Hi I'm trying to solve a 'triangle' problem, for some times but I can't really figure it out.
I have 2 triangles as following:
What informations do I have ?
I know the angles of $B_0$ and $B_1$ and
$c_0 + c_1 = l$
$a_0 = a_1 = a$
$b_0 + b_1 = 2a$
I want to find the value of $a$ (so the length of $a_1$ and $a_0$)
Do you have any idea how to solve this problem ?
Let $B_2$ be the common angle. Then from Sine Theorem we have:
$$\frac{\sin B_0}{b_0} = \frac{\sin B_2}{a_0} \quad \quad \text{and} \quad \quad \frac{\sin B_1}{b_1} = \frac{\sin B_2}{a_1}$$
Now both right sides are same and as you will get a relation for $b_0$ and $b_1$. Namely $\frac{\sin B_0}{b_0} = \frac{\sin B_1}{b_1}$. Now you have:
$$2a = b_0 + b_1 = b_0 + \frac{\sin B_1}{\sin B_0}\cdot b_0 \implies \frac{a}{b_0} = \frac{1 + \frac{\sin B_1}{\sin B_0}}{2}$$
Now from the very first equation you should be able to find the value of $B_2$
Now similarly from Sine Theorem we have:
$$\frac{\sin C_0}{c_0} = \frac{\sin B_2}{a_0} \quad \quad \text{and} \quad \quad \frac{\sin C_1}{c_1} = \frac{\sin B_2}{a_1}$$
So similarly express find a relation for $c_0$ and $c_1$ and you will get:
$$l = c_0 + c_1 = c_0 + \frac{\sin C_1}{\sin C_0}\cdot c_0 \implies c_0 = \frac{l}{1+ \frac{\sin C_1}{\sin C_0}}$$
Once you find $c_0$, the rest should be easy.