In triangle $\triangle{ABC}$, $BH$ is the height of $AC$. $|AB|=|AC|=10$. $|BC|=x$. And inequality $2\cdot\measuredangle{BAC}>\measuredangle{ABC}+\measuredangle{ACB}$ is given. How many possible integer values $x$ can be?
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My attempt: I know that $\measuredangle{ABC}=\measuredangle{ACB}$ due to the fact that $|AB|=|AC|$. Let $\alpha=\measuredangle{ABC}=\measuredangle{ACB}$. So i have $\measuredangle{BAC}=180-2\alpha$. According to given inequality i have:
$$ 360-4\alpha>\alpha+\alpha \\ 6\alpha<360 \\ \alpha<60 $$ So $$ \measuredangle{BAC}=180-2\alpha>60 $$ I can conclude $x>10$ according to last fact since we have isosceles triangle. I get a lower limit. If i write an inequality in triangle $\triangle{ABC}$: $$ 10-10<x<10+10 \\ x<20 $$ And this the upper bound i've found. I can combine them and write: $$ 10<x<20 $$ But seems like i can't compress the inequality enough because the book says that the answer is $4$.
If $H$ between $A$ and $C$ then $\measuredangle A<90^{\circ}$, which gives $10<x<10\sqrt2$ and four values of $x$.