Question

Given $\tan A=\frac{7}{24}$, find the other trigonometric ratios of angle $A$.

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 $\tan A=\frac{7}{24}$ and we know that, in a right angled triangle, $\tan \theta$ is equal to opposite side over adjacent side that is $\tan \theta =\frac{Oppositeside}{Adjacentside}$, therefore, opposite side $PQ=7$ and adjacent side $QR=24$.

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

$P{R}^2=P{Q}^2+Q{R}^2=7^2+{24}^2=49+576=625\Rightarrow PR=\sqrt{625}=25$

Therefore, the hypotenuse $PR=25$.

We know that, in a right angled triangle,

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

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

Here, we have opposite side $PQ=3$, adjacent side $QR=4$ and the hypotenuse $PR=5$, therefore, the trignometric ratios of angle $A$ can be determined as follows:

$\sin A=\frac{Oppositeside}{Hypotenuse}=\frac{PQ}{PR}=\frac{7}{25}$

$\cos A=\frac{Adjacentside}{Hypotenuse}=\frac{QR}{PR}=\frac{24}{25}$

$\csc A=\frac{1}{\sin A}=\frac{1}{\frac{7}{25}}=1\times \frac{25}{7}=\frac{25}{7}$

$\sec A=\frac{1}{\cos A}=\frac{1}{\frac{24}{25}}=1\times \frac{25}{24}=\frac{25}{24}$

$\cot A=\frac{1}{\tan A}=\frac{1}{\frac{7}{24}}=1\times \frac{24}{7}=\frac{24}{7}$

Hence, $\sin A=\frac{7}{25}$, $\cos A=\frac{24}{25}$, $\tan A=\frac{7}{24}$, $\csc A=\frac{25}{7}$, $\sec A=\frac{25}{24}$ and $\cot A=\frac{24}{7}$.