Çözüldü Semicircle - Euclid's Formula on Right Triangles - Trigonometry

SAT adaptation of a solved problem from an old book:

If a tangent line is drawn to the radius of a semicircle and another line passing from the other endpoint of the radius crosses semicircle and tangent line at the points C and D respectively so that |BD| = 4·|BC|, what is the acute angle between the radius and [BD]?
A) 15°
B) 30°
C) 45°
D) 60°
E) 75°
("Çözülmüş Trigonometri Problemleri", Cilt 3, Hasan Fehmi Ergin, 1962, page 171)

Author's Solution:
∡ABD = α
cos(α) = |AB| / |BD| ⇒ |BD| = |AB| / cos(α)....(1)
For the right-angled triangle ABC, cos(α) = |BC| / |AB| ⇒ |BC| = |AB|·cos(α)....(2)
Substituting the equations (1) and (2) to the relation given in the problem, |AB| / cos(α) = 4·|AB|·cos(α) and simplifying,
[ cos(α) ]^2 = 1 / 4 ⇒ cos(α) = ∓1 / 2 and for the right angle (which is less than 90°) from the table showing the trigonometric values of the degrees between 0° and 90° on page 27, α = 60°

Solution - 2
|BC| = x ⇒ |BD| = 4x ⇒ |CD| = |BD| - |BC| = 4x - x = 3x
Recalling an Euclidean formula for the right triangles, |AC|^2 = |BC|·|CD| = 3(x^2) ⇒ |AC| = x·√3
In the right triangle ACB, tan(α) = |AC| / |BC| = (x·√3) / x = √3 ⇒ α = arctan(√3) = 60°