How to make correct system of equations to solve for the angles in this triangle?

Question

I'm trying to solve this triangle for $X$. Thereby, I've tried to make correct system of equations. What would be the correct equations?

Here are the equations I can find

• In $\triangle ABC$, recalling that $\angle ACD = y$

$$48 + 24 + x + 12 + y = 180$$

• In $\triangle ADC$

$$84+x + y = 180$$

However, I'm getting the same equations.

Regards

Answer 1

The key insight is to observe that $\angle(BDA)=\angle(BDC)=138$. From here in triangle $BDC$ we have $AB/\sin(138)=BD/\sin(24)$ and from triangle BCD we get $BC/\sin(138)=BD/\sin(12)$. Divingig both equations gives $$BC=\frac{AB\sin(24)}{\sin(12)}.$$

Now use law of cosine in $ABC$ to get $$AC^2=AB^2+BC^2-2AB\cdot BC\cos(48) =BC^2\left(\frac{\sin^2(12)}{\sin^2(24)}+1 -2\frac{\sin(12)}{\sin(24)}{\cos(48)}\right).$$

Finally use law of sine in triangle $ABC$: $$\frac{\sin(x+24)}{BC}=\frac{\sin(48)}{AC}$$ plug in $AB$ to get $$\sin(x+48)=\frac{\sin(48)}{\sqrt{\dfrac{\sin^2(12)}{\sin^2(24)}+1 -2\dfrac{\sin(12)}{\sin(24)}{\cos(48)}}}$$

Answer 2

I use degrees, but omit the notation.

With the use of the law of sines we have $$\frac{AD}{\sin 18}=\frac{AB}{\sin 138}=\frac{BD}{\sin 24}.$$

One lenght can be chosen arbitrarily, because convenient triangles are similar. Set $AD=1.$ We get the lenghts $AB, BD.$

Similarly we obtain the sides of $\triangle BCD.$

It remains the side $AC,$ common to two triangles. If I have not mistaken, $$AC=\frac{CD\sin 84}{\sin X}=\frac{BC \sin 48}{\sin(24+X)}$$ From this one gets easily $\tan X.$