Answer:
1128cm
Step-by-step explanation:
The diagram of the problem is annotated and attached.
In the diagram, we are required to find the distance between Aliaa and the far corner, b.
In Triangle ABC
[tex]\angle A+\angle B+\angle C=180^\circ\\53^\circ+100^\circ+\angle C=180^\circ\\\angle C=180^\circ-153^\circ\\\angle C=27^\circ[/tex]
Using Law of Sines
[tex]\dfrac{b}{sin B} =\dfrac{c}{sin C} \\\dfrac{b}{sin 100} =\dfrac{520}{sin 27}\\b=(520 X sin 100)\div sin 27\\b=1127.997\\b\approx 1128$ cm (to the nearest centimeter.)[/tex]
Therefore, Aliaa should hit the ball a distance of 1128cm so that it lands perfectly in the far corner of the court.