## Zadatak B-3.3.

Trokutu $A B C$ upisana je kružnica. Diralište sa stranicom $\overline{A B}$ dijeli tu stranicu na dijelove kojima su duljine u omjeru 2 : 3. Polumjer kružnice jednak je petini duljine stranice $\overline{A B}$. Odredite mjeru kuta nasuprot stranice $\overline{A \bar{B}}$ ?


## Rješenje.

![](https://cdn.mathpix.com/cropped/2024_05_30_8e0ddf1d7bf4bfb115dcg-11.jpg?height=391&width=668&top_left_y=621&top_left_x=614)

$|A D|=2 k,|D B|=3 k \Rightarrow c=|A B|=5 k \Rightarrow r=k$

$$
\begin{array}{rlr}
\operatorname{tg} \frac{\alpha}{2}=\frac{k}{2 k}=\frac{1}{2} & 1 \text { bod } \\
\operatorname{tg} \frac{\beta}{2}=\frac{k}{3 k}=\frac{1}{3} & 1 \text { bod } \\
\operatorname{tg} \alpha & =\frac{2 \operatorname{tg} \frac{\alpha}{2}}{1-\operatorname{tg}^{2} \frac{\alpha}{2}}=\frac{4}{3} & 1 \text { bod } \\
\operatorname{tg} \beta & =\frac{2 \operatorname{tg} \frac{\beta}{2}}{1-\operatorname{tg}^{2} \frac{\beta}{2}}=\frac{3}{4} & 1 \text { bod } \\
\operatorname{tg} \alpha=\frac{1}{\operatorname{tg} \beta}=\operatorname{ctg} \beta \Rightarrow \alpha+\beta=90^{\circ}, \gamma=90^{\circ} . & 2 \text { boda }
\end{array}
$$

