This thing is given by our Math teacher and our "team"(The best 5 students in the grade) does not have any idea how to solve that. But all techniques used are 2 grades lower that our grade...(That is provided by our teacher) :(.

Find x.

MaxWong
Jun 7, 2017

#2**+2 **

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.

.

Alan
Jun 7, 2017

#2**+2 **

Best Answer

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.

.

Alan
Jun 7, 2017