In right angle tr

To find QR:

Since, tan θ = perpendicular/base

We know that,

⇒ [∵, PQ = 6 cm & tan 60° = √3]

⇒ QR = 6√3

Now, PR can be found by two ways -

1^st method: In ∆PQR, using Pythagoras theorem,

PR² = PQ² + QR² [∵, (hypotenuse)² = (perpendicular)² + (base)²]

⇒ PR² = 6² + (6√3)²

⇒ PR² = 36 + 108 = 144

⇒ PR = √144 = 12

2^nd Method:

Since, cos θ = base/hypotenuse

We know that,

⇒

⇒

⇒ PR = 2 × PQ

⇒ PR = 2 × 6 = 12

Thus, QR = 6√3 cm and PR = 12 cm.