Question

Please help me out with these questions. It's trigonometry

Answer 1

Step-by-step explanation:

LOOK AT THE PICTURE

In ΔABC:

$\sin\alpha=\dfrac{x}{y+p}\\\\\cos\alpha=\dfrac{q}{y+p}\\\\\tan\alpha=\dfrac{x}{q}$

In ΔADC

$\sin\alpha=\dfrac{y}{x}\\\\\cos\alpha=\dfrac{z}{x}\\\\\tan\alpha=\dfrac{y}{z}$

In ΔBCD

$\sin\alpha=\dfrac{z}{q}\\\\\cos\alpha=\dfrac{p}{q}\\\\\tan\alpha=\dfrac{z}{p}$