[ The area of the triangle $ABC$ has side $BC$ equal to 6, side $AC$ equal to 5, and the angle at vertex $B$ equal to $30^{\circ}$. Find the area of the triangle if the distance from vertex $A$ to the line $BC$ is less than $\frac{1}{\sqrt{2}}$.

#
Apply the Law of Cosines.

## Solution

Let $A B=x$. By the Law of Cosines in triangle $A B C$, we find that

$A C^{2}=A B^{2}+B C^{2}-2 A B \cdot B C \cos 30^{\circ}$, or

$25=36+x^{2}-2 \cdot 6 \cdot x \cdot \frac{\sqrt{3}}{2}$, or $x^{2}-$

$$
6 \sqrt{3} x+11=0
$$

From this, we find that $x_{1}=3 \sqrt{3}-4$ or $x_{2}=3 \sqrt{3}+4$.

We will find the distance from vertex $A$ to line $B C$ in each of these cases:

$$
d_{1}=A B \cdot \sin 30^{\circ}=\frac{x_{1}}{2}=\frac{3 \sqrt{3}-4}{2}
$$

$$
d_{2}=\frac{3 \sqrt{3}+4}{2}
$$

The condition $d<\frac{1}{\sqrt{2}}$ is satisfied only by $x_{1}$. Therefore,

$S_{\triangle \mathrm{ABC}}=\frac{1}{2} A B \cdot B C \sin 30^{\circ}=\frac{1}{2}(3$ $\sqrt{3}-4) 6 \cdot \frac{1}{2}=\frac{3(3 \sqrt{3}-4)}{2}$.

![](https://cdn.mathpix.com/cropped/2024_05_06_5e9166ba6ad1f1b1f21fg-11.jpg?height=258&width=564&top_left_y=820&top_left_x=61)

## Answer

$\frac{3(3 \sqrt{3}-4)}{2}$.

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