Can anyone help with geometry (area with an unknown length) question? I would really appreciate it. 
**Note - the problem I'm struggling with is how to calculate the area of APBQ (the last question)
Figure 1 on the right shows a right-angled
triangle ABC where AB ＝ 1 cm, AC ＝ 2 cm, and
angle BAC ＝ 90°. Triangle PAB is an
isosceles triangle where AP ＝ AB and sides PA
and BC are parallel.
Assume point P is located opposite to point C
with respect to line AB.
Answer the following questions.
〔Question 1 〕 Consider the case in Figure 1 where the magnitude of angle APB is a°.
Find the magnitude of angle ACB in terms of a.
〔Question 2 〕 
Figure 2 on the right shows the case in
Figure 1 where a perpendicular line to
side BC is drawn from vertex A.
Let Q be the intersection of side BC and
the perpendicular line.
Answer （１）and （２）.
（１） Prove triangle ABQ is similar to triangle CAQ.
（２） Calculate the area of quadrilateral APBQ.

Following on from your comment:
$\angle CAB = \tan^{-1} (2)$, and since $PA$ and $BC$ are parallel, $\angle PAB = \tan^{-1} (2)$ as well.
Now if you split $\Delta PAB$ in half where $M$ is the midpoint of $PB$, you will have $\sin PAM = \frac{PM}{PA} = \frac{PM}{1}$. This gives me a value of $PM = \sqrt{\frac{2}{5+\sqrt5}}$ and $PB= \sqrt{2 - \frac{2}{\sqrt5}}$.
