In right angle triangle ∆PQR, right angle is at Q and PQ = 6cms \angle RPQ = 60\degree. Determine the lengths of QR and PR.

Step 1: According to the given values construct a triangle
CONSTRUCTION:
Step 2: Find the suitable trigonometric ratio to fin the unknown lengths
NOTE: \sin 60\degree = \frac{opposite}{hypotenuse} = \frac{QR}{PR}
\cos 60\degree = \frac{adjacent}{hypotenuse} = \frac{PQ}{PR}
Step 3: Find the unknown lengths
EXAMPLE: \cos 60\degree = \frac{PQ}{PR} = \frac{6}{PR}
\frac{1}{2} = \frac{6}{PR} (since \cos 60\degree = \frac{1}{2})
PR = 12 .........................(1)
\sin 60\degree = \frac{QR}{PR} = \frac{QR}{12}
\frac{\sqrt{3}}{2} = \frac{QR}{12}
QR = \frac{12*\sqrt{3}}{2}
QR = 6 \sqrt{3}
PR = 12 cm, QR = 6 \sqrt{3}