In a $\triangle ABC$, right angled at $B, AB = 24\ cm, BC = 7\ cm$. Determine$sin\ C, cos\ C$

Given:

In a $\triangle ABC$, right angled at $B, AB = 24\ cm, BC = 7\ cm$.

To do:

We have to determine $sin\ C, cos\ C$.

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,

$sin\ C=\frac{Opposite}{Hypotenuse}=\frac{AB}{AC}$

$cos\ C=\frac{Adjacent}{Hypotenuse}=\frac{BC}{AC}$

Here,

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

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

$\Rightarrow AC^2=576+49$

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

Therefore,

$sin\ C=\frac{AB}{AC}=\frac{24}{25}$

$cos\ C=\frac{BC}{AC}=\frac{7}{25}$

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Updated on: 10-Oct-2022

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