utefdxd的部落格

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trigonometry question

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trigonometry please show step clearly 圖片參考：http://imgcld.yimg.com/8/n/HA00167623/o/701205310018513873410020.jpg

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最佳解答:

(a)(i) the angle required is ∠ABD sin∠ABD = AD/AB = 5/13 thus ∠ABD = 22.6199' =21.3' # (a)(ii) cos∠ABC = (AB2 + BC2 - AC2) / (2*AB*BC) = (132)/ (2*13*8) ∠ABC = 35.6591' let E be a point on BC such that AE⊥BC sin∠ABC = AE / AB AE = (13)*sin35.6591 = 7.57850 sin∠AED = AD/AE = 5/7.57850 ∠AED = 41.2817 = 41.3' # so the required angle is 41.3' (b)(i) AD2 + BD2 = AB2 (pyth. thm.) 52 + BD2 = 132 BD = 12, AD2 + DC2 = AC2 (pyth. thm.) 52 + DC2 = 82 DC = sqrt39 = 6.24500 area of ΔBCD can be found by Heron's Formula: let s = (BC + CD + DB)/2 = (8+6.24500+12)/2 = 13.1225 area ofΔBCD = √[s(s-BC)(s-CD)(s-DB)] = √[(13.1225)(13.1225-8)(13.1225-6.24500)(13.1225-12)] = 22.7802 = 22.8cm2 # (b)(ii) if vertex A is further lifted up, ie, AD increases BD and DC decreased as AB and AC are constants (by pyth. thm.) thus area of the projection will decreases (by Heron's formula)

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