How to solve triangle side length with unknown variable?

Question

How do I solve this problem? I think it has to do with sinus, but I'm not sure. I'm quite new to this area so I would appreciate a detailed explanation. Thanks!

Answer 1

Here is a method that does not use the cosine's law but an observation : whatever $y$, we have

$$CA+CB=13.5$$

Therefore, $C$ belongs to the ellipse with foci $A$ and $B$ and parameters $2a=13.5$ and interfocal distance $2f=4.5$ giving :

$$b^2=a^2-f^2=81/2$$

Taking (see figure) coordinate axes such that $A(-2.25 ; 0)$ and $B(2.25 ; 0)$, we get the following equation :

$$\dfrac{x^2}{(27/4)^2}+\dfrac{y^2}{(81/2)}=1 \tag{1}$$

Besides, straight line passing through $A$ with a $70°$ slope angle has equation

$$y=\tan(70°)(x+2.25).$$

It suffices now to find the intersection point of this ellipse and the straight line.

We find the point $C$ with coordinates

$$(x_0= 0.066181051712292,\ \ y_0=6.363655138948179)$$

Now, one must take care : the value of $y_0$ found here isn't the same as the looked for $y$. One has to solve $9-y=CA=\sqrt{(x_0+2.25)^2+(y_0-0)^2}$ finally giving :

$$\color{red}{y=2.227939649429236}$$

Remark : This method is not the shortest ; its interest is that it brings a better understanding of the issue, especially in connection with the value of the angle which can be taken in the whole range $(0,180°)$.