Please help<3What is the length of SR?9 units12 units15 units18 units
Accepted Solution
A:
From ΔRTQ :
cos (RQT) = TQ / RQ = 16/20 = 4/5 = 0.8
∴ sin (RQT) = √(1-sin²(RQT)) = 0.6
∴ tan (RQT) = sin (RQT)/ cos (RQT) = 0.6/0.8 = 0.75 ⇒⇒⇒ (1)
But tan (RQT) = SR/RQ ⇒⇒⇒ (2)
From (1) and (2):
∴ SR/RQ = 0.75 ∴ SR = 0.75 * RQ = 0.75 * 20 = 15 ============================================================== Another solution: ------------------------ ∵ ΔSRQ is a right triangle and RT⊥SQ
RT = [tex] \sqrt{RQ^2 - TQ^2} [/tex] ∴ RT = [tex] \sqrt{20^2 - 16^2} [/tex] = 12 And RT² = ST * TQ ∴ ST = RT² / TQ = 12²/16 = 9 ∴ SR = [tex] \sqrt{ST^2 + TR^2} [/tex] ∴ SR = [tex] \sqrt{9^2 + 12^2} [/tex] = 15 ============================================================== Another solution: ------------------------ ∵ ΔSRQ is a right triangle and RT⊥SQ