Question

In $\vartriangle $PQR, right angled at Q, PR+QR =25 cm and PQ = 5 cm. Determine the values of sinP, cosP and tanP.
${\text{A}}{\text{. }}\sin {\text{P}} = \dfrac{{11}}{{13}},\cos {\text{P}} = \dfrac{4}{{13}}$ and $\tan {\text{P}} = \dfrac{{12}}{5}$
${\text{B}}{\text{. }}\sin {\text{P}} = \dfrac{{11}}{{13}},\cos {\text{P}} = \dfrac{5}{{13}}$ and $\tan {\text{P}} = \dfrac{{17}}{5}$
${\text{C}}{\text{. }}\sin {\text{P}} = \dfrac{{12}}{{13}},\cos {\text{P}} = \dfrac{4}{{13}}$ and $\tan {\text{P}} = \dfrac{{17}}{5}$
${\text{D}}{\text{.}}$ None of these

Answer 1

Hint: Here, we will proceed by finding the remaining two sides i.e., QR and PR using the Pythagoras Theorem i.e., ${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$ and then using the formulas $\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$, $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$ and $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$.

Complete step-by-step answer:
Given, In right angled $\vartriangle $PQR, PR + QR =25 cm and PQ = 5 cm
According to Pythagoras Theorem in any right angled triangle, we can write
${\left( {{\text{Hypotenuse}}} \right)^2} = {\left( {{\text{Perpendicular}}} \right)^2} + {\left( {{\text{Base}}} \right)^2}$
Using the above formula in right angled triangle PQR, we have
$
   \Rightarrow {\left( {{\text{PR}}} \right)^2} = {\left( {{\text{PQ}}} \right)^2} + {\left( {{\text{QR}}} \right)^2} \\
   \Rightarrow {\left( {{\text{PR}}} \right)^2} = {\left( {\text{5}} \right)^2} + {\left( {{\text{QR}}} \right)^2}{\text{ }} \to {\text{(1)}} \\
 $
PR + QR = 25
$ \Rightarrow $QR = (25 - PR) $ \to (2)$
By substituting equation (2) in equation (1), we get
\[
   \Rightarrow {\left( {{\text{PR}}} \right)^2} = {\left( {\text{5}} \right)^2} + {\left( {{\text{25}} - {\text{PR}}} \right)^2} \\
   \Rightarrow {\left( {{\text{PR}}} \right)^2} = 25 + {\left( {{\text{25}}} \right)^2} + {\left( {{\text{PR}}} \right)^2} - 2\left( {25} \right)\left( {{\text{PR}}} \right) \\
   \Rightarrow {\left( {{\text{PR}}} \right)^2} = 25 + 625 + {\left( {{\text{PR}}} \right)^2} - 50\left( {{\text{PR}}} \right) \\
   \Rightarrow {\left( {{\text{PR}}} \right)^2} - {\left( {{\text{PR}}} \right)^2} + 50\left( {{\text{PR}}} \right) = 650 \\
   \Rightarrow 50\left( {{\text{PR}}} \right) = 650 \\
   \Rightarrow {\text{PR}} = \dfrac{{650}}{{50}} = 13{\text{ cm}} \\
 \]
Put PR = 13 in equation (2), we get
$ \Rightarrow $QR = (25 - 13) = 12 cm
According to the definitions of sine, cosine and tangent trigonometric functions in any right angled triangle, we can write
$\sin \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Hypotenuse}}}}$, $\cos \theta = \dfrac{{{\text{Base}}}}{{{\text{Hypotenuse}}}}$ and $\tan \theta = \dfrac{{{\text{Perpendicular}}}}{{{\text{Base}}}}$
In right angled triangle PQR,
$\sin {\text{P}} = \dfrac{{{\text{QR}}}}{{{\text{PR}}}} = \dfrac{{12}}{{13}}$, $\cos {\text{P}} = \dfrac{{{\text{PQ}}}}{{{\text{PR}}}} = \dfrac{5}{{13}}$ and $\tan {\text{P}} = \dfrac{{{\text{QR}}}}{{{\text{PQ}}}} = \dfrac{{12}}{5}$
Therefore, $\sin {\text{P}} = \dfrac{{12}}{{13}}$, $\cos {\text{P}} = \dfrac{5}{{13}}$ and $\tan {\text{P}} = \dfrac{{12}}{5}$
Hence, option D is correct.

Note- In any right angled triangle, the side opposite to the right angle is the hypotenuse, the side opposite to the considered angle is the perpendicular and the remaining side is the base. In this particular problem, side PR is the hypotenuse, side QR is the perpendicular for angle P and side PQ is the base for angle P.