After leaving camp for a three-day journey into the frozen tundra, Daniela has just returned to camp. The
three days of her journey can be described by displacement (distance and direction) vectors d , d), and dz.
where each vector indicates Daniela's displacement from the start of her day to the end of her day.
di = 8i +3
d, is not given
az = i +47
(Distances are given in kilometers, km.)
What distance did Daniela travel on day two?
kilometers
(Round your final answer to the nearest tenth.)
In what direction did Daniela travel on day two? Assume 0° is the rightward direction.
(Round your final answer to the nearest degree. Your answer should be between 0 and 360°.)

Respuesta :

Answer:

  a) 8.7 km

  b) 187.6°

Step-by-step explanation:

Given

Travel vectors (distance)∠(direction) ...

  • d₁ = 8∠3°
  • d₂ = unknown
  • d₃ = 1∠47°
  • total travel was a round trip

Find

(Rounded to the nearest tenth ...)

  a) |d₂|

  b) ∠d₂

Solution

In order for the sum of the three travel vectors to represent a round trip, their sum must be zero. That is, the value of d₂ must be the opposite of the sum of the other two vectors. A vector calculator can find the desired value easily. Here, we will show the calculation using the law of cosines and the law of sines.

__

a) A diagram can be helpful. Since we want to find the opposite of the sum of the given vectors, we can start by drawing their sum. In our diagram, d₁ is segment AB, and d₃ is segment BC. Their sum is the segment AC.

The internal angle ABC is the supplement of the difference of the angles for d₁ and d₃, so is 180° -(47° -3°) = 136°. Then the law of cosines gives length AC as ...

  AC² = AB² +BC² -2AB·BC·cos(B)

  AC² = 8² +1² -2·8·1·cos(136°) = 65 -16cos(136°) ≈ 76.5094

  AC ≈ √76.5094 ≈ 8.747

The distance traveled on day 2 is about 8.7 km.

__

b) Given the side lengths of a triangle and one angle, the law of sines can be used to find the other angles. Here, it is convenient to find the internal angle at A. The law of sines tells us ...

  sin(A)/BC = sin(B)/AC

  sin(A) = BC/AC·sin(B)

  A = arcsin(BC/AC·sin(B)) = arcsin(1/8.747·sin(136°)) ≈ 4.555°

The direction angle from C to A will be the sum of the 3° direction angle of AB and the internal triangle angle A we just found, added to 180°. That sum is ...

  ∠CA = 180° +3° +4.6° = 187.6°

  ∠d₂ = 187.6°

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Answer:

Distance traveled 8.6 kilometers on day two

Daniela traveled in a direcrion of 234degrees on day two

Step-by-step explanation:

I had the same exact question and found the answers after having it incorrect