[ Similar triangles (further). Application of trigonometric formulas (geometry). The area of a figure is equal to the sum of the areas of the figures into which it is divided [ The area of a triangle (using the semiperimeter and the radius of the inscribed or exscribed circle).

The radius of the circle inscribed in triangle $A B C$ is 4, and $A C = B C$. A point $D$ is taken on the line $A B$, which is at a distance of 11 and 3 from the lines $A C$ and $B C$ respectively. Find the cosine of angle $D B C$.

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Let $CM$ be the altitude of triangle $ABC$, and $P$ and $Q$ be the projections of point $D$ onto lines $AC$ and $BC$, respectively. Clearly, point $D$ cannot lie on the extension of $AB$ beyond point $A$.

If point $D$ lies on the extension of side $AB$ beyond point $B$, then the difference $DQ - DP = 8$ equals the height $BH$. However, this is impossible because the height is greater than twice the radius of the inscribed circle.

Therefore, point $D$ lies on the segment $AB$ and $\cos \angle DBC = \cos \angle B$.

First method. Let $AC = BC = a$, $AB = 2c$.

$S_{ABC} = S_{ADC} + S_{BDC} = \frac{1}{2} AC \cdot DP + \frac{1}{2} BC \cdot DQ = \frac{1}{2}(11a + 3a) = 7a$.

On the other hand, $S_{ABC} = \frac{1}{2}(a + a + 2c) \cdot 4 = 4(a + c)$.

From the equation $7a = 4(a + c)$, we find that $\frac{c}{a} = \frac{3}{4}$. Therefore, $\cos \angle B = \frac{3}{4}$.

Second method. Let $CM$ be the altitude of triangle $ABC$, $O$ be the center of the inscribed circle of triangle $ABC$, and $\angle A = \angle B = \alpha$. Then $AB = 2AM = 2OM \operatorname{ctg} \angle OAB = 8 \operatorname{ctg} \frac{\alpha}{2} = \frac{8(1 + \cos \alpha)}{\sin \alpha}$.

Moreover, $AB \sin \alpha = BH = DQ + DP = 14$. From the equation $8(1 + \cos \alpha) = 14$, we find that $\cos \alpha = \frac{3}{4}$.

## Answer

$\frac{3}{4}$.