a) \(\sin {40^0}54'\)
Ta có: \(\sin {40^0}54' = 0,6547408137 \approx 0,655\)
b) \(\cos {52^0}15'\)
Ta có: \(\cos {52^0}15' = 0,61221728 \approx 0,612\)
c) \(\tan {69^0}36'\)
Ta có: \(\tan {69^0}36' = 2,688918967 \approx 2,689\)
d) \(\cot {25^0}18'\)
Ta có: \(\tan {25^0}18' = 0,4726978344\) nên \(\cot {25^0}18' = \frac{1}{{\tan {{25}^0}18'}} = 2,115516356 \approx 2,116\)
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