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.
 A: Question (1): 


*

*If $\angle ~ APB = a^{\circ}$, then $\angle ~ PBA = a^{\circ} $, because $\Delta APB$ is isosceles and $AP = AB$. 

*If $\angle ~ APB = a^{\circ} $ and $\angle ~ PBA = a^{\circ} $, then $\angle ~ BAP = 180^{\circ}-2a^{\circ}$.

*If $PA\mathbin{\|} BC$, then $ \angle ~CBA  = 180^{\circ}-2a^{\circ}$ (alternate interior angles).

*If $\angle ~CBA  = 180^{\circ}-2a^{\circ}$ and $\angle~ BAC = 90^{\circ} $, then $\angle ~ ACB = 2a^{\circ}-90^{\circ}$.


Question (2): 



*

*The triangle $\Delta ~ CAQ$ is similar to the triangle $\Delta ABQ$ because all three angles in $\Delta ~ CAQ$ are also present in $\Delta ABQ$.

*In $\Delta ABQ$ we have $AQ = \cos(2a^{\circ}-90^{\circ})$ and $BQ = \sin(2a^{\circ}-90^{\circ})$ by the definitions of the sin and cos functions in a right triangle

*By the definition of the area of a trapezoid we get for the area $A_{APBQ}$ of the trapezoid $APBQ$ the following result: $$  A_{APBQ} = (1/2)\cdot (PA+BQ)\cdot AQ$$
$$A_{APBQ}=(1/2)\cdot (1+\sin(2a^{\circ}-90^{\circ}))\cdot \cos(2a^{\circ}-90^{\circ})$$
Another way of doing things is $AQ/AB=AC/BC$, because $\Delta BQA \sim \Delta BAC $ are similar. With $BC=\sqrt{AB^2+AC^2}=\sqrt{5}$ we get $AQ = AB \cdot (AC/BC) =  2/\sqrt{5}$. From there we can calculate $BQ = \sqrt{AB^2-AQ^2}=\sqrt{1-(2/\sqrt{5})^2}=1/\sqrt{5}$ and the trapezoid area according to the above formula
$$  A_{APBQ} =  2\Phi/5  $$
where $\Phi=(1+\sqrt{5})/2$ is the golden ratio.
A: 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 $PB= \sqrt{2 - \frac{2}{\sqrt5}}$.
