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trigonometry question
發問:
trigonometry please show step clearly 圖片參考：http://imgcld.yimg.com/8/n/HA00167623/o/701205310018613873410030.jpg
最佳解答:
(a) let F' be the projection of F on BC so the required angle is ∠FAF' in triangle BFF', sin 45 = FF'/8 => FF' = 8sin 45 in triangle ABF, AF^2 = 7^2 + 8^2 => AF = sqrt113 in triangle AFF', sin∠FAF' = FF'/AF => sin∠FAF' = 8sin 45/sqrt113 => ∠FAF' = 32.2 Therefore, the angle between FA and plane ABCD is 32.2 (b) in triangle ABF, AF^2 = 7^2 + 8^2 => AF = sqrt113 in triangle ABC, AC^2 = 7^2 + 8^2 => AC = sqrt113 in triangle FBC, FC^2 = 8^2 + 8^2 - 2(8)(8)cos 45=> FC = 6.12 in triangle AFC, cos∠FAC = (AF^2 + AC^2 - FC^2)/[2(AF)(AC)] =>∠FAC = 33.5 Therefore, the angle between FA and AC is 33.5
trigonometry question
發問:
trigonometry please show step clearly 圖片參考：http://imgcld.yimg.com/8/n/HA00167623/o/701205310018613873410030.jpg
最佳解答:
(a) let F' be the projection of F on BC so the required angle is ∠FAF' in triangle BFF', sin 45 = FF'/8 => FF' = 8sin 45 in triangle ABF, AF^2 = 7^2 + 8^2 => AF = sqrt113 in triangle AFF', sin∠FAF' = FF'/AF => sin∠FAF' = 8sin 45/sqrt113 => ∠FAF' = 32.2 Therefore, the angle between FA and plane ABCD is 32.2 (b) in triangle ABF, AF^2 = 7^2 + 8^2 => AF = sqrt113 in triangle ABC, AC^2 = 7^2 + 8^2 => AC = sqrt113 in triangle FBC, FC^2 = 8^2 + 8^2 - 2(8)(8)cos 45=> FC = 6.12 in triangle AFC, cos∠FAC = (AF^2 + AC^2 - FC^2)/[2(AF)(AC)] =>∠FAC = 33.5 Therefore, the angle between FA and AC is 33.5
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