Find the length of a triangle's side

Question

I have the following triangle and I'm supposed to find the value of x.

First thought that came to mind is to use the the following tangent equation

$$\tan(y)=\frac{x}{27}$$ and $$\tan(19+y) = \frac{2x}{27}$$ which implies that $$\tan(19+y) =2\tan(y)$$ and solve for $y$ and once I've found $y$ I can easily find $x$.

I don't have an easy solution for the last equation and I feel that I'm complicating things unnecessarily. (It's a grade 10 geometry question).

Edit: Using the fact $\tan(a+b) = \dfrac{\tan(a)+\tan(b)}{1-\tan(a)\tan(b)}$ yields the following quadratic equation $$2\tan(19)\tan^2(y)-\tan(y)+\tan(19) = 0$$which gives you that $\tan(y) = 0.890833$ or $\tan(y)= 0.561272$ which means $x = 27\times0.890833$ or $x= 27\times 0.561272$.

So the question is now which $x$ should I pick and why? and is this more complicated than it should be?

Answer 1

As you noticed $2\tan(y)=\tan(19+y)$. Hence $$2\tan(y)=\frac{2x}{27}=\tan(19+y)=\frac{\tan(19)+\tan(y)}{1-\tan(y)\tan(19)}=\frac{\tan(19)+\frac{x}{27}}{1-\tan(19)·\frac{x}{27}}=\frac{27\tan(19)+x}{27-\tan(19)x}$$ $$\iff \frac{2x}{27}=\frac{27\tan(19)+x}{27-\tan(19)x}\iff 54x-2\tan(19)·x^2=729\tan(19)+27x$$ $$\iff 2\tan(19)·x^2-27x+729\tan(19)=0$$ This gives $$x=15.15...\lor x=24.05...$$

Now, as you can see in the following image (made with geogebra), both values are correct:

Answer 2

I think you can just use the sinus rule and pythagorean theorem, then: $$\frac{x}{\sin(19)} = \frac{\sqrt{x^2+27^2}}{\sin(71-y)}$$ And we see: $$\sin(71-y)=\frac{27}{\sqrt{4x^2+27^2}}$$ Thus, if we substitute: $$\frac{x}{\sin(19)} = \frac{\sqrt{x^2+27^2}\sqrt{4x^2+27^2}}{27}$$ Solving for $x$: $$x=15.15...\lor x=24.05...$$ ... witch is in compliance with your answer.

PS. Both values of $x$ comply, because they satisfy the equation. And is that tan property known in 10th grade? Otherwise use the above.