Question

In $Δ$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
PQ = 5 Let PR be x.
Therefore, QR = 25 - x

Applying Pythagoras theorem in $Δ$PQR, we obtain
$\text{PR}^ 2 = \text{PQ}^ 2 + \text{QR}^ 2$
$x^2= (5)^ 2 + (25 - x)^ 2$
$x^2= 25 + 625 + x^ 2 - 50x$
$50x=650$
$x=13$

Therefore, PR = 13 cm
QR = (25 - 13) cm = 12 cm

$\text{sin p} =\frac{\text{ Opposite Side}}{\text{Hypotenuse }}= \frac{QR}{PR} = \frac{12}{13}$

$\text{sin p} = \frac{\text{Opposite Side}}{\text{Hypotenuse }}= \frac{QR}{PR} =\frac{ 12}{13}$

$\text{tan p} =\frac{\text{Opposite Side}}{\text{Adjacent side }}= \frac{QR}{PQ} = \frac{12}{5}$