Question

If in $\Delta ABC,\angle B={90}^{\circ },a=12cm,b=13cm,$ then write all the trigonometric ratios for $\angle A$ and $\angle C$

Answer 1

Given that $\angle B={90}^{\circ },AC=13cm,BC=12cm$
$\because$ $A{B}^2+B{C}^2=A{C}^2$
$\Rightarrow$ $AB=\sqrt{A{C}^2-B{C}^2}$
$=\sqrt{(13{)}^2-(12{)}^2}$
$=\sqrt{169-144}=\sqrt{25}$
$=5cm$
For acute angle $A$ , the value of trigonometric ratios are
$\sin A=\frac{BC}{AC}=\frac{12}{13},\cos A=\frac{AB}{AC}=\frac{5}{13}$,
$\tan A=\frac{BC}{AB}=\frac{12}{5}$
$cosecA=\frac{AC}{BC}=\frac{13}{12},\sec A=\frac{AC}{AB}=\frac{13}{5}$
$\cot A=\frac{AB}{BC}=\frac{5}{12}$