In Figure, ABCDE is a regular pentagon and CDFG is a square. BG produced meets AE at P.
(a) Find ∠BCG, ∠ABP and ∠APB.
(b) Using the fact that AP/ sin∠ABP = AB/sin∠APB, or otherwise, determine which line segment, AP or PE, is longer.
Figure http://server7.pictiger.com/img/160499/picture-hosting/--.jpg
F4 math problems ( Trigonometry )
2007-05-25 3:23 am
回答 (1)
2007-05-25 3:47 am
✔ 最佳答案
a)∵∠BCD=(5-2)180°/5(∠sum of Δ)=108°
∴∠BCG=108°-90°
=18°
∵BC=CG(given)
∴∠CBG=∠CGB(opp.∠s eq.sides)
2∠CBG=180°-18°(∠sum of Δ)
∠CBG=81°
∠ABP=108°-81°
=27°
∠APB=180°-27°-108°(∠sum of Δ)
=45°
b)∵AP/ sin∠ABP = AB/sin∠APB
AP/AB=sin27°/sin45°
=0.642(corr. to 3 sign. fig.)
∴AP is longer than PE
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