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In $\triangle ABC$, $D$ is on $\overline{CA}$, $\angle B=120^\circ$, $\angle C=2x$, $\angle ADB=3x$, $\overline{AD}$=$\overline{BC}$

How do I find $x$? The solution is $x=10^\circ$, but I can't find it.

I tried to rotate triangle BDC and make a equilateral, but I thought that I missed some clues...

Michael Rozenberg
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sim
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2 Answers2

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With Euclidean geometry it is even simpler. Here's a possible path. Consider the Figure below where $AC$ has been produced to $E$ so that $AD\cong ED$.

enter image description here

  1. Angle chasing gives $\angle CAB = 60^\circ -2x$. Hence $\angle AED = 60^\circ-2x$.
  2. Since $\angle ECD = 60^\circ+x$, we have that triangle $\triangle CDE$ is isosceles, yielding $EC\cong ED$.
  3. From the hypothesis $AD\cong CB$ we therefore get that $\triangle CBE$ is equilateral, and $\triangle DBE$ isososceles.
  4. From $$\angle EDB + \angle EDC + \angle CDA = 180^\circ$$ we get the equation $$60^\circ+2x + 60^\circ+x+3x = 180^\circ$$ leading directly to the expected solution $$\boxed{x=10^\circ}.$$
dfnu
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enter image description here

Note $\angle A = 180 - 3x - (120-x) = 60-2x$. Apply the sine rule to the triangles ABD and BCD,

$$\frac ab = \frac{\sin(60-2x)}{\sin(120-x)}=\frac{\sin 2x}{\sin 3x}$$

Rearrange and then factorize the equation as follows,

$$\sin(120-x)\sin 2x = \sin(60-2x)\sin 3x$$ $$\cos(120-3x)-\cos(120+x)=\cos(60-5x)-\cos(60+x)$$ $$\cos(120-3x)+\cos(60+x)=\cos(120+x)+\cos(60-5x)$$ $$\sin x\cos(30-2x)= \sin 2x\cos(30+3x)$$ $$\sin x[\cos(30-2x)-2\cos x\cos(30+3x)]=0$$ $$\sin x[\cos(30-2x)-\cos(30+2x)-\cos(30+4x)]=0$$ $$\sin x[\sin 2x-\sin(60-4x)]=0$$ $$\sin x\cos(30-x)\sin(30-3x)=0$$

which leads to $\sin(30-3x)=0$ and the solution $x=10$.

Quanto
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