## Task B-3.5.

The lengths of two sides of a triangle are $7 \mathrm{~cm}$ and $4 \mathrm{~cm}$. The angle opposite the longer side is twice as large as the angle opposite the shorter side. What is the length of the third side of the triangle?
## Solution.

Let $a=4 \text{ cm}, b=7 \text{ cm}, \beta=2 \alpha$.

Using the Law of Sines, we have $\frac{4}{\sin \alpha}=\frac{7}{\sin 2 \alpha}$

from which it follows that $4 \sin 2 \alpha=7 \sin \alpha$

or $8 \sin \alpha \cos \alpha=7 \sin \alpha$.

The sine of an angle in a triangle cannot be zero, so $\cos \alpha=\frac{7}{8}$.

We will determine the length of the third side $c$ using the Law of Cosines for side $a$

\[
\begin{gathered}
a^{2}=b^{2}+c^{2}-2 b c \cos \alpha \\
4^{2}=7^{2}+c^{2}-2 \cdot 7 \cdot c \cdot \frac{7}{8}
\end{gathered}
\]

from which we obtain the quadratic equation

\[
c^{2}-14 c+33=0
\]

The solutions to this equation are $c=4$ and $c=\frac{33}{4}$.

However, the solution $c=4$ must be discarded because for a triangle with sides $a=c=4, b=7$, $\beta=2 \alpha$ does not hold. Indeed, from $a=c$ and $\alpha=\gamma$ and $\beta=2 \alpha$ it follows that $\alpha=\gamma=45^{\circ}, \beta=90^{\circ}$, which clearly does not hold in this triangle.

The length of the third side of the triangle is $\frac{33}{4}=8.25 \text{ cm}$.