Trigonometry! 77

Using trigonometric version of Ceva's Theorem to $\Delta CDB$ ,

$\dfrac{\sin(DCE)}{\sin(ECB)} \times\dfrac{\sin(BDE)}{\sin(EDC)} \times \dfrac{\sin(CBE)}{\sin(EBD)} \\ \Rightarrow \dfrac{\sin(30)}{\sin(50)} \times\dfrac{\sin(-x)}{\sin(x+30)} \times \dfrac{\sin(80)}{\sin(-10)} =1 \\ \Rightarrow \dfrac{\sin(30)}{\sin(50)} \times\dfrac{\sin(-x)}{\sin(x+30)} \times \dfrac{8\cos(40)\cos(20)\sin(10)\cos(10)}{\sin(-10)}=1 \\ \because \sin(50)=\cos(40) \Rightarrow 8\sin(30)\sin(x)\cos(10)\cos(20)=\sin(x+30) \\ \Rightarrow 4\sin(x)\cos(10)\cos(20)=\sin(x+30) \\ \Rightarrow 2\cos(20)[sin(x+10)+\sin(x-10)] =\sin(x+30) \\ \Rightarrow 2\cos(20)sin(x+10)+2\cos(20)sin(x-10) = \sin(x+30) \\ \Rightarrow \sin(x+30)+\sin(x-30)+\sin(x+10)+\sin(x-30)=\sin(x+30) \\ \Rightarrow \sin(x-30)+\sin(x+10)=-\sin(x-30) \\ \Rightarrow 2\sin(x)\cos(10)=\sin(30-x)$

We know that $2\sin\theta\cos\theta = sin(2\theta)$ . Compare this equation with the above equation that we have got. To get a solution , $\theta = x=10$ and $2\theta=30-x$ . So by cross check , $30-x=30-10=20 = 2x$ . Thus $\large\boxed{x=10}$ .

$\angle BED = 170^{\circ} - x$

In $\Delta BED$ , by sine rule,

$\frac{\sin (170^{\circ} - x)}{\sin(x)} = \frac{BD}{BE}$

And in $\Delta BCD$ , by sine rule

$\frac{\sin (80^{\circ})}{\sin(30^{\circ})} = \frac{BD}{BC}$

Now, $\Delta BEC$ is isosceles, and hence $BE=BC$ .

Hence we have the following equality.

$\frac{\sin(170^{\circ}-x)}{\sin(x)}=\frac{\sin(80^{\circ})}{\sin(30^{\circ})}$

$\frac{\sin(x+10^{\circ})}{\sin(x)}=\frac{\cos(10^{\circ})}{\frac{1}{2}}$

$\sin(x)\cos(10^{\circ}) + \cos(x)\sin(10^{\circ}) = 2\sin(x)\cos(10^{\circ})$

$\cos(x)\sin(10^{\circ}) = \sin(x) \cos(10^{\circ})$

$\tan(x)=\tan(10^{\circ})$

$\therefore \boxed{x=10^{\circ}}$

Could someone try and prove this for me using Trigonometric Version of Ceva's Theorem?

Kudou Shinichi Here you go.