35. Given $\triangle A B C$ with side $\overline{A B}=2 \mathrm{~cm}, \alpha=30^{\circ}$ and $\beta=45^{\circ}$. Calculate the lengths of sides $B C$ and $A C$.
Solution. Let the intersection point of the height $h_{c}$ and the side $A B$ be point $D$. From $\beta=45^{\circ}$ and $h_{C} \perp A B$ it follows that $\measuredangle B C D=45^{\circ}$, i.e., $\triangle B C D$ is an isosceles right triangle and $\overline{B D}=h_{c}$. Then from $\sin 45^{\circ}=\frac{h_{c}}{a}$ it follows that $a=\sqrt{2} h_{c}$. From $\sin 30^{\circ}=\frac{h_{c}}{b}$ it follows that $b=2 h_{c}$. From $\triangle A D C$ and $\triangle B D C$ we have

$$
\left(2 h_{c}\right)^{2}-\left(2-h_{c}\right)^{2}=h_{c}^{2} \text { and }\left(\sqrt{2} h_{c}\right)^{2}-h_{c}^{2}=h_{c}^{2} .
$$

By equating the last equalities we have

$$
\left(2 h_{c}\right)^{2}-\left(2-h_{c}\right)^{2}=\left(\sqrt{2} h_{c}\right)^{2}-h_{c}^{2} \text { i.e. } 2 h_{c}^{2}+4 h_{c}-4=0 \text {. }
$$

By solving the quadratic equation we get $h_{c}=\sqrt{3}-1 \mathrm{~cm}$. It follows that $a=\sqrt{2}(\sqrt{3}-1) \mathrm{~cm}$, and $b=2(\sqrt{3}-1) \mathrm{~cm}$.

## Third Year