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Two boats leave the same port at the same time. One travels at a speed of 31 mi/h in the direction N 50° E and the other travel...Asked by JOHN
1) Two boats leave the same port at the same time. One travels at a speed of 27 mi/h in the direction N 50° E and the other travels at a speed of 22 mi/h in a direction S 70° E (see the figure). How far apart are the two boats after one hour? (Round your answer to the nearest mile.)
2) A communications tower is located at the top of a steep hill, as shown. The angle of inclination of the hill is 58°. A guy wire is to be attached to the top of the tower and to the ground, 150 m downhill from the base of the tower. The angle α in the figure is determined to be 7°. Find the length of cable required for the guy wire. (Round your answer to the nearest meter.)
Thank you!
2) A communications tower is located at the top of a steep hill, as shown. The angle of inclination of the hill is 58°. A guy wire is to be attached to the top of the tower and to the ground, 150 m downhill from the base of the tower. The angle α in the figure is determined to be 7°. Find the length of cable required for the guy wire. (Round your answer to the nearest meter.)
Thank you!
Answers
Answered by
Steve
#1
convert to rectangular coordinates and subtract to get a distance of 24.9
#2
assuming the angle α is the angle between the ground and the cable, then using the law of sines, the distance d of the cable is
d/sin148° = 150/sin35°
convert to rectangular coordinates and subtract to get a distance of 24.9
#2
assuming the angle α is the angle between the ground and the cable, then using the law of sines, the distance d of the cable is
d/sin148° = 150/sin35°
Answered by
kuai
1. x^2 = 22^2+ 27^2 -2(22)(27)cos60
x^2 =1213 -1188cos60
x^2 =619
x= 24.9 miles
Answered by
Steve
kuai's solution is better, because the assignment is evidently involving the law of sines/cosines.
also, less work!
also, less work!
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