Question

In the triangle $\Delta PQR$ right angled at Q, PR+RQ = 25cm and PQ = 5cm. Determine the value of sin P, cos P and tan p

Answer 1

Hint: First we will assume the first variable from PR and QR as x then we can easily calculate other in terms of other. Then, we will use Pythagora's theorem to calculate all sides stated as, "In a right angled triangle the square of hypotenuse sides is equal to sum of squares of other two sides". Then, we will use the given formula to calculate the result.
\[\sin \theta =\dfrac{\text{Perpendicular}}{\text{Hypotenuse}},\cos \theta =\dfrac{\text{Base}}{\text{Hypotenuse}},\tan \theta =\dfrac{\text{Perpendicular}}{\text{Base}}\]

Complete step-by-step answer:
Let us first understand the scenario by drawing a figure of $\Delta PQR$ right angled at Q.

Given that PR + QR = 25 cm.
Let the length of side PR be x, then by substituting this value in above equation we get:
\[\begin{align}
  & x+QR=25 \\
 & QR=25-x \\
\end{align}\]
Now, we have three sides of $\Delta PQR$ as PQ = 5, PR = x and QR = 25-x.
Now, because $\Delta PQR$ is a right angled triangle, we can use Pythagora's theorem to find the rest two sides. It is stated as,
"In a right angled triangle the square of hypotenuse sides is equal to sum of squares of other two sides"
Here we have hypotenuse as side PR and right angled at Q. Using Pythagoras theorem in $\Delta PQR$ we get:
\[{{\left( PR \right)}^{2}}={{\left( PQ \right)}^{2}}+{{\left( QR \right)}^{2}}\]
Substituting the value of PR = x, QR = 25-x and PQ = 5 in above we get:
\[{{\left( x \right)}^{2}}={{\left( 5 \right)}^{2}}+{{\left( 25-x \right)}^{2}}\]
Now, using the identity ${{\left( a-b \right)}^{2}}={{a}^{2}}-2ab+{{b}^{2}}$ in ${{\left( 25-x \right)}^{2}}$ we get:
\[{{x}^{2}}={{5}^{2}}+{{\left( 25 \right)}^{2}}+{{x}^{2}}-50x\]
Cancelling ${{x}^{2}}$ from both sides, we get:
\[\begin{align}
  & 0={{5}^{2}}+{{25}^{2}}-50x \\
 & 625+25=50x \\
 & 650=50x \\
 & 5x=65 \\
 & x=\dfrac{65}{5}=13 \\
\end{align}\]
Hence, length of side PR = 13 cm.
Now the length of QR = 25-x = 25-13 = 12 cm.
Hence, we have obtained in $\Delta PQR$
PQ = 5, PR = 13 and QR = 12.

Finally, we will consider sin P, cos P and tan P.
If ABC is a right angled triangle, right angled at B then for $\angle A$

\[\begin{align}
  & \sin A=\dfrac{\text{Perpendicular}}{\text{Hypotenuse}}=\dfrac{BC}{AC} \\
 & \cos A=\dfrac{\text{Base}}{\text{Hypotenuse}}=\dfrac{AB}{AC} \\
 & \tan A=\dfrac{\text{Perpendicular}}{\text{Base}}=\dfrac{BC}{AB} \\
\end{align}\]
Using these formula in $\Delta PQR$ we have:
\[\begin{align}
  & \sin P=\dfrac{\text{QR}}{\text{PR}}=\dfrac{12}{13} \\
 & \cos P=\dfrac{\text{PQ}}{\text{PR}}=\dfrac{5}{13} \\
 & \tan P=\dfrac{\text{QR}}{\text{PQ}}=\dfrac{12}{5} \\
\end{align}\]
Therefore, value of $\sin P=\dfrac{12}{13},\cos P=\dfrac{5}{13}\text{ and }\tan P=\dfrac{12}{5}$ is the required answer.

Note: Another method to calculate the value of tan P is by using the trigonometric formula:
\[\dfrac{\sin \theta }{\cos \theta }=\tan \theta \]
We can substitute Q = P and use this formula to calculate tan P.
Values of $\sin P=\dfrac{12}{13}\text{ and }\cos P=\dfrac{5}{13}\text{ }$ then \[\tan P=\dfrac{\sin P}{\cos P}=\dfrac{\dfrac{12}{13}}{\dfrac{5}{13}}\]
Cancelling 13 both sides, we get:
$\tan P=\dfrac{12}{5}$ Value of tan P is anyway the same.