In Matemáticas / Universidad | 2025-08-25
Asked by juanavalos1530
Hipotenusa del triángulo.y = √[(√3)² + (2 √5)²] = √(3 + 20) = √23senα = 2 √5 / √23 = 2 √115 / 23cosα = √3 / √23 = √69 / 23tgα = 2 √5 / √3 = 2 √15 / 3cotgα = 1 / tgα = √15 / 10secα = 1 / cosα = √69 / 3cosecα = 1 / senα = √115 / 10
Answered by mateorinaldi | 2025-08-25