Question

In a $\Delta$ABC, right angled at B, if AB = 12 and BC = 5, find:
$\cos C$ and $\cot C$

Answer 1

By Pythagoras theorem, we have
$A{C}^2=A{B}^2+B{C}^2$
TRIGONOMETRICRATIOS
$\Rightarrow A{C}^2={12}^2+5^2$
$\Rightarrow A{C}^2=169$
$\Rightarrow AC=13$
When we consider t-ratios of $\angle C$, we have
Base = BC = 5, Perpendicular = AB = 12 and, Hypotenuse = AC = 13
$\therefore cosC=\frac{Base}{Hypotenuse}=\frac{5}{13}andcotC=\frac{Base}{Perpendicular}=\frac{Base}{Perpendicular}=\frac{5}{12}$