make CE=CD=y AE=x-y 4/sin5=AD/sin90 4/sin5=AD

If the use of the Law of Sines is not forbidden, then the problem only takes a few steps. The hypotenuse of the left triangle [line segment AD] is 45.89; since angle BAD is 5 degrees and the base = 4 [using Law of Sines] Since all the angles of triangle ACD are known and the length of side AD = 45.89, the other two sides can be determined using the Law of Sines. Side X = 263.26 and side Y = 255.26. X - Y = 8

I must agree the second method a lot easier than the first. I will have to use that as a practice!

I like the second method ! Thank you sir.

In general, X-Y=2*BD, as long as angle BCA is twice the angle BAD.

there is something wrong: the Angle BAC has to be 90-10=80 degrees. Therefore angle BAD =5 degrees. But the angle ADC=95 degrees whiich is not correct. Correct me if I am wrong, please.

Second method is brilliant

Second method was classic

8 Use the law of sines twice

Difícil, mas muito bonita questão.

bro.... i solve this..

Second method 👍

A variant of Method 2: Produce _CB_ to _E_ such that _|EC| = X._ _ΔECA_ is isosceles with apex angle _10°. ∴ ∠AEC = 85°_ But also _∠EDA = 75⁰ + 10⁰ = 85⁰_ (exterior angle to _ΔADC)._ ∴ _ΔAED_ is isosceles. ∴ _AB_ bisects _ED._ ∴ _|EB| = |BD| = 4_ ∴ *_X - Y = |ED| = 8_*

If you actually draw this figure out accurately, the dimensions are surprising. X is 263.29, and AB is 45.72. It's surprising to see how narrow a 5° or 10° angle actually is after looking at all these videos with angles and triangles that look similarly for the sake of clarity.