Point $D$ lies on side $CB$ of the right triangle $ABC\left(\angle C=90^{\circ}\right)$, and $AB=5, \angle ADC=\arccos$ $\frac{1}{\sqrt{10}}, DB=\frac{4 \sqrt{10}}{3}$. Find the area of triangle $ABC$.
Compose an equation with respect to $C D$.

## Solution

Let $C D=x, \angle A D C=\alpha$. Since $\cos \alpha=\frac{1}{\sqrt{10}}$, then $\operatorname{tg} \alpha=3$. Therefore, $A C=3 x$.

By the Pythagorean theorem from triangle $A B C$, we have

$$
A C^{2}+B C^{2}=A B^{2}, \text { or } 9 x^{2}+\left(x+\frac{4 \sqrt{10}}{3}\right)^{2}=25
$$

From this equation, we find that $x=\frac{\sqrt{10}}{6}$. Then

$$
3 x=\frac{\sqrt{10}}{2}, B C=C D+D B=\frac{3 \sqrt{10}}{2}
$$

Therefore,

$$
S_{\triangle \mathrm{ABC}}=\frac{1}{2} B C \cdot A C=\frac{15}{4}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_aa2960437c2efca8c53fg-03.jpg?height=415&width=606&top_left_y=707&top_left_x=726)

## Answer

$\frac{15}{4}$.

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