F.5 maths trigo&geometry

Please check:

http://i.imgur.com/d2GJCDF.jpg

18. C
45. B

2015-06-16 17:21:31 補充：
明顯 AMNC 中兩塊的面積是一樣的，由於將所有條件上下調轉整體情況並沒有改變，所以推測陰影面積為 AMNC 的面積的 1/2。

至於要嚴謹證明還是較為繁複，故忽略掉算了。

Q18
Note that triangle ADN and triangle CMB are congruent
join MN, obviously area of //gram ANCM = 2* area of triangle ADN
therefore area of triangle ADN = 128 / 4 = 24cm²

BD is diagonal, so it bisects //gram ANCM
so area of shaded region
= 0.5 * //gram ANCM
= area of triangle ADN
= 24cm² (A)

Q45
sum of roots = sin@ + cos@ + 2 = 2
this implies sin@ + cos @ = 0
sin@ + cos @ = 0
tan@ + 1= 0
@ = -45°

1 + sin@ +cos@ - sin(90+@)cos(90+@)
= 1 + cos@sin@
= 1 + sin(-45°)cos(-45°)
= 1 - 1/2
=1/2 (B)

2015-06-16 20:05:04 補充：
Q18
Note that triangle ADN and triangle CMB are congruent
join MN, obviously area of //gram ANCM = 2* area of triangle ADN
therefore area of triangle ADN = 128/4 = 32cm²
BD is diagonal, so it bisects //gram ANCM
so area of shaded region
= 0.5 * //gram ANCM
= area of triangle ADN
= 32cm²(C)