Answered

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In ΔPQR, q = 590 cm, r = 310 cm and ∠P=24°. Find the area of ΔPQR, to the nearest square centimeter.

Sagot :

MrRoyalidn

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

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