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uranmaximum [27]
3 years ago
10

???????????????????????????????????????????????????????????????????????????????

Mathematics
1 answer:
garik1379 [7]3 years ago
3 0

Answer:

Step-by-step explanation:

This problem is just really long- it's trying to confuse you with complicated wording. Here's it demystified:

a)

t = 0 degrees Celsius

d = 1500 meters

Find the temperature, T.

b)

d = 300 meters

T = 26 degrees Celsius

Find the ground temperature, t.

NOTE: The temperatures are different! T is the final temperature and t is the ground temperature.

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The cosine law
romanna [79]
\bf \textit{Law of sines}
\\ \quad \\
\cfrac{sin(\measuredangle A)}{a}=\cfrac{sin(\measuredangle B)}{b}=\cfrac{sin(\measuredangle C)}{c}\\\\
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\bf \cfrac{sin(88^o)}{39.9}=\cfrac{sin(A)}{31}\implies \cfrac{31\cdot  sin(88^o)}{39.9}=sin(A)
\\\\\\
sin^{-1}\left[ \cfrac{31\cdot  sin(88^o)}{39.9} \right]=sin^{-1}\left[ sin(A) \right]
\\\\\\
sin^{-1}\left[ \cfrac{31\cdot  sin(88^o)}{39.9} \right]=\measuredangle A
\\\\\\
50.93841^o\approx \measuredangle A\qquad thus\qquad \measuredangle C\approx 41.0619^o
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