Question

Question 10
In $\Delta$ PQR, right-angled at Q, PR + QR = 25 cm and PQ = 5 cm. Determine the values of sin P, cos P and tan P.

Answer 1

Given that, PR + QR = 25 and PQ = 5
Let PR be x.
$\therefore$ QR = 25 - x

By Pythagoras theorem,
$P{R}^2=P{Q}^2+Q{R}^2$
$x^2=(5{)}^2+(25-x{)}^2$
$x^2=25+625+x^2-50x$
$50x=650$
$x=13$
$\therefore$ PR = 13 cm
QR = (25 - 13) cm = 12 cm
sin P = $\frac{QR}{PR}$ = $\frac{12}{13}$
cos P = $\frac{PQ}{PR}$ = $\frac{5}{13}$
tan P = $\frac{QR}{PQ}$ = $\frac{12}{5}$