Question
In the given figure, find tan P and cot R. Is tan P = cot R?
✓
Answer
The given figure is below:
To Find: tan P, cot R In the given right angled? $PQR$, length of side $QR$ is unknown.
Therefore, to find length of side QR we use Pythagoras Theorem
Hence, by applying Pythagoras theorem in $\triangle\text{PQR},$
We get, $PR^2 = PQ^2 + QR^2$
Now, we substitute the length of given side $PR$ and $PQ$ in the above equation
$13^2=12^2+\text{QR}^2$
$\text{QR}^2=13^2-12^2$
$\text{QR}^2=169-144$
$\text{QR}^2=25$ $\text{QR}^2=\sqrt{25}$
$\text{QR}^2=5$ By definition, we know that $\tan\text{P}=\frac{\text{Perpendicular side opposite to }\angle \text{P}}{\text{Base side adjacent to }\angle \text{P}}$
$\tan\text{P}=\frac{\text{QR}}{\text{PQ}}$
$\tan\text{P}=\frac{5}{12}\ \dots(1)$
Also, by definition, we know that $\cot\text{P}=\frac{\text{Base side adjacent to }\angle \text{R}}{\text{Perpendicular side opposite to }\angle \text{R}}$
$\cot\text{R}=\frac{\text{QR}}{\text{PQ}}$
$\cot\text{R}=\frac{5}{12}\ \dots(2)$
Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal
Therefore, L.H.S of both the equation are also equal
$\tan\text{P}=\cot\text{R}$ Yes, $\tan\text{P}=\cot\text{R}=\frac{5}{12}$
To Find: tan P, cot R In the given right angled? $PQR$, length of side $QR$ is unknown.
Therefore, to find length of side QR we use Pythagoras Theorem
Hence, by applying Pythagoras theorem in $\triangle\text{PQR},$
We get, $PR^2 = PQ^2 + QR^2$
Now, we substitute the length of given side $PR$ and $PQ$ in the above equation
$13^2=12^2+\text{QR}^2$
$\text{QR}^2=13^2-12^2$
$\text{QR}^2=169-144$
$\text{QR}^2=25$ $\text{QR}^2=\sqrt{25}$
$\text{QR}^2=5$ By definition, we know that $\tan\text{P}=\frac{\text{Perpendicular side opposite to }\angle \text{P}}{\text{Base side adjacent to }\angle \text{P}}$
$\tan\text{P}=\frac{\text{QR}}{\text{PQ}}$
$\tan\text{P}=\frac{5}{12}\ \dots(1)$
Also, by definition, we know that $\cot\text{P}=\frac{\text{Base side adjacent to }\angle \text{R}}{\text{Perpendicular side opposite to }\angle \text{R}}$
$\cot\text{R}=\frac{\text{QR}}{\text{PQ}}$
$\cot\text{R}=\frac{5}{12}\ \dots(2)$
Comparing equation (1) and (2) , we come to know that R.H.S of both the equation are equal
Therefore, L.H.S of both the equation are also equal
$\tan\text{P}=\cot\text{R}$ Yes, $\tan\text{P}=\cot\text{R}=\frac{5}{12}$
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