Question

in triangle PQR right angled at Q, PQ=3cm & PR=6cm. Determine angle QPR & angle PRQ.

Answer 1

∆PQR is a right angle triangle
by Pythagoras theorem
(PR)^2=(PQ)^2+(QR)^2
6^2=3^2+((QR)^2
36=9+(QR)^2
36-9=(QR)^2
(QR)^2=27
QR=✓9×3
QR=3✓2
AngleQPR =60° because side opposite to P is ✓3/2 times the hypo(PR)
Angle PRQ=30° because side opposite to R is 1/2 of the hypo(PR)

Answer 2

Answer: The measurement of angle PRQ = 30°

And, The measurement of angle QPR = 60°

Step-by-step explanation:

Here, PQR is a right triangle,

In which m∠Q = 90°, PQ = 3 cm and PR = 6 cm,

By the law of sine,

$\frac{sinR}{PQ}=\frac{sinQ}{PR}$

$\implies \frac{sinR}{3}=\frac{sin90^{\circ}}{6}$

$\implies sin R = 3\times \frac{sin90^{\circ}}{6}$

$sin R = \frac{3}{6}=\frac{1}{2}$

$\implies \angle R = 30^{\circ}$

⇒ $\implies \angle PRQ = 30^{\circ}$

By the property of a triangle,

$\angle P+\angle Q+\angle R = 180^{\circ}$

$\implies \angle P + 90^{\circ}+30^{\circ}=180^{\circ}$

$\implies \angle P = 180^{\circ}-120^{\circ}=60^{\circ}$

$\implies \angle QPR = 60^{\circ}$