In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.$ cos \theta=\frac{12}{15} $

Given:

\( cos \theta=\frac{12}{15} \)

To do:

We have to find the values of the other trigonometric ratios.

Solution:  

We know that,

In a right-angled triangle $ABC$ with right angle at $B$,

By Pythagoras theorem,

$AC^2=AB^2+BC^2$

By trigonometric ratios definitions,

$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$

$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$

$tan\ A=\frac{Opposite}{Adjacent}=\frac{BC}{AB}$

$cosec\ A=\frac{Hypotenuse}{Opposite}=\frac{AC}{BC}$

$sec\ A=\frac{Hypotenuse}{Adjacent}=\frac{AC}{AB}$

$cot\ A=\frac{Adjacent}{Opposite}=\frac{AB}{BC}$

Here,

Let $cos\ \theta=\frac{AB}{AC}=\frac{12}{15}$

$AC^2=AB^2+BC^2$

$\Rightarrow (15)^2=(12)^2+(BC)^2$

$\Rightarrow BC^2=225-144$

$\Rightarrow BC=\sqrt{81}=9$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{9}{15}=\frac{3}{5}$

$tan\ \theta=\frac{BC}{AB}=\frac{9}{12}=\frac{3}{4}$

$cosec\ \theta=\frac{AC}{BC}=\frac{15}{9}=\frac{5}{3}$

$sec\ \theta=\frac{AC}{AB}=\frac{15}{12}=\frac{5}{4}$

$cot\ \theta=\frac{AB}{BC}=\frac{12}{9}=\frac{4}{3}$ 

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Updated on: 10-Oct-2022

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