A decagon has 10 sides.
It it is regular you can build 10 isosceles triangles from the center of the decagon to the 10 sides.
Each triangle has a common vertex where the angle of each triangle is 360° / 10 = 36°.
So each time that you rotate the decagon a multiple of 36° around the center you get an image that coincides with the original decagon.
If the letters are given clockwise:
- when you rotate 36° counter clockwise, the point A' (the image of A) will coincide with the point J.
- when you rotate 72° (2 times 36°) counter clockwise, the point A' will land on I.
- when you rotate 108° (3 times 36°) counter clockwise, the point A' will land on H.
- when you rotate 144° (4 times 36°) counter clockwise, the point A' will land on G.
- when you rotate 180° (5 times 36°) counter clockwise, the point A' will land on I.
- when you rotate 216° (6 times 36°) counter clockwise, the point A' will land on E.
- whn you rotate 252° (7 times 36°) counterclockwise, the point A' will coincide with D.
Add other 36° each time and A' will coincide successively with C, B and the same A.
The answer would be 20 im pretty sure
Let m∠CLN = x. Then m∠ALM = 3x, and m∠A = 90°-x, m∠C = 90°-3x.
The sum of angles of ∆ABC is 180°, so we have
... 180° = 40° + m∠A + m∠C
Using the above expressions for m∠A and m∠C, we can write ...
... 180° = 40° + (90° -x) + (90° -3x)
... 4x = 40° . . . . . . . . . add 4x-180°
... x = 10°
From which we conclude ...
... m∠C = 90°-3x = 90° - 3·10° = 60°
The ratio of CN to CL is
... CN/CL = cos(∠C) = cos(60°)
... CN/CL = 1/2
so ...
... CN = (1/2)CL
