Questions: trigonometic ratios

(1)
A)
Consider DABM: tan45o = (h cm)/BM
Hence, BM = h/tan45o cm ...... (*)

Consider DDNC: tan70o = (h cm)/NC
Hence, NC = h/tan70o cm ...... (**)

MN = AD = 3 cm
BM + MN + NC = 8 cm
Hence, BM + NC = 5 cm ...... (***)

Substitute (*) and (**) into (***):
h/tan45o + h/tan70o = 5
h = 5/(1/tan45o + 1/tan70o)
h = 3.67

B)
(**)
NC = 3.67/tan70 cm
NC = 1.34 cm


(2)
A)
Consider DABC: cos60o = AB/10
Hence, AB = 10cos60o cm
AB = 5 cm

Consider DABD: cos30o = AB/BD
BD = AB/cos30o
BD = 5/cos30o
BD = 10/√3 cm
BD = 5.77 cm

B)
Consider DABC: sin60o = AC/(10 cm)
AC = 10sin60o cm
AC = 5√3 cm

Consider DABD: tan30o = AD/(5 cm)
AD = 5tan30o cm
AD = (5√3)/3 cm

CD = AC - AD
CD = 5√3 - (5√3)/3
CD = 10/√3 cm
CD = 5.77 cm
=

1

圖片參考：http://g.imagehost.org/0101/ScreenHunter_04_Apr_10_20_27.gif

2(a)
AB=10cos60。

圖片參考：http://g.imagehost.org/0736/ScreenHunter_06_Apr_11_00_59.gif



2009-04-11 14:37:33 補充：
2(b)錯左,sorry~!
http://g.imagehost.org/0264/ScreenHunter_03_Apr_11_14_35.gif