Asked by Molly
Triangle PQR has vertices P(0, 1), Q(0, -4), and R(2, 5). Find the coordinates of R' to the nearest hundredth after rotating triangle PQR counterclockwise about the origin 45º.
Answers
Answered by
Steve
the radius to (2,5) is at an angle θ such that tanθ = 5/2
So
cosθ = 2/√29
sinθ = 5/√29
R' is at an angle (θ+45), so
x = √29 cos(θ+45)
y = √29 sin(θ+45)
x = √29 (cosθ cos45 - sinθ sin45)
= √29 (2/√29 * 1/√2 - 5/√29 * 1/√2)
= 2/√2 - 5/√2
= -3/√2
y = √29 (sinθ cos45 + cosθ sin45)
= √29 (5/√29 * 1/√2 + 2/√29 * 1/√2)
= 7/√2
So, R' = (-3/√2, 7/√2)
So
cosθ = 2/√29
sinθ = 5/√29
R' is at an angle (θ+45), so
x = √29 cos(θ+45)
y = √29 sin(θ+45)
x = √29 (cosθ cos45 - sinθ sin45)
= √29 (2/√29 * 1/√2 - 5/√29 * 1/√2)
= 2/√2 - 5/√2
= -3/√2
y = √29 (sinθ cos45 + cosθ sin45)
= √29 (5/√29 * 1/√2 + 2/√29 * 1/√2)
= 7/√2
So, R' = (-3/√2, 7/√2)
Answered by
jhoy
wow
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