In ∆PQR. ∠PQR is a right angle. |QR| = 2cm and ∠PRQ = 60o. Find |PR|
- 4√3cm
-
4cm
- 2√3cm
- \(\frac{4}{3}cm\)
- 1cm
The correct answer is: B
Explanation
cos60 = \(\frac{adj}{hyp}\)
cos60 = \(\frac{2}{x}
x = \(\frac{2}{cs60}\) = 2 x 2/1 = 4cm