Here's an alternative, vector type solution.
Let \(\displaystyle \hat{\imath}\) be a unit vector in the direction PQ and \(\displaystyle \hat{\jmath}\) a unit vector perpendicular (upwards in the diagram), to this.
Then, \(\displaystyle \underline{PB}=2\cos(60)\hat{\imath}+2\sin(60)\hat{\jmath}=\hat{\imath}+\sqrt{3}\hat{\jmath}\) , and \(\displaystyle \underline{PC}=3\hat{\imath}\) . (All angles in degrees.)
In the triangle PBC, \(\displaystyle \underline{PB}+\underline{BC}=\underline{PC}\), so \(\displaystyle \underline{BC}=3\hat{\imath}-(\hat{\imath}+\sqrt{3}\hat{\jmath})=2\hat{\imath}-\sqrt{3}\hat{\jmath}\) .
Let the vector \(\displaystyle \underline{CA} = \alpha\hat{\imath}+\beta\hat{\jmath}\) , then since CA is at right angles to BC, the scalar product \(\displaystyle \underline{CA}.\underline{BC}=0\) ,
so, \(\displaystyle 2\alpha-\sqrt{3}\beta = 0\) ...................... (1).
The vector \(\displaystyle \underline{AQ} \) will be some fraction of the vector \(\displaystyle \underline{RQ}\) , \(\displaystyle \underline{AQ} = k\underline{RQ}\), say,
so \(\displaystyle \underline{AQ}=k(5\cos(60)\hat{\imath}-5\sin(60)\hat{\jmath})=k \left(\frac{5}{2}\hat{\imath}-\frac{5\sqrt{3}}{2}\hat{\jmath}\right)\).
Finally, (almost), in the triangle CAQ, \(\displaystyle \underline{CA} +\underline{AQ}= \underline{CQ}\) ,
so, \(\displaystyle \alpha\hat{\imath}+\beta\hat{\jmath}+k\left(\frac{5}{2}\hat{\imath}-\frac{5\sqrt{3}}{2}\hat{\jmath}\right)=2\hat{\imath}\) .
Equating coefficients, \(\displaystyle \alpha +\frac{5k}{2}=2\quad \text{ so, }\quad\alpha = 2-\frac{5k}{2}\),
and \(\displaystyle \beta -\frac{5k\sqrt{3}}{2}=0 \quad \text{so}\quad \beta = \frac{5k\sqrt{3}}{2}\).
Substituting into (1), \(\displaystyle 2\left(2-\frac{5k}{2}\right)-\sqrt{3}\left(\frac{5k\sqrt{3}}{2}\right)=0\) , so \(\displaystyle 4-\frac{25k}{2}=0,\quad k=\frac{8}{25},\)
so \(\displaystyle AQ = \frac{8}{25}\times5=\frac{8}{5}=1.6\) .
Tiggsy
