ construct in exterior equilateral triangle $ADE$. Connect $BE$.
Then $AB=AC, AE=BC, \angle BAE=\angle ABC$ gives $\Delta BAE =\Delta ABC$ and hence $AB=BE$.
But then $$AB=BE, BD=BD, DA=DE \Rightarrow ADB =EDB$$
Hence $\angle ADB=\angle EDB$. Since the two angles add to $300^\circ$ they are each $150^\circ$. Then $\angle ABD + \angle ADB+ \angle BAD=180^\circ$ gives $ABD=10^\circ$.
$GE=1/2*CE (opposite 30), ACE isosceles (angles 70,70),draw perpendicular to CE, there are 2 congruent right triangles , angle 20, common hypotenuse. So, GE=EI.
Best Answer
The triangle is completely determined by its side lengths $\overline{AB}$, $\overline{AC}$, $\overline{BC}$, which you say are all known (the knowledge of $h$ and $ABC$ being isosceles is unnecessary).
To find $\angle BAC$, apply the law of cosines to the side lengths:
$$\overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2\times\overline{AB}\times\overline{AC}\times\cos\angle BAC$$