A metal ring with diameter 1.94 metres is suspended in a horizontal plane by four support ropes each 5.40 m long, which are attached to points equidistant around the ring, and joined together at a swivel above the ring a) What is the angle between two ropes attached to opposite sides of the ring, at the point that they attach to the swivel, in degrees? b)What is the angle between two adjacent ropes, at the point that they attach to the swivel, in degrees?
+130071 A metal ring with diameter 1.94 metres is suspended in a horizontal plane by four support ropes each 5.40 m long, which are attached to points equidistant around the ring, and joined together at a swivel above the ring a) What is the angle between two ropes attached to opposite sides of the ring, at the point that they attach to the swivel, in degrees? b)What is the angle between two adjacent ropes, at the point that they attach to the swivel, in degrees?
a) For the first one, we can use the Law of Cosines..... we have an isosceles triangle whose base = 1.94m and sides = 5.4m......and we're lookig for the apex angle....so we have.....
(1.94)^2 = 2(5.4)^2 - 2(5.4)^2cos(theta)
So....using the cosine inverse, we have.....
cos-1 [ ( 1.94^2 - 2(5.4)^2 ) / (-2(5.4)^2) ] = about 20.7°
b) We can also use the Law of Cosines, again........we have an isosceles triangle whose base = 1.94/sqrt(2)m and whose sides = 5.4m.....and we're looking for the apex angle.....so we have......
cos-1 [ ( 1.94^2/2 - 2(5.4)^2 ) / (-2(5.4)^2) ] = about 14.59°
+130071 A metal ring with diameter 1.94 metres is suspended in a horizontal plane by four support ropes each 5.40 m long, which are attached to points equidistant around the ring, and joined together at a swivel above the ring a) What is the angle between two ropes attached to opposite sides of the ring, at the point that they attach to the swivel, in degrees? b)What is the angle between two adjacent ropes, at the point that they attach to the swivel, in degrees?
a) For the first one, we can use the Law of Cosines..... we have an isosceles triangle whose base = 1.94m and sides = 5.4m......and we're lookig for the apex angle....so we have.....
(1.94)^2 = 2(5.4)^2 - 2(5.4)^2cos(theta)
So....using the cosine inverse, we have.....
cos-1 [ ( 1.94^2 - 2(5.4)^2 ) / (-2(5.4)^2) ] = about 20.7°
b) We can also use the Law of Cosines, again........we have an isosceles triangle whose base = 1.94/sqrt(2)m and whose sides = 5.4m.....and we're looking for the apex angle.....so we have......
cos-1 [ ( 1.94^2/2 - 2(5.4)^2 ) / (-2(5.4)^2) ] = about 14.59°
