F2 數 ... 10點

(1)Discount
= $280 x 20%
= $56

Cost
= $56 / 35%
= 160

Profit
= ($280 - $56 - 160) /$160 x 100%
=40%

(2) h/tan40 + h/tan30 = 100
h(1/tan40 + 1/tan30) = 100
h = 34.2 (corr. to 1 d. p.)

(3)Let h be the height of D.
h/tan25 - h/tan70 = 80
h(1/tan25 - 1/tan70) = 80
h = 44.9303
The height of D above the water surface is 45 cm. (corr. to nearset cm)

(4) Let P be a point under A where OAP is a st. line.
(a)(i)angle DAP = 180 - 90 - (180 - 90 - theta) = theta
So, distance between point D and the wall OA
= (45 sin theta) cm
= (45 sin 60) cm
= 39.0 cm (corr.to 3 sig. fig.)
(ii) Distance between point D and the wall OB
= (120 sin theta + 45 cos theta) cm
= (120 sin 60 + 45 cos 60) cm
= 126 cm (corr. to 3 sig. fig.)

(b)Distance between point D and the wall OB will be largest if it equals BD such that
theta + angle DBA = 90.

tan (angle DBA) = 45/120
angle DBA = 20.5560 degree

so theta
=90 - 20.5560 degree
=69.4 degree (corr. to 3 sig. fig.)

The largest distance
=BD
= sqrt(45^2 + 120^2) cm
= 128 cm (corr. to 3 sig. fig.)

(5)Don't know ...

(6)
Area of triangle CDE = lr/2

Area of sector OAB: area of circle with radius r
=arc AB: circumference of the circle with radius r
=l : (2 x Pi x r)
Area of sector OAB
= r^2 x Pi x l/(2 x Pi x r)
= lr/2

So the areas of them are the same.

1) Selling price 280 x (1-0.2)=(0.8)280
Cost = 280 x (1-0.35) = (0.65)280
Profit = selling/cost-1 = 0.8/0.65 - 1 = 23.077%

2) angle C = 180-40-30=110

Line AB=100=h/tan 40 + h/tan30

h=100 /(1/tan 30 + 1/tan 40)=34.2m

3) Let the lenght of AB and AD be x and h respectivily

h/x=tan70, h/(x+80)=tan 25
x/h=1/tan70, x/h+80/h=1/tan25

80/h=1/tan25-1/tan70
h=80 / (1/tan25-1/tan70)=44.93cm

4) thetra = 60, angle DAX = 60 (X is the extension pt on the wall)
Let the distance of D with the wall be d1
AD=45cm
d1/AD=sin60, d1=ADsin60=39.0 cm

OA=ABsin60=103.923
AX=ADcos60=22.5

distance with OB is OA+AX=126.4cm

d1=ADsin t
d2=ABsin t + AD cos t

To max d2, d d2/dt=0

ABcos t - ADsin t = 0
tan t =AB/AD=120/45
t=69.44

d1=ADsin t=42.13cm
d2=ABsin t + AD cos t=128.16cm

2007-07-21 12:34:24 補充：
6) The arc lenght l = r thetra (in radian)Area = r^2 thetra/2 [Area of a circle =r^2 pi), thetra = 2 pi in this case]=r^2/2 x l/r=rl/2Area of the triangle = rl/2Two area are the same.