Come on in, junior high school math finale.

Explanation:

The answers upstairs didn't solve the real problem, but they all used the knowledge of high school triangle. It is not too difficult to use the knowledge of junior high school.

Learn about this conclusion of junior high school mathematics first:

In Rt△ABC, ∠ c = 90.

Then Sina = BC/AB = COSB = COS (90-A).

Answer the prompt:

Situation 1:

Let ∠ DAB' = α, ∠ BAE = ∠ B 'AE = β.

Then α+2β = 90.

So sin ∠ dab' = sinα = sin (90-2β) = cos2β.

As long as an acute angle in a right triangle is equal to 2β, the problem will be solved.

Extend AB to m, make BM = AB, connect EM, make MN⊥ ray AF, and let the vertical foot be n.

Then ∠ men = 2β.

Obviously, AB = BM = 3, BE = 2, AE = ME = √ 13.

According to s △ AEM = am * be/2 = AE * Mn/2 (or by similarity)

Calculate Mn = 12/√ 13.

So en = 5/√ 13.

So sin ∠ dab' = sinα = sin (90-2β) = cos2β.

= cos∠MEN = EN/EM =(5/√ 13)√ 13 = 5/ 13

(Actually, we don't need the conclusion of Sina = COS (90-A) to solve the problem: in Rt△AMN, it is not difficult to prove ∠ CME = ∠ DAB' first, and then find the conclusion in △MEF. The following situation 2 is the same)

Case 2:

Let ∠ dab' = α, ∠ daf = γ.

Then BAE = α+γ.

So α+2γ = 90.

So sin ∠ dab' = sinα = sin (90-2γ) = cos2γ.

Similarly, as long as an acute angle equal to 2γ is constructed in a right triangle, the problem will be solved.

Obviously, the midpoint of AE is F, connecting BF, BP⊥AF, and the vertical foot is P.

Then ∠ AFB = 2 ∠ E = 2γ.

Obviously, AB = 3, BE = 6, AE = 3 √ 5, BF = AE/2 = 3 √ 5/2.

According to s △ aeb = ab * be/2 = AE * BP/2.

Calculate BP = 6/√ 5.

So FP = 9/(2 √ 5)