8,9

In triangle $ABC$, the lengths of the sides are given as $AB=\sqrt{2}$, $BC=\sqrt{5}$, and $AC=3$. Compare the measure of angle $BOC$ and $112.5^{\circ}$, where $O$ is the center of the inscribed circle in triangle $ABC$.
Apply the Law of Cosines and the formula $\angle B O C=90^{\circ}+\frac{1}{2} \angle B A C$.

## Solution

By the Law of Cosines

$$
\cos \angle B A C=\frac{A B^{2}+A C^{2}-B C^{2}}{2 \cdot A B \cdot A C}=\frac{2+9-5}{2 \cdot \sqrt{2} \cdot 3}=\frac{1}{\sqrt{2}}
$$

therefore $\angle B A C=45^{\circ}$. Consequently,

$$
\angle B O C=180^{\circ}-\frac{1}{2}\left(180^{\circ}-\angle B A C\right)=90^{\circ}+\frac{1}{2} \angle B A C=90^{\circ}+22.5^{\circ}=112.5^{\circ}
$$

![](https://cdn.mathpix.com/cropped/2024_05_06_1b17058527caf5b5d058g-29.jpg?height=375&width=640&top_left_y=2208&top_left_x=709)

Answer

$\angle B O C=112.5^{\circ}$.

In triangle $A B C$, the lengths of the sides are given as $A B=\sqrt{7}, B C=4$ and $A C=\sqrt{3}$. Compare the measure of angle $A O B$ and $105^{\circ}$, if $O$ is the center of the inscribed circle in triangle $A B C$.

## Hint

Apply the Law of Cosines and the formula $\angle A O B=90^{\circ}+\frac{1}{2} \angle A C B$.

## Answer

$\angle A O B=105^{\circ}$.