Respuesta :

gmany

Step-by-step explanation:

LOOK AT THE PICTURE

In ΔABC:

[tex]\sin\alpha=\dfrac{x}{y+p}\\\\\cos\alpha=\dfrac{q}{y+p}\\\\\tan\alpha=\dfrac{x}{q}[/tex]

In ΔADC

[tex]\sin\alpha=\dfrac{y}{x}\\\\\cos\alpha=\dfrac{z}{x}\\\\\tan\alpha=\dfrac{y}{z}[/tex]

In ΔBCD

[tex]\sin\alpha=\dfrac{z}{q}\\\\\cos\alpha=\dfrac{p}{q}\\\\\tan\alpha=\dfrac{z}{p}[/tex]

Ver imagen gmany