Question

razones trigonometricas 37°/2

Answer 1

Respuesta:

En la imagen tenemos al triangulo de ahí

sen 37°/2 = $\frac{k}{\sqrt{10}k } = \frac{k}{\sqrt{10}.k }.\frac{\sqrt{10} }{\sqrt{10}} = \frac{k\sqrt{10} }{10}$

cos 37°/2 = $\frac{3k}{\sqrt{10}k } = \frac{3k}{\sqrt{10}.k }.\frac{\sqrt{10} }{\sqrt{10}} = \frac{3k\sqrt{10} }{10}$

tan 37°/2 = $\frac{k}{3k}$

cot 37°/2 = $\frac{3k}{k}$

sec 37°/2 = $\frac{\sqrt{10}k }{3k}$

csc 37°/2 = $\frac{\sqrt{10}k }{k}$