Discover answers to your questions with Westonci.ca, the leading Q&A platform that connects you with knowledgeable experts. Discover in-depth solutions to your questions from a wide range of experts on our user-friendly Q&A platform. Discover in-depth answers to your questions from a wide network of professionals on our user-friendly Q&A platform.
Sagot :
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
We appreciate your time. Please revisit us for more reliable answers to any questions you may have. Thank you for your visit. We're dedicated to helping you find the information you need, whenever you need it. We're glad you chose Westonci.ca. Revisit us for updated answers from our knowledgeable team.