12.183. The area of an isosceles obtuse triangle is 8, and the median drawn to its lateral side is $\sqrt{37}$. Find the cosine of the angle at the vertex.
## Solution.

Let $\triangle A B C A B=B C, A D-$ median, $A D=\sqrt{37}$ (Fig. 12.52), $\angle B=\alpha, B D=x$. Then $A B=2 x$.

From $\triangle A B D:$

$A D^{2}=A B^{2}+B D^{2}-2 A B \cdot B D \cos \angle B$

$37=4 x^{2}+x^{2}-4 x^{2} \cos \alpha$

$37=x^{2}(5-4 \cos \alpha)$.

$S_{\triangle A B C}=\frac{1}{2} A B^{2} \sin \angle B$. Then $2 x^{2} \sin \alpha=8 ; x^{2}=\frac{4}{\sin \alpha} \Rightarrow$

$\frac{4}{\sin \alpha}(5-4 \cos \alpha)=37 \Leftrightarrow 37 \sin \alpha+16 \cos \alpha+20=0 \Leftrightarrow$

$\Leftrightarrow \frac{74 \operatorname{tg} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}+\frac{16\left(1-\operatorname{tg}^{2} \frac{\alpha}{2}\right)}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}+20=0 \Leftrightarrow 18 \operatorname{tg}^{2} \frac{\alpha}{2}-37 \operatorname{tg} \frac{\alpha}{2}+2=0 ;$

1) $\left.\operatorname{tg} \frac{\alpha}{2}=2 ; 2\right) \operatorname{tg} \frac{\alpha}{2}=\frac{1}{18}$.

Then $\cos \alpha=\frac{1-\operatorname{tg}^{2} \frac{\alpha}{2}}{1+\operatorname{tg}^{2} \frac{\alpha}{2}}=-\frac{3}{5}$ when $\operatorname{tg} \frac{\alpha}{2}=2, \cos \alpha=\frac{323}{325}$ when $\operatorname{tg} \frac{\alpha}{2}=\frac{1}{18}$.

Since by the condition $\alpha$ is an obtuse angle, then $\cos \alpha=-3 / 5$.

Answer: $-3 / 5$.