If $ \tan \theta=\frac{24}{7} $, find that $ \sin \theta+\cos \theta $

Given:

$tan\ \theta = \frac{24}{7}$.

To do:

We have to find the value of \( \sin \theta+\cos \theta \).

Solution:  

Let, in a triangle $ABC$ right-angled at $B$ and $\ tan\ \theta = tan\ A=\frac{24}{7}$.

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,

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

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

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

Here,

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

$\Rightarrow AC^2=(7)^2+(24)^2$

$\Rightarrow AC^2=49+576$

$\Rightarrow AC=\sqrt{625}=25$

Therefore,

$sin\ \theta=\frac{BC}{AC}=\frac{24}{25}$

$cos\ \theta=\frac{AB}{AC}=\frac{7}{25}$

$sin \theta+\cos \theta=\frac{24}{25}+\frac{7}{25}$

$=\frac{31}{25}$

The value of \( \sin \theta+\cos \theta \) is \( \frac{31}{25} \).  

Tutorialspoint

Updated on: 10-Oct-2022

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