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In a $\triangle ABC$, right angled at $B, AB = 24\ cm, BC = 7\ cm$. Determine$sin\ A, cos\ A$
Given:
In a $\triangle ABC$, right angled at $B, AB = 24\ cm, BC = 7\ cm$.
To do:
We have to determine $sin\ A, cos\ A$.
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,
$sin\ A=\frac{Opposite}{Hypotenuse}=\frac{BC}{AC}$
$cos\ A=\frac{Adjacent}{Hypotenuse}=\frac{AB}{AC}$
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,
$sin\ A=\frac{BC}{AC}=\frac{7}{25}$
$cos\ A=\frac{AB}{AC}=\frac{24}{25}$
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