Question

If $\cos$ $\theta =\frac{8}{17}$, find the other five trigonometric ratios.

Answer 1

We have,

cos B = $\frac{Base}{Hypotenuse}=\frac{8}{17}$

So, we draw a triangle ABC, right angled at A such that

Base $=AB=8$ unit and, Hypotenuse $=AC=17$ units. and $\angle BAC=\theta$

By Pythagoras theorem, we have

$A{C}^2=A{B}^2+B{C}^2$

$\Rightarrow {17}^2=8^2+B{C}^2$

$\Rightarrow B{C}^2=289-64=225$

$\Rightarrow BC=\sqrt{225}=15$

When we consider the t-ratios of $\angle BAC=\theta$, we have

Base $=AB=8$, Perpendicular $=BC=15$, and Hypotenuse $=AC=17$

$\therefore sin\theta =\frac{Perpendicular}{Hypotenuse}=\frac{15}{17}$

$tan\theta =\frac{Perpendicular}{Base}=\frac{15}{8}$

$cosec\theta =\frac{Hypotenuse}{Perpendicular}=\frac{17}{15}$

$sec\theta =\frac{Hypotenuse}{Base}=\frac{17}{8}$

$cot\theta =\frac{Base}{Perpendicular}=\frac{8}{15}$