Question

Given $\tan A=\frac{3}{4}$, find the value of $\sin A$ and $\cos A$.

Answer 1

Let $△PQR$ be a right angled triangle where $\angle Q={90}^0$ and $\angle R=A$ as shown in the above figure:

Now it is given that $\tan A=\frac{3}{4}$ and we know that, in a right angled triangle, $\tan \theta$ is equal to opposite side over adjacent side that is $\tan \theta =\frac{\text{Opposite side}}{\text{Adjacent side}}$, therefore, opposite side $PQ=3$ and adjacent side $QR=4$.

Now, using pythagoras theorem in $△PQR$, we have

$P{R}^2=P{Q}^2+Q{R}^2$

$=3^2+4^2$

$=9+16=25$

$\Rightarrow PR=\sqrt{25}=5$

Therefore, the hypotenuse $PR=5$.

We know that, in a right angled triangle,

$\sin \theta$ is equal to opposite side over hypotenuse that is $\sin \theta =\frac{\text{Opposite side}}{\text{Hypotenuse}}$ and

$\cos \theta$ is equal to adjacent side over hypotenuse that is $\cos \theta =\frac{\text{Adjacent side}}{\text{Hypotenuse}}$

Here, we have opposite side $PQ=3$, adjacent side $QR=4$ and the hypotenuse $PR=5$.

Therefore, $\sin A$ and $\cos A$ can be determined as follows:

$\sin A=\frac{\text{Opposite side}}{\text{Hypotenuse}}=\frac{PQ}{PR}=\frac{3}{5}$

$\cos A=\frac{\text{Adjacent side}}{\text{Hypotenuse}}=\frac{QR}{PR}=\frac{4}{5}$

Hence, $\sin A=\frac{3}{5}$ and $\cos A=\frac{4}{5}$.