Doubling The Angle Does What Exactly?

Relevant wiki: Tangent - Sum and Difference Formulas

As the areas are equal, we get:

$\frac{1}{2}\times2\times{a}=\frac{1}{2}\times4\times{b}$

$a=2b$

Now let's take a look at the qiven angles and use the following formula $\tan(2\alpha)=\frac{2\tan(\alpha)}{1-\tan^2(\alpha)}$ :

Judging by the triangles, we get: $\tan(\alpha)=\frac{2}{a}$ and $\tan(2\alpha)=\frac{b}{4}$ , which leads into

$\frac{b}{4}=\frac{2\times{\frac{2}{a}}}{1-(\frac{2}{a})^2}$

Substituting $a=2b$ leads into:

$\frac{b}{4}=\frac{\frac{2}{b}}{1-(\frac{1}{b})^2}$

$b-\frac{1}{b}=\frac{8}{b}$ , as $1-(\frac{1}{b})^2=0$ would mean $b=1$ and $a=2$ , so $\alpha=\frac{\pi}{4}$ and $2\alpha=\frac{\pi}{2}$ , so the triangle with $2\alpha$ wouldn't exist.

$b^2-1=8$

$b=\pm 3$

$a=2b=2\times3=6$ which means $a+b=6+3=\boxed{9}$ .

Do triângulo I, temos que:

$\\ \mathsf{\tan \alpha = \frac{2}{a} \Leftrightarrow \frac{\sin \alpha}{\cos \alpha} = \frac{2}{a}}$

Desse modo, $\mathsf{\exists \ k \in \mathbb{N}}$ tal que $\mathsf{\sin \alpha = 2 \cdot k}$ e $\mathsf{\cos \alpha = a \cdot k}$ .

Do triângulo II, temos:

$\\ \large \mathsf{\tan (2\alpha) = \frac{b}{4}} \\\\\\ \mathsf{\frac{\sin (2\alpha)}{\cos (2\alpha)} = \frac{b}{4}} \\\\\\ \mathsf{\frac{2 \cdot \sin \alpha \cdot \cos \alpha}{\cos^2 \alpha - \sin^2 \alpha} = \frac{b}{4}} \\\\\\ \mathsf{\frac{2 \cdot 2k \cdot ak}{a^2k^2 - 4k^2} = \frac{b}{4}}$

$\\ \large \mathsf{\frac{4a\diagup \!\!\!\! k^2}{(a^2 - 4)\diagup \!\!\!\! k^2} = \frac{b}{4}} \\\\\\ \mathsf{\frac{4a}{(a^2 - 4)} = \frac{b}{4}}$

Mas, de acordo com o enunciado as áreas são iguais, então:

$\\ \mathsf{\frac{a \cdot 2}{2} = \frac{4 \cdot b}{2}} \\\\\\ \mathsf{2a = 4b} \\\\\\ \mathsf{a = 2b}$

Isto posto,

$\\ \large \mathsf{\frac{4a}{a^2 - 4} = \frac{b}{4}} \\\\\\ \mathsf{\frac{4 \cdot 2b}{4b^2 - 4} = \frac{b}{4}} \\\\\\ \mathsf{\frac{8b}{4(b^2 - 1)} = \frac{b}{4}} \\\\\\ \mathsf{b^2 - 1 = 8} \\\\\\ \mathsf{b = \pm 3} \\\\\\ \boxed{\mathsf{b = 3}}$

Com efeito, uma vez que $\mathsf{a = 2b}$ , tiramos que: $\boxed{\mathsf{a = 6}}$

Por fim,

$\\ \mathsf{a + b = 6 + 3} \\\\ \boxed{\boxed{\mathsf{a + b = 9}}}$

There are some special right triangles, one of them has sides of 3,4,5. Since the both triangles has the same are and taking b as 3 we have an area of 6 we would need a 6 on a to get an area of 6 on that triangle. Therfore 6+3=9