Please help<3What is the length of SR?9 units12 units15 units18 units
Question
Answer:
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
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Another solution:
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∵ 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
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Another solution:
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∵ ΔSRQ is a right triangle
and RT⊥SQ
RQ² = QT * QS
∴ QS = RQ²/QT = 20²/16 = 400/16 = 25
∴ SR = [tex] \sqrt{SQ^2 - RQ^2} [/tex]
∴ SR = [tex] \sqrt{25^2 - 20^2} [/tex] = 15
solved
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10 months ago
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