8.2 The Six Trigonometric Ratios

First, find the hypotenuse $AC$ using Pythagoras Theorem.

$AC^2 = AB^2 + BC^2 = 12^2 + 5^2 = 144 + 25 = 169$.

$AC = \sqrt{169} = 13$.

For angle A: Base ($B$) is Adjacent side $AB=12$. Perpendicular ($P$) is Opposite side $BC=5$. Hypotenuse ($H$) is $AC=13$.

Calculate Primary Ratios: $\sin A = P/H = 5/13$ $\cos A = B/H = 12/13$ $\tan A = P/B = 5/12$

Calculate Reciprocal Ratios: $\text{cosec } A = 13/5$ $\sec A = 13/12$ $\cot A = 12/5$