1. **Problem statement:** Given a right-angled triangle with an angle $\Omega = 55^\circ$ and a circle of diameter $D = 20$ mm inscribed such that it is tangent to the hypotenuse and the base line $L2$, find the lengths $L1$ and $L2$. 2. **Understanding the problem:** The circle is inscribed touching the hypotenuse and the base $L2$. The diameter $D$ is the diameter of the inscribed circle. 3. **Key formulas and rules:** - The radius of the circle is $r = \frac{D}{2} = 10$ mm. - The circle is tangent to $L2$ (the base) and the hypotenuse, so the distance from the center of the circle to these lines is $r$. - Using trigonometry, if $L2$ is the base and $L1$ is the height, then $\tan(\Omega) = \frac{L1}{L2}$. 4. **Step-by-step solution:** - Let the center of the circle be at a distance $r$ from $L2$ and the hypotenuse. - The hypotenuse makes an angle $\Omega = 55^\circ$ with $L2$. - The distance from the center to the hypotenuse line is $r$, so the vertical distance from the base to the center is $r$. - The horizontal distance from the base to the center along $L2$ is $r / \tan(\Omega)$. 5. **Expressing $L1$ and $L2$ in terms of $r$ and $\Omega$:** - $L1 = r + r / \tan(\Omega)$ - $L2 = r / \sin(\Omega)$ 6. **Calculate values:** - $r = 10$ - $\tan(55^\circ) \approx 1.4281$ - $\sin(55^\circ) \approx 0.8192$ Calculate $L1$: $$L1 = r + \frac{r}{\tan(\Omega)} = 10 + \frac{10}{1.4281} \approx 10 + 7.0 = 17.0 \text{ mm}$$ Calculate $L2$: $$L2 = \frac{r}{\sin(\Omega)} = \frac{10}{0.8192} \approx 12.2 \text{ mm}$$ **Final answer:** $$L1 \approx 17.0 \text{ mm}, \quad L2 \approx 12.2 \text{ mm}$$