WAEC - Mathematics (1998 - No. 32)
In ∆PQR. ∠PQR is a right angle. |QR| = 2cm and ∠PRQ = 60o. Find |PR|
4√3cm
4cm
2√3cm
\(\frac{4}{3}cm\)
1cm
Објашњење
cos60 = \(\frac{adj}{hyp}\)
cos60 = \(\frac{2}{x}
x = \(\frac{2}{cs60}\) = 2 x 2/1 = 4cm
In ∆PQR. ∠PQR is a right angle. |QR| = 2cm and ∠PRQ = 60o. Find |PR|
cos60 = \(\frac{adj}{hyp}\)
cos60 = \(\frac{2}{x}
x = \(\frac{2}{cs60}\) = 2 x 2/1 = 4cm