In the right angle triangle PQS,
[tex]\begin{gathered} sin62.58^{\circ}=\frac{PS}{PQ} \\ \sin 62.58^{\circ}=\frac{PS}{4.5} \\ PS=4.5\sin 62.58^{\circ} \\ PS=3.99 \end{gathered}[/tex]Thus, value of PS is 3.99 units.
The value of QS can be determined as,
[tex]\begin{gathered} \cos 62.58=\frac{QS}{PQ} \\ QS=4.5\cos 62.58 \\ QS=2.07 \end{gathered}[/tex]Thus, the value of QS is 2.07 units.
The area of triangle PQR can be determined as,
[tex]\begin{gathered} \Delta=\frac{1}{2}\times QR\times PS \\ =\frac{1}{2}\times(5+2.07)\times3.99 \\ =14.10 \end{gathered}[/tex]Thus, the required area is 14.10 square units.
