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application of trigonometry 2題

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1,http://i529.photobucket.com/albums/dd333/riverdog1994/Q16.jpg 我唔識(c) @@ 2,http://i529.photobucket.com/albums/dd333/riverdog1994/Q17.jpg Thx!!!!! 更新: 2樓:tan∠YBF > tan∠XBC = ∠YBF > ∠XBC??

最佳解答:

14. (a) In ΔCBE : tan x = (5 cm)/BE BE = 5 / tan x cm ...... [1] In ΔADE : tan x = (10 cm)/EA EA = 10/ tan x cm ...... [2] [1] + [2] : BE + EA = (5 / tan x) + (10 / tan x) cm 15 / tan x = 12 x = 51.34° (b) In ΔADE : sin 51.34° = (10 cm)/DE DE = 10 / sin 51.34° cm DE = 12.81 cm (c) In ΔCDE : sin 51.34° =(5 cm)/CE CE = 5 / sin 51.34° CE = 6.40 cm CD is the diagonal of the cuboid. CD = √(52 + 102 + 122) cm = 16.40 cm In ΔCED : cos∠CED = (CE2 + DE2 - CD2) / (2 x CE x DE) (cosinelaw) cos∠CED = (6.402 + 12.812 - 16.402) / (2 x 6.40 x 12.81) ∠CED = 112.9° ∠ between CE and ED = 112.9° ===== 17. (a) tan∠YBF = YF/BF and tan∠XBC= XC/BC Hence, tan∠YBF / tan∠XBC = (YF/BF) / (XC/BC) Since YF = XC, then tan∠YBF / tan∠XBC = BC/BF = 1/cos24° > 1 Hence, tan∠YBF > tan∠XBC (b) From (a) : tan∠YBF / tan∠XBC = 1/cos24° tan∠YBF / tan56° = 1/cos24° tan∠YBF = tan56°/cos24° ∠YBF = 58.36° 2012-10-27 04:15:10 補充： Both ∠YBF and ∠XBC are greater than 0° but less than 90°. In the range etween 0° and 90°, the greater the angle, the greater the tangent of the angle. Since tan∠YBF > tan∠XBC, then ∠YBF > tan∠XBC.

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