Question

If $13\sin A=5$ and $A$ is acute, find the value of $\frac{5\sin A-2\cos A}{\tan A}$.

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 $13\sin A=5$ that is$\sin A=\frac{5}{13}$

We know that, in a right angled triangle, $\sin \theta$ is equal to opposite side over hypotenuse that is $\sin \theta =\frac{Oppositeside}{Hypotenuse}$, therefore, opposite side $PQ=5$ and hypotenuse $PR=13$.

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

$P{R}^2=P{Q}^2+Q{R}^2\Rightarrow {13}^2=5^2+Q{R}^2\Rightarrow 169=25+Q{R}^2\Rightarrow Q{R}^2=169-25=144\Rightarrow QR=\sqrt{144}=12$

Therefore, the adjacent side $QR=12$.

We know that, in a right angled triangle,

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

$\tan \theta$ is equal to opposite side over adjacent side that is $\tan \theta =\frac{Oppositeside}{Adjacentside}$

Here, we have opposite side $PQ=5$, adjacent side $QR=12$ and the hypotenuse $PR=13$, therefore, the trignometric ratios of angle $A$ can be determined as follows:

$\cos A=\frac{Adjacentside}{Hypotenuse}=\frac{QR}{PR}=\frac{12}{13}$

$\tan \theta =\frac{Oppositeside}{Adjacentside}=\frac{PQ}{QR}=\frac{5}{12}$

Now, we find

$\frac{5\sin A-2\cos A}{\tan A}=\frac{(5\times \frac{5}{13})-(2\times \frac{12}{13})}{\frac{5}{12}}=\frac{\frac{25}{13}-\frac{24}{13}}{\frac{5}{12}}=\frac{\frac{1}{13}}{\frac{5}{12}}=\frac{1}{13}\times \frac{12}{5}=\frac{12}{65}$

Hence, $\frac{5\sin A-2\cos A}{\tan A}=\frac{12}{65}$.