In PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.
Step-by-Step Solution
Key Concept: Use Pythagoras theorem to find the unknown side of the right‑angled triangle and then apply the definitions of sine, cosine and tangent for the acute angle P (opposite side / hypotenuse, adjacent side / hypotenuse, opposite side / adjacent side).
1. Identify the sides:\
- Right angle at Q ⇒ PR is the hypotenuse.\
- Given: \(PQ = 5\) cm, \(PR + QR = 25\) cm.\
- Let \(QR = x\) cm, then \(PR = 25 - x\) cm.\
2. Apply Pythagoras theorem:\
\[PQ^{2} + QR^{2} = PR^{2}\]\
Substituting the known values:\
\[5^{2} + x^{2} = (25 - x)^{2}\]\
\[25 + x^{2} = 625 - 50x + x^{2}\]\
Cancel \(x^{2}\) from both sides and solve for \(x\):\
\[25 = 625 - 50x\]\
\[50x = 600\]\
\[x = 12\]\
Hence, \(QR = 12\) cm and \(PR = 25 - 12 = 13\) cm.\
3. Compute the trigonometric ratios for angle \(P\):\
- Opposite side to \(P\) = \(QR = 12\) cm\
- Adjacent side to \(P\) = \(PQ = 5\) cm\
- Hypotenuse = \(PR = 13\) cm\
\[\sin P = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{12}{13}\]\
\[\cos P = \frac{\text{adjacent}}{\text{hypotenuse}} = \frac{5}{13}\]\
\[\tan P = \frac{\text{opposite}}{\text{adjacent}} = \frac{12}{5} = 2.4\]\
4. State the answers:\
\(\sin P = \frac{12}{13},\; \cos P = \frac{5}{13},\; \tan P = \frac{12}{5} = 2.4\).
Correct Answer: sin P = 12/13, cos P = 5/13, tan P = 12/5 = 2.4