Given triangle ABC, where ∠C is a right angle.
[tex]\begin{gathered} \text{Using }\angle A \\ \\ \sin A=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{a}{c} \\ \cos A=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{b}{c} \\ \tan A=\frac{\text{opposite}}{\text{adjacent}}=\frac{a}{b} \\ \\ \text{Using }\angle B \\ \sin B=\frac{\text{opposite}}{\text{hypotenuse}}=\frac{b}{c} \\ \cos B=\frac{\text{adjacent}}{\text{hypotenuse}}=\frac{a}{c} \\ \tan B=\frac{\text{opposite}}{\text{adjacent}}=\frac{b}{a} \end{gathered}[/tex]The law of sine states that:
[tex]\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}[/tex]The law of cosine states that:
[tex]c=\sqrt[]{a^2+b^2-2ab\cdot\cos C}[/tex]