急!!2math

14.
Let AE//BF//DC.
∠BCD = 50° (alt.∠s, BF//DC)
tan 50° = BD÷DC
DC = BD ÷ tan50° ...(1)
∠ACD = 63°(alt.∠s, BF//DC)
tan 63° = (BD+15)÷DC
DC = (BD+15) ÷ tan63° ...(2)
(1) – (2),
0 = BD÷tan50° – (BD+15)÷tan63°
BD÷tan50° = (BD+15)÷tan63°
tan63°BD = tan50°(BD+15)
tan63°BD = tan50°BD + 15tan50°
BD (tan63°- tan50°) = 15tan50°
BD = 15tan50°÷(tan63°- tan50°)
= 23.2 m (3 sig. fig.)

15.
Draw BF⊥AC.

tan44° = 65÷CM
CM = (65 ÷ tan44°) m,

tan32° = 43÷MD
MD = (43 ÷ tan32°) m,

BF = CD (property of rectangle)
= CM + MD
= (65 ÷ tan44° + 43 ÷ tan32°) m
= 136.12 m (5 sig. fig.)
AF = (65 – 43) m (property of rectangle)
= 22m
AB² = AF² + BF² (Pyth. theorem)
= 22² + 136.12²
AB = 138m (3 sig. fig.)

14.
BD=AD-AB=>BD=CDtan63˚-15}

BD=CDtan50˚}

=> CDtan63˚-15=CDtan50˚

.....CDtan63˚-CDtan50=15˚

.....CD(tan63˚-tan50˚)=15

.....CD=15/(tan63˚-tan50˚)}

BD=CDtan50˚}

=>BD=[15/(tan63˚-tan50˚)]* tan50˚

.... BD=23.19(m)

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15.AB=√[CD^2+(AC-BD)^2]

…AB=√[(CM+MD)^2+(AC-BD)^2]

…AB=√[(ACcot44˚+BDcot32˚)^2+(65-43)^2

…AB=√[(65*cot44˚+43*cot32˚)^2+(22)^2

…AB=137.89(m)

2007-03-07 14:59:57 補充：
其實第15題還有第二個方法AB^2=AM^2 MB^2-2*AM*MB*cos∠AMBAB^2=(AC/sin44˚)^2 (BD/sin32˚)^2-2*(AC/sin44˚)*(BD/sin32˚)*cos∠(180˚-44˚-32˚)AB^2=(65/sin44˚)^2 (43/sin32˚)^2-2*(65/sin44˚)*(43/sin32˚)*cos(104˚)AB=√[(65/sin44˚)^2 (43/sin32˚)^2-2*(65/sin44˚)*(43/sin32˚)*cos(104˚)]AB=137.89(m)