中二升中三 數學問題 急急急***************

For Q1,
b)
∠ABC=∠BDC=60˚
∠C=∠C (common ∠)
So △ABC~△BDC (A.A.)
AC^2=AB^2-BC^2
AC=4√3
BD/AB=BC/AC (corr. ∠s,~△s)
BD=4.619

c)
∠C=90˚ (given)
2cos∠CBD=2(BC/BD)
2cos30˚ =2√3
cos∠ABC=BC/AB
cos60˚=1/2
cos60˚≠2cos30˚
So cos2ɵ = 2cosɵ is not always true

For 2)
area of △CDE=lr/2
Considered OAB as a cone,
Radius of the cone = l/2π
Curved surface area of the cone = area of OAB
Area of OAB = π(l/2π)(r) = lr/2 = Area of △CDE

For 3),
I cannot open it as it cannot be displayed. Can you please re-upload the image?

2013-08-02 14:57:37 補充：
NO.2 explanation:

你將個圙接成一個CONE,
個CONE嘅圓周係l咁長(細階L)
因為圓周=2πX半徑
半徑=圓周/2π

2013-08-02 14:59:18 補充：
你當個ɵ係30度
cos(2ɵ)=cos(2x30度)=cos60

2013-08-02 15:15:54 補充：
For 3ai),

Draw a horizontal line from D to a point E on the wall
∠OAB=180˚-ɵ-90˚=30˚(∠ sum of △)
∠BAD=90˚
∠DAE=180-∠BAD-∠OAB=60˚ (adj. ∠s on st.line)
∠DEA=90˚
DA=CB=45cm
DE=DAsin∠DAE
DE=39.0 cm
So the required distance = 39.0cm

2013-08-02 15:15:58 補充：
3aii),
Draw DF such that DF=EO and DF⊥BO
∠O=90˚
OA=ABsinɵ=103.923cm
∠DEA=90˚
AE=DAcos∠DAE=22.5cm
DF=EO=OA+AE=126cm

3b),
When ɵ=90˚
The required distance which is longest = CD = AB = 120cm

2013-08-02 15:57:06 補充：
Sorry that the previous answer of 3b is wrong. Now I type you the modified one.

Adjust the ɵ until DB⊥BO
∠BCD=90˚
CD=BA=120cm
BD^2=CD^2+BC^2 (Pyth. thm.)
BD=128cm
∠BAD=90˚
cos∠DBA=BA/BD
∠DBA=20.6˚
∠DBO=90˚ #(DB⊥BO is adjusted at first)
ɵ+∠DBA=90˚
ɵ=69.4˚

請問cos2ɵ是否代表 cosɵ??

2013-08-01 18:48:09 補充：
請問Q.2 的Radius of the cone = l/2π是如何得出的@@?