JAMB - Mathematics (2006 - No. 34)

In the diagram above, PQ = 10 cm, PS = 8 cm and ∠PSR is 60^o while ∠SRQ is a right angle. Find SR

14 cm

14√3 cm

10 cm

10√3 cm

In ΔPST; cos 60 = ^ST/₈
ST = 8cos 60
ST = 8 x ¹/₂ = 4
TR = 10 cm (opp. sides of a rectangle PQRT)
SR = ST + TR
SR = 4 + 10
SR = 14 cm