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F.5 Maths

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Picture: http://postimg.org/image/mixflp3fj/In the figure, PQR is a right-angled triangle, where angle R=90° and PR=5a cm. RR1(RR1 means the length RR1 in the picture) and R1R2 are the perpendiculars drawn from R to PQ and from R1 to QR respectively, where PR1= 3a cm.a) express the lengths of RR1 and... 顯示更多 Picture: http://postimg.org/image/mixflp3fj/ In the figure, PQR is a right-angled triangle, where angle R=90° and PR=5a cm. RR1(RR1 means the length RR1 in the picture) and R1R2 are the perpendiculars drawn from R to PQ and from R1 to QR respectively, where PR1= 3a cm. a) express the lengths of RR1 and R1R2 in terms of a. b) the perpendiculars R2R3, R3R4, R4R5, ... are drawn in the same way as RR1 and R1R2. Express the length of Rn R(n+1) in terms of a and n. I did (a) by using similar triangles but the method is complicated, so I cannot find the pattern of the length Rn R(n+1). Is there any easier methods to do (a) so that can do (b) simply? Please help. Thank you very much!! 更新: RE 土扁 : This question has (c) part that ''show that the length of RR1, R1R2, R2R3, ... form a geometric sequence. Hence, find the common ration.''. Is there any methods to find Rn R(n+1) without proving it is a geometric sequence?

最佳解答:

Let ∠PRR? be θ, therefore, ∠RR?R?＝∠R?R?R?＝∠R?R?R?＝??＝∠Rn-1RnRn+1＝θand, sin θ＝PR?/PR＝3/5＝0.6, therefore, cos θ＝4/5＝0.8As RR?＝PR cos θ＝0.8 PRR?R?＝RR? cos θ＝0.8 RR?＝0.82 PRR?R?＝R?R? cos θ＝0.8 R?R?＝0.83 PRR?R?＝R?R? cos θ＝0.8 R?R?＝0.8? PR??RnRn+1＝0.8??1 PR (a) RR?＝0.8 PR＝4aR?R?＝0.82 PR＝3.2a (b) RnRn+1＝0.8??1 PR＝5a*0.8??1 (or 4a*0.8?) (c) Yes, it is a G.S. with common ratio 0.8, first term RR? is 3.2a. 2015-02-10 19:07:06 補充： Sorry, the last line is wrongly written, it should be first term R?R? is 3.2a.

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