Answer:
[tex]Area= 37193\ cm^2[/tex]
Step-by-step explanation:
Given
[tex]q = 590cm[/tex]
[tex]r = 310cm[/tex]
[tex]\angle P =24^o[/tex]
Required
The area of [tex]\triangle PQR[/tex]
This is calculated as:
[tex]Area= \frac{1}{2} * qr* \sin(P)[/tex]
So, we have:
[tex]Area= \frac{1}{2} * 590cm * 310cm* \sin(24^o)[/tex]
[tex]Area= \frac{1}{2} * 590cm * 310cm* 0.4067[/tex]
[tex]Area= 37192.715cm^2[/tex]
[tex]Area= 37193\ cm^2[/tex] ----- approximated
