12.031. In the square $A B C D$, an isosceles triangle $A E F$ is inscribed; point $E$ lies on side $B C$, point $F$ - on side $C D$, and $A E=A F$. The tangent of angle $A E F$ is 3. Find the cosine of angle $F A D$.
Solution.

Given: $ABCD$ is a square, $AEF$ is an isosceles triangle, $AE = AF, E \in BC, F \in CD, \operatorname{tg} \angle AEF = 3$ (Fig. 12.34); we need to find $\cos \angle FAD$. Let $\angle AEF = \alpha, \angle FAD = \beta$; then $\angle EAF = 180^\circ - 2\alpha$. Since $\triangle ABE = \triangle ADF$ (by leg and hypotenuse), we have $\beta = \frac{1}{2}(90^\circ - 180^\circ + 2\alpha) = \alpha - 45^\circ$. Therefore, $\cos \beta = \cos(\alpha - 45^\circ) = \frac{\sqrt{2}}{2}(\cos \alpha + \sin \alpha)$. Considering,

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-738.jpg?height=390&width=400&top_left_y=157&top_left_x=192)

Fig. 12.34

![](https://cdn.mathpix.com/cropped/2024_05_22_fa9db84b44b98ec0a5c7g-738.jpg?height=452&width=390&top_left_y=94&top_left_x=761)

Fig. 12.35

that $\operatorname{tg} \alpha = 3$, we find $1 + 9 = \frac{1}{\cos^2 \alpha}$, from which $\cos \alpha = \frac{1}{\sqrt{10}}, \sin \alpha = \operatorname{tg} \alpha \cos \alpha = \frac{3}{\sqrt{10}}, \cos \beta = \frac{\sqrt{2}}{2} \cdot \frac{4}{\sqrt{10}} = \frac{2 \sqrt{5}}{5}$.

Answer: $\frac{2 \sqrt{5}}{5}$.