(1) Lay out segment QR with length 7
(2) Construct 60-degree angle at Q, giving a ray with initial point Q
(3) With center at R, draw an arc with radius 8, intersecting the ray from step (2)
Since RP = 8 is greater than RQ = 7, the arc from step (3) will intersect the ray from step (2) in only one place; that point of intersection is point P.
(4) draw segment PR to complete the triangle
Note if RP had been less than RQ, the arc in step (3) would have intersected the arc from step (2) in two places. That would have given two possibilities for the location of point P, leading to two different triangles with the prescribed features.
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