After leaving camp for a three-day journey into the frozen tundra, Daniela has just returned to camp. The
2 answers:
Answer:
a) 8.7 km
b) 187.6°
Step-by-step explanation:
<u>Given</u>
Travel vectors (distance)∠(direction) ...
- d₁ = 8∠3°
- d₂ = unknown
- d₃ = 1∠47°
- total travel was a round trip
<u>Find</u>
(Rounded to the nearest tenth ...)
a) |d₂|
b) ∠d₂
<u>Solution</u>
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.
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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.
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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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Allow me to prove the result: odd numbers come in the form 2n-1, because 2n is always even, and the number immediately before an even number is always odd.
So, if we sum the first N odd numbers, we have
The first sum is the sum of all integers from 1 to N, which is N(N+1)/2. We want twice this sum, so we have
The second sum is simply the sum of N ones:
So, the final result is
which ends the proof.