HKCEE - 2006 - Paper 2 -

Ans: A

Since the bearing of $Q$ from $A$ is $\text{S}28^\cir

Ans: D
$\begin{array}{cl}
& 2\sin(90^\circ –

Ans: D
I may be false. For $\theta= 40^\circ$, $\tan 40^\cir

Ans: C
Add a straight line from the upper base to the lower bas

Ans: A
$\begin{array}{rcl}
3\cos^2 x -4\cos x +1 & = &#

Ans: C

$\begin{array}{rcl}
\alpha + 110^\circ & = �

Ans: D
Since $x+y=90^\circ$, we have $x = 90^\circ – y

Ans: D
$\begin{array}{cl}
& \dfrac{\cos A}{\sin A} +

Ans: D
$\begin{array}{rcl}
\sin\theta & = & \sq

Ans: B
In $\Delta ABC$,

$\begin{array}{rcl}
\angle ABC �

Ans: B
By applying the sine law to the triangle, we have

$\beg

Ans: A
By applying the cosine law to $\Delta ABD$, we have

$\b

Ans: $220^\circ$
Note that $\angle PRQ=90^\circ$, then we

Ans: C
Consider $\Delta ABD$.

$\begin{array}{rcl}
\tan 30

Ans: A
$\begin{array}{rcl}
\dfrac{\cos A}{\tan (90^\cir

Ans: A

Note that

$\begin{array}{rcl}
AC^2 + BC^2 & = &#

Ans: B
Let $y = \sin \theta$. Then the equation becomes

$\beg

Ans: B
For $0^\circ \le \theta \le 360^\circ$, we have

$\beg

Ans: D

Add a point $E$ on $AB$ such that $AC//ED$. Then $AEDC$

Ans: A

Let $\theta$ be the angle of depression of $Q$ from $P$