Asked by Emily
consider the interior angles of the triangle with vertices
p=(=4.39,-0,948) q=(3.03,8.31) r=(4.21,6.54)
find the angle of the vertex of p answer in units of degrees
find the angle of vertex q answer in units of degrees
p=(=4.39,-0,948) q=(3.03,8.31) r=(4.21,6.54)
find the angle of the vertex of p answer in units of degrees
find the angle of vertex q answer in units of degrees
Answers
Answered by
Steve
I'll do p; you can do q.
assuming p=(-4.39,-0.948),
pq = √((3+4.39)^2 + (8.31+0.948)^2) = 11.845
pr = √((4.21+4.39)^2 + (6.54+0.948)^2) = 11.403
qr = √((4.21-3.03)^2 + (6.54-8.31)^2) = 2.172
so, using the law of cosines,
2.172^2 = 11.845^2 + 11.403^2 - 2(11.845)(11.403) cos P
cosP = (11.845^2 + 11.403^2 - 2.172^2)/(2(11.845)(11.403)) = 0.983
P = 10.5°
assuming p=(-4.39,-0.948),
pq = √((3+4.39)^2 + (8.31+0.948)^2) = 11.845
pr = √((4.21+4.39)^2 + (6.54+0.948)^2) = 11.403
qr = √((4.21-3.03)^2 + (6.54-8.31)^2) = 2.172
so, using the law of cosines,
2.172^2 = 11.845^2 + 11.403^2 - 2(11.845)(11.403) cos P
cosP = (11.845^2 + 11.403^2 - 2.172^2)/(2(11.845)(11.403)) = 0.983
P = 10.5°
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