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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