Challenging Maths Problem [Geometry]?

Question

http://img.photobucket.com/albums/v627/nunx/findx.jpg

find the angle of x

what is the answer? how do u know?

Answer 1

x is 30°

The other four points are already labeled (the point in the lower-right corner is C, although there's a line there that partially obscures the letter), so let us label the point in the middle where the two lines intersect E. Now we have:

1: ∠BCE + ∠EBC + ∠CEB = 180° (since CEB is a triangle)
2: ∠CEB=70° (from the previous equation and the angles given on the figure)
3: ∠AED=70° (since ∠AED is vertical to ∠CEB)
4: ∠BEA = ∠DEC (since ∠BEA is vertical to ∠DEC)
5: ∠CEB + ∠BEA + ∠AED + ∠DEC = 360° (since they together form a complete circle)
6: ∠BEA = ∠DEC = 110° (from the previous two lines)
7: ∠EAB + ∠BEA + ∠ABE = 180° (since ABE is a triangle)
8: ∠EAB = 50° (from the previous two lines and the angle given on the figure)
9: ∠CDE + ∠DEC + ∠ECD = 180° (since CDE is a triangle)
10: ∠CDE = 40° (from lines 9 and 6)
11: ∠EDA + ∠DAE + ∠AED = 180° (since AED is a triangle)

Let us denote ∠DAE as y to speed up the proof. Now, moving on:

12: sin x/AE = sin y/DE (law of sines)
13: sin 50°/BE = sin 20°/AE (law of sines, line 8 and the angle given on the figure)
14: sin 50°/BE = sin 60°/CE (law of sines and the angles given on the figure)
15: sin 30°/DE = sin 40°/CE (law of sines, line 10 and the angle given on the figure)

So we have:

16: sin 20°/AE = sin 60°/CE (lines 13 and 14)
17: CE = sin 60°/sin 20° AE (from the previous line)
18: sin 30°/DE = sin 40°/(sin 60°/sin 20° AE) (from previous two lines)
19: sin 30°/sin 40° * sin 60°/sin 20° = DE/AE (from the previous line)
20: sin y/sin x = DE/AE (from line 12)
21: sin y/sin x = sin 30°/sin 40° * sin 60°/sin 20° (from the previous two lines)
22: x + y + 70° = 180° (from lines 11 and 3)

Through these considerations, we have reduced the problem to a system of two equations and two unknowns, which we may solve algebraically.

23: y=110°-x (from line 22)
24: sin (110°-x)/sin x = sin 30°/sin 40° * sin 60°/sin 20° (from lines 23 and 21)
25: (sin 110° cos x - sin x cos 110°)/sin x = sin 30°/sin 40° * sin 60°/sin 20° (angle addition formula)
26: sin 110° cot x - cos 110° = sin 30°/sin 40° * sin 60°/sin 20° (from previous line)
27: cot x = (sin 30°/sin 40° * sin 60°/sin 20° + cos 110°)/sin 110°
28: tan x = sin 110°/(sin 30°/sin 40° * sin 60°/sin 20° + cos 110°)
29: x = arctan (sin 110°/(sin 30°/sin 40° * sin 60°/sin 20° + cos 110°))

30: x = 30°

Answer 2

Angle B = 80
Angle C = 80

Answer 3

Draw the line as reported. DC is a million/4 of the total length, FG is a million/32, FD is a million/sixteen. you will locate that H is 5/8 of how from A to B, so... GH = a million/32 + a million/sixteen + a million/4 + a million/8 of the total length. you apart from mght understand that DC = a million/4 = sixteen, so... (a million/8)*sixteen + (a million/4)*sixteen + sixteen + (a million/2)*sixteen = 30.

Answer 4

x = 10 degrees.

Where AC crosses BD call point P
You can find all angles except x and angle DAP

but x must have value that satisfies internal angles of triangle ADC and internal angles of triangle APD = 180 degrees.

Answer 5

20 DEGREES. It's the same as the angle directly opposite it. Don't llok at the entire problem, divide it into triangle

Answer 6

i say its 30