Math Olympiad Geometry Problem | How to solve for "x" in this problem? @MathOlympiad0

[@key_board_x]

tan(x) = AC/AB
tan(x) = (3 + a)/b
Triangle ABD → AB² + AD² = BD²
b² + a² = 5²
a² + b² = 25
Triangle ABC → AB² + AC² = BC²
b² + (3 + a)² = 7²
b² + 9 + 6a + a² = 49
a² + b² + 6a = 40 → recall: a² + b² = 25
25 + 6a = 40
6a = 40 - 25
6a = 15
a = 15/6
a = 5/2
Recall: a² + b² = 25
b² = 25 - a²
b² = 25 - (5/2)²
b² = (100/4) - (25/4)
b² = 75/4
b = ± √(75/4) ← as b represents a length, we keep only the positive value
b = √(75/4)
b = √[3 * (25/4)]
b = √[3 * (5/2)²]
b = (5/2).√3
b = (5√3)/2
Recall:
tan(x) = (3 + a)/b → where: a = 5/2
tan(x) = [3 + (5/2)]/b
tan(x) = [(6/2) + (5/2)]/b
tan(x) = (11/2)/b → where: b = (5√3)/2
tan(x) = (11/2)/[(5√3)/2]
tan(x) = (11/2) * [2/(5√3)]
tan(x) = 11/(5√3)
tan(x) = (11 * √3)/(5√3 * √3)
tan(x) = (11√3)/15
→ x = 51.7868 °

[@MathOlympiad0]

You are a genius and so sweet my friend ❤️❤️