Suppose that you wish to cross a river that is 3900 feet wide and flowing at a rate of 5 mph from north to south. Starting on th
e eastern bank, you wish to go directly across the river to a point on the western bank opposite your current position. You have a boat that travels at a constant rate of 11 mph.
a) In what direction, measured clockwise from north, should you aim your boat? Include appropriate units in your answer.
b) How long will it take you to make the trip? Include appropriate units in your answer.\
Please show your work so I may understand. Thank you so much!
a) In what direction, measured clockwise from north, should you aim your boat? Include appropriate units in your answer.
b) How long will it take you to make the trip? Include appropriate units in your answer.\
Please show your work so I may understand. Thank you so much!
1 answer:
Answer:
a) 297°
b) 4.52 minutes
Step-by-step explanation:
a) Consider the attached figure. The boat's actual path will be the sum of its heading vector BA and that of the current, vector AC. The angle of BA north of west has a sine equal to 5/11. That is, the heading direction measured clockwise from north is ...
270° + arcsin(5/11) = 297°
__
b) The "speed made good" is the boat's speed multiplied by the cosine of the angle between the boat's heading and the boat's actual path. That same value can be computed as the remaining leg of the right triangle with hypotenuse 11 and leg 5.
boat speed = √(11² -5²) = √96 ≈ 9.7980 . . . . miles per hour
Then the travel time will be ...
time = distance/speed
(3900 ft)×(1 mi)/(5280 ft)×(60 min)/(1 h)/(9.7980 mi/h) ≈ 4.523 min
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Answer:
Part A none
Part B drawn
Step-by-step explanation:
Suppose the transversal intersects a pair of parallel lines it creates two pairs of alternate exterior angles.
Now if we look at the first figure we find that the alternate exterior angle 1 and 4 are equal . Also 3 and 2 are equal.
But <1 and < 2 are not equal . They are supplementary angles. Their sum is equal to 180 degrees.
Similarly angles 3 and4 are supplementary angles.
This situation does not support Ricky's claim. So you will select none in part A.
Now in part B we see that angle 5 &8 are equal and 6 & 7 are congruent.
This refutes Ricky's claims and this can be proved mathematically.
Angles 5 & 6 are supplementary angles. Similarly 7 & 8 are supplementary therefore they cannot be equal as they are unequal and a transversal is drawn not a perpendicular.
