The base of an isosceles triangle is 40, the cosine of the apex angle is 15/17. Two vertices of the rectangle

The base of an isosceles triangle is 40, the cosine of the apex angle is 15/17. Two vertices of the rectangle lie on the base of the triangle, and the other two lie on the sides. Find the area of a rectangle if you know that one of its sides is twice the size of the other.

Let’s designate an isosceles triangle ABC, rectangle NLTK, N and K lie on the AC.
Let АN = KC = x, NK = b = 2a (by condition), then we write the АС as:
AC = 2 * x + 2 * a.
By condition, AC = 40, which means
2 * x + 2 * a = 40,
x + a = 20.
Find the segment x = AN.
To do this, consider similar triangles ALN and ABD, where BD is the bisector (in the isosceles also the height and median).
By assumption, α = arcsin 15/17
Then the angle L is equal to the angle α / 2, that is:
Angle L = angle α / 2 = (arcsin (15/17)) / 2.
Means,
AN = NL * tgL, where NL = a is the smaller side of the rectangle.
AN = x = a * tg ((arcsin (15/17)) / 2).
Thus, we got a system of equations:
x + a = 20;
x = a * tg ((arcsin (15/17)) / 2).
Hence,
a * (tg ((arcsin (15/17)) / 2) + 1) = 20,
a = 20 / (tg ((arcsin (15/17)) / 2) + 1).
Rectangle area:
S = a * b = 2 * a ^ 2.
tg ((arcsin (15/17)) / 2) = 0.0087.
a = 20 / (1 + 0.0087) ≈ 19.83
This means that the area of ​​the rectangle is S = 2 * a ^ 2 = 786.4578 ≈786.46.