Easy hard geometry problem

[Guest]

Please, help me with this one. The answer is 30[tex]^\circ[/tex]. Could someone explain why?

[shyamjayakannan]

[tex]\angle EAC+ \angle ECA=40 ^\circ[/tex] by the exterior angle property and further, [tex]\angle EAC= \angle ECA[/tex] because [tex]\triangle EAC[/tex] is isosceles. So, [tex]\angle ECA=\frac{40 ^\circ }{2}=20 ^\circ[/tex]. Also, [tex]\angle AEC=180 ^\circ - 40 ^\circ=140 ^\circ[/tex]

Now, using the sine rule in [tex]\triangle EAC[/tex], we have [tex]\frac{EA}{ \sin{\angle ECA} }=\frac{AC}{\sin{\angle AEC}} \Rightarrow \frac{EA}{ \sin{20 ^\circ } }=\frac{AC}{\sin{140 ^\circ }} \Rightarrow \frac{EA}{ AC}=\frac{\sin{20 ^\circ }}{ \sin{140 ^\circ }}\ldots(1)[/tex]

Now, in [tex]\triangle BEC[/tex], let [tex]\angle EBC=x[/tex]. So, [tex]\angle BCE=180 ^\circ -40 ^\circ -x=120 ^\circ-x[/tex]. Using the sine rule, we have
[tex]\frac{EC}{ \sin{\angle EBC} }=\frac{EB}{\sin{\angle ECB}} \Rightarrow \frac{EC}{ \sin{x } }=\frac{EB}{\sin{(120 ^\circ-x )}} \Rightarrow \frac{EC}{ EB}=\frac{\sin{x}}{ \sin{(120 ^\circ-x) }}\ldots(2)[/tex]

Since [tex]EA=EC[/tex] and [tex]AC=EB[/tex], from [tex](1)[/tex] and [tex](2)[/tex], we have [tex]\frac{\sin{20 ^\circ }}{ \sin{140 ^\circ }}=\frac{\sin{x}}{ \sin{(120 ^\circ-x )}}[/tex]
[tex]\Rightarrow\sin{20 ^\circ }\sin{(120 ^\circ-x )}=\sin{140 ^\circ }\sin{x} \Rightarrow \sin{20 ^\circ }(\sin{120 ^\circ }\cos{x}-\cos{120 ^\circ}\sin{x})=\sin{140 ^\circ }\sin{x}[/tex]
[tex]\Rightarrow\sin{20 ^\circ }\sin{120 ^\circ }\cos{x}-\sin{20 ^\circ }\cos{120 ^\circ}\sin{x}=\sin{140 ^\circ }\sin{x}[/tex]
[tex]\Rightarrow \sin{20 ^\circ }\sin{120 ^\circ }\cos{x}=\sin{x}(\sin{140 ^\circ }+\sin{20 ^\circ }\cos{120 ^\circ})[/tex]
[tex]\Rightarrow\frac{\sin{x}}{\cos{x}}=\tan{x}=\frac{\sin{20 ^\circ }\sin{120 ^\circ }}{\sin{140 ^\circ }+\sin{20 ^\circ }\cos{120 ^\circ}}=\frac{\sin{20 ^\circ }\sin{120 ^\circ }}{\sin{(20 ^\circ +120 ^\circ )}+\sin{20 ^\circ }\cos{120 ^\circ}}[/tex]
[tex]\Rightarrow\tan{x}=\frac{\sin{20 ^\circ }\sin{120 ^\circ }}{\sin{20 ^\circ }\cos{120 ^\circ }+\cos{20 ^\circ }\sin{120 ^\circ }+\sin{20 ^\circ }\cos{120 ^\circ}}=\frac{\sin{20 ^\circ }\sin{120 ^\circ }}{\cos{20 ^\circ }\sin{120 ^\circ }+2\sin{20 ^\circ }\cos{120 ^\circ}}[/tex]
[tex]\displaystyle\Rightarrow\tan{x}=\frac{\displaystyle\sin{20 ^\circ }\times\frac{\sqrt{3}}{2}}{\displaystyle\cos{20 ^\circ }\times\frac{\sqrt{3}}{2}-2\sin{20 ^\circ }\times\frac{1}{2}}=\frac{\sqrt{3}\sin{20 ^\circ }}{\sqrt{3}\cos{20 ^\circ }-2\sin{20 ^\circ }} \Rightarrow\boxed{x \approx 32.12 ^\circ }[/tex]