HKCEE - 2009 - Paper 2 -

Ans: D
$\begin{array}{rcl}
2AB & = & 3 BC \\
\dfra

Ans: B

Add a point $M$ on $AB$ such that $CM\perp AB$, and add a

Ans: C
Since $A+B=90^\circ$, then $B = 90^\circ – A$.

Ans: D
Since $AB:BC:AC = 3:4:5$, then let $AB=3k$, $BC= 4k$ a

Ans: B
$\begin{array}{rcl}
\cos^2 x – \sin^2 x &

Ans: B
Since $G$ is the centroid of $\Delta ABC$, then $BL = LC

Ans: B
In $\Delta ABD$,

$\begin{array}{rcl}
\tan \alpha &#

Ans: D
$\begin{array}{cl}
& \tan \theta + \tan(90^\c

Ans: B
$\begin{array}{cl}
& \cos^2 1^\circ + \cos^2 2

Ans: D
Let $AB=BC=AC=2x$. Since $AB=BC=AC$, then $\Delta A

Ans: A
Add a point $X$ on the figure as shown.

In $\Delta ABX$,

Ans: D
Note that $x+y+z=180^\circ$ and $x+y=90^\circ$. Th

Ans: D
Let $r$ be the length of the hypotenuse.

By Pythagorus

Ans: C
In $\Delta ABD$,

$\begin{array}{rcl}
\sin 40^\circ

Ans: A
By Heron’s formula, we have

$\begin{array}{r