Question

In $△ABC$, right angled at $B,15sinA=12$. Find the other five trigonometric ratios of the angle $A$. Also find the six ratios of the angle $C$

Answer 1

Given that $15\sin A=12$, so $\sin A=\frac{12}{15}$. Let us consider $DABC$ (see fig), right angled at $B$, with $BC=12$ and $AC=15$. By Pythagoras theorem
$A{C}^2=A{B}^2+B{C}^2$
${15}^2=A{B}^2+{12}^2$
$A{B}^2={15}^2-{12}^2=225-144=81$
$\therefore AB=\sqrt{81}=9$
We now use the three sides to find the six trigonometric ratios of angle $A$ and angle $C$.
$\cos A=\frac{AB}{AC}=\frac{9}{15}=\frac{3}{5}\sin C=\frac{AB}{AC}=\frac{9}{15}=\frac{3}{5}$
$\tan A=\frac{BC}{AB}=\frac{12}{9}=\frac{4}{3}\cos C=\frac{BC}{AC}=\frac{12}{15}=\frac{4}{5}$
$cosecA=\frac{AC}{BC}=\frac{15}{12}=\frac{5}{4}\tan C=\frac{AB}{BC}=\frac{9}{12}=\frac{3}{4}$
$\sec A=\frac{AC}{AB}=\frac{15}{9}=\frac{5}{3}cosecC=\frac{AC}{AB}=\frac{15}{9}=\frac{5}{3}$
$\cot A=\frac{AB}{BC}=\frac{9}{12}=\frac{3}{4}\sec C=\frac{AC}{BC}=\frac{15}{12}=\frac{5}{4}$
$\cot C=\frac{BC}{AB}=\frac{12}{9}=\frac{4}{3}$