What on earth is this!?

Here's one approach:

Without loss of generality let AB = 1

Angle ADB = 180-80-50 = 50 so triangle ABD is isosceles and AD = AB = 1

Angle AEB = 180-60-80 = 40,  so AE/sin(80) = 1/sin(40) or AE = sin(80)/sin(40)

Angle BDE = 180-30-(40+x) = 110-x

Angle ADE = ADB + BDE = 50 + 110 - x = 160 - x

Sine rule on triangle ADE:  

 sin(x)/AD = sin(160-x)/AE or sin(x)/1 = sin(160-x)*sin(40)/sin(80)

Rearrange:  sin(80)*sin(x) = sin(40)*sin(160-x)

 sin(80)*sin(x) = sin(40)*(sin(160)*cos(x) - cos(160)*sin(x))

I'm sure you can take it from here (collect sin(x) terms on one side, cos(x) on the other, divide to get tan( x) ...etc.)

As Guest #1 notes, you should find x = 30 degrees.

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hectictar: You can do this!. Hint: angle AEB=40 degrees, and x=30 degrees !!.

yo tengo una preg pata ti haber sie eres inteligente

dos angulos coeterminales estan en relacion de 11a 3 halla el valor del menor de eellos si  el mayor se encuentra en el intervalo de 450 a500