In each of the following, one of the six trigonometric ratios is given. Find the values of the other trigonometric ratios.$ \sin \mathrm{A}=\frac{2}{3} $

Given:

\( \sin \mathrm{A}=\frac{2}{3} \)

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,

$\sin \mathrm{A}=\frac{BC}{AC}=\frac{2}{3}$

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

$\Rightarrow (3)^2=AB^2+(2)^2$

$\Rightarrow AB^2=9-4$

$\Rightarrow AB=\sqrt{5}$

Therefore,

$cos\ A=\frac{AB}{AC}=\frac{\sqrt5}{3}$

$tan\ A=\frac{BC}{AB}=\frac{2}{\sqrt5}$

$cosec\ A=\frac{AC}{BC}=\frac{3}{2}$

$sec\ A=\frac{AC}{AB}=\frac{3}{\sqrt5}$

$cot\ A=\frac{AB}{BC}=\frac{\sqrt5}{2}$

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Updated on: 10-Oct-2022

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