Question

In $△$ PQR, right angled at Q, PR + QR = 25 cm and PQ = 5 cm. Find the value of sin P.

• A. $\frac{12}{13}$
• B. $\frac{12}{14}$
• C. $0.9231$
• D. 0.8571

Answer 1

Answer: (A), (C)

The correct options are
A $\frac{12}{13}$
C $0.9231$

We are given

PR + QR = 25 cm

or PR = (25 – QR) cm

By Pythagoras theorem,

$P{R}^2$ = $Q{R}^2$ + $P{Q}^2$

or $(25-QR{)}^2$ = $Q{R}^2$ + $5^2$

or 625 + $Q{R}^2$ – 50 QR = $Q{R}^2$ + 25

or 50 QR = 625 – 25 = 600

or QR = 12 cm.

and PR = (25 – 12) cm = 13 cm

Now sin P = $\frac{QR}{PR}$ = $\frac{12}{13}$