Cute Math problem

Find x , the angle marked “ ? “ in the diagram

Let O be the inside point of ABCD so there are 4 triangles

AOB, AOD, BOC and COD

Let x= angle CDO, y= angle DCO

From triangle AOB get angle AOB= 110 so angle BOC= 70

and so angle BCO = 40 and x+y= 70

Using the law of sine in triangle AOB,

let AO= u sin40 , BO= u sin30 , AB= u sin110

in triangle AOD, let AO= w sin80 , DO= w sin30

so AO = u sin40 = w sin80 (1)

in triangle COD, let DO= v sin y , CO= v sin x

so DO = w sin 30 = v sin y (2)

in triangle BOC, let CO= z sin 70 , BO= z sin 40

so CO= v sin x = z sin 70 (3)

and BO= z sin 40 = u sin 30 (4)

from (4) get z= u sin30 / sin40 (5)

from (1) get w= u sin40 / sin80 (6)

from (2) get v= w sin30 / sin y (7)

from (3) get z sin70 = u sin30 (sin70) / sin40 = v sin x using (5)

= w sin30 (sin x) / sin y using (7)

= u sin40 (sin30 sin x) / (sin80 sin y) by (6)

so cancel u sin30 and cross multiply from line 1 and 3 get

sin70 sin80 sin y = sin x sin40 sin40

so 2 cos40 sin70 sin y = sin x sin40 since sin80= 2 sin40 cos40

now sin y = sin (70 — x) = sin70 cos x — cos70 sin x

so 2 cos40 sin70 (sin70 cos x — cos70 sin x) = sin x sin40

2 cos40 sin70 (sin70 — cos70 tan x) = tan x sin40

tan x = 2 cos40 sin70 sin70 / (sin40 + 2 sin70 cos70 cos40)

= 2 cos40 cos20 cos20 / (sin40 + sin40 cos40)

since sin70 = cos(90–70) and 2 sin70 cos70 = sin140 =sin(180- 140)

now sin40 = 2 sin20 cos20 so tan x = cos20 cos40 / sin20 (1+ cos40)

now 1 + cos 2A = 2 cosA cos A so 1+cos40 = 2 cos20 cos20

so tan x = cos40 / (2 sin20 cos20) = cos40 / sin40 = cot 40 = tan(90–40)

therefore tan x = tan 50 and x = 50