1條A. Maths Compound Angles問題

2007-01-19 4:28 am

*附圖在http://aerodrive.twghwfns.edu.hk/~4s227/27..bmp

ABCDE is a regular pentagon of sides 1 unit. M, P and N are points in BE such that AP, CM and DN are all perpendicular to BE.

(a) Find ∠ABP and ∠CBM.
(b) By considering △ABE, prove that BE = 2cos36°.
(c) By considering quadrilateral BCDE, prove that BE = 2cos72° + 1.
(d) Hence, express cos36° in surd form.

回答 (1)

用戶21831

2007-01-19 4:56 am

✔ 最佳答案

(a) pentagon inner angle = 108°
∠PAB= 108°/2 = 54°, ∠APB=90°, so ∠ABP=(180-90-54)°=36°
∠CBM= (108-36)°= 72°

(b) Because BP=PE (symmetry),
so BE= BP+PE= 2BP= 2AB cos∠ABP = 2cos36° unit.

(c) Because BM=NP (symmetry) and MN=CD=1 unit
BE= BM+MN+NP= 1+ 2NP= 1+2cos72° unit.

(d) Consider BE, and from result of (b) and (c)
2cos36° = 2cos72° + 1 = 2cos(36 x 2)° + 1
Let x be cos36°. From compound angle formula,
2x = 2 (2x^2-1)+1
x^2-1/2x-1/4 =0
(x-1/4)^2=5/16
consider 0<=x<=1,
x= sqrt(5/16)+1/4 = (sqrt(5)+1)/4
(another solution (1-sqrt(5))/4 should be disgarded)

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