In ΔPQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine sin P, cos P and tan P.
sin P = \(\dfrac{12}{13}\)
cos P = \(\dfrac{5}{13}\)
tan P = \(\dfrac{12}{5}\)
Given: In \(\triangle PQR\), \(\angle Q = 90^\circ\). Also, \(PR + QR = 25\) cm and \(PQ = 5\) cm.
To determine: \(\sin P\), \(\cos P\) and \(\tan P\).
Step 1: Identify the sides in the right triangle
Since \(\angle Q\) is the right angle, the side opposite \(Q\) is the hypotenuse.
So, hypotenuse \(= PR\).
Step 2: Use the given sum to set up a variable
We know \(PR + QR = 25\). Let \(QR = x\) cm.
Then \(PR = 25 - x\) cm.
Step 3: Apply Pythagoras theorem
In a right triangle,
\(PR^2 = PQ^2 + QR^2\)
Substitute \(PR = 25 - x\), \(PQ = 5\), \(QR = x\):
\((25 - x)^2 = 5^2 + x^2\)
Expand the left side:
\(625 - 50x + x^2 = 25 + x^2\)
Now cancel \(x^2\) from both sides:
\(625 - 50x = 25\)
\(-50x = 25 - 625 = -600\)
\(x = \dfrac{-600}{-50} = 12\)
So, \(QR = 12\) cm.
Then \(PR = 25 - 12 = 13\) cm.
Student Note: Once you get one side, immediately find the other using the given sum. Here, the triangle becomes a neat \(5\text{-}12\text{-}13\) right triangle.
Step 4: Mark sides with respect to angle \(P\)
For angle \(P\):
Opposite side = \(QR = 12\) cm
Adjacent side = \(PQ = 5\) cm
Hypotenuse = \(PR = 13\) cm
Step 5: Use trigonometric definitions
\(\sin P = \dfrac{\text{opposite}}{\text{hypotenuse}} = \dfrac{QR}{PR} = \dfrac{12}{13}\)
\(\cos P = \dfrac{\text{adjacent}}{\text{hypotenuse}} = \dfrac{PQ}{PR} = \dfrac{5}{13}\)
\(\tan P = \dfrac{\text{opposite}}{\text{adjacent}} = \dfrac{QR}{PQ} = \dfrac{12}{5}\)
Quick Check: Since \(\sin P\) and \(\cos P\) are both fractions with denominator \(13\), that confirms \(PR\) (hypotenuse) is \(13\) cm, matching our result.