The coordinates of the vertices of triangle PQR are P(-3, 3), Q(2, 3), and R(-3, -4).
Find the side lengths to the nearest hundredth and the angle measures to the nearest degree.
Answer (2)
Answer: PQ=5, QR=radical 61= 7.81, angle: 50degrees, Why:PQ=|-3|+2=5PR=6 angle Alfa, so there is right angle triangle PQR so I can use following formula: PQ^2 + RP^2=QR^2;25 + 36=QR^2;QR=radical 61 also I can use: sinus (angle)=PR/QR; sin(angle)=6/radical61=0.76822 which gives angle to be little bit more than 50 degrees{sinus 50 degrees= 0,7660 }.
The side lengths of triangle PQR are approximately PQ = 5, QR ≈ 8.60, and PR = 7. The angles are approximately angle P = 90°, angle Q = 50°, and angle R = 40°. This uses the distance formula and Law of Cosines for calculations.
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