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AB,AC,BC and h are known and its a isosceles trianglehow to find angle BAC?
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$\begingroup$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:
$$\overline{BC}^2=\overline{AB}^2+\overline{AC}^2-2\times\overline{AB}\times\overline{AC}\times\cos\angle BAC$$
$\endgroup$ 0 $\begingroup$Let D be the point where the altitude touches segment BC. Then $\frac{AD}{AB}$=$\cos\theta$, or $\theta=\text {Angle BAD}=\arccos(\frac{AD}{AB})$. The same logic apples to angle CAD=$\arccos(\frac{AD}{AC}).$ Adding these 2 values gives the value of Angle CAB (Angle Addition Postulate.)
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