Question

In $△ABC$, right angled at $B,AB=24cm,BC=7cm$. Determine
$\sin A,\cos A$

Answer 1

We have, Hypotenuse $=$ side opposite to right angle $=AC$

We have to find value of $AC$ with the help of Pythagoras theorem

According to Pythagoras theorem

${\left(Hypotenuse\right)}^2={\left(Base\right)}^2+{\left(Perpendicular\right)}^2$

$\Rightarrow {\left(AB\right)}^2+{\left(BC\right)}^2={\left(AC\right)}^2$

$\Rightarrow {\left(24\right)}^2+{\left(7\right)}^2={\left(AC\right)}^2$

$\Rightarrow 576+49={\left(AC\right)}^2$

$\Rightarrow {\left(AC\right)}^2=625$

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

But side can't be negative, so $AC=25cm$

Now, $BC=7$ and $AC=25cm$

We know that

$\sin A=\frac{sideoppositetoangleA}{hypotenuse}$

and, $\cos A=\frac{sideadjacenttoangleA}{hypotenuse}$

side opposite to $\angle A=BC=7cm$

side adjacent to $\angle A=AB=24cm$

Hypotenuse $=AC=25cm$

So, $\cos A=\frac{AB}{AC}=\frac{24}{25}$

and, $\sin A=\frac{BC}{AC}=\frac{7}{25}$