Question

In PQR, right-angled at Q, PR + QR = 25 and PQ = 5. The value of sinR is:

[1 mark]

• A. $\frac{5}{13}$
• B. $\frac{12}{13}$
• C. $\frac{3}{5}$
• D. $\frac{4}{5}$

Answer 1

The correct option is A $\frac{5}{13}$


PR + QR = 25 = x + y …..(1)

In $\Delta PQR,$ using the Pythagoras theorem,

$(PQ{)}^2=(PQ{)}^2+(QR{)}^2$

$P{R}^2=5^2+Q{R}^2$

$P{R}^2-Q{R}^2=25$

$(PR+QR)(PR-QR)=25$

$PR-QR=1$…...(2) (using (1)

(1) + (2) = 2y = 26

y = 13
SinR $=\frac{PQ}{PR}=\frac{5}{13}$