cooper230 cooper230
  • 13-01-2022
  • Mathematics
contestada

Please help me solve this question

Please help me solve this question class=

Respuesta :

dt33183 dt33183
  • 17-01-2022

Answer:

sin(0) = opposite/hypotenuse = (3√3)/√31

cos(0) = adjacent/hypotenuse = 2/√31

tan(0) = opposite/adjacent = (3√3)/2

cot(0) = 2/(3√3)

sec(0) = 1/cos(0) = 1/(2/√31) = √31/2

Step-by-step explanation:

Hypertenuse = √[2^2 + (3√3)^2] = √[4 + 27] = √31

sin(0) = opposite/hypotenuse = (3√3)/√31

cos(0) = adjacent/hypotenuse = 2/√31

tan(0) = opposite/adjacent = (3√3)/2

cot(0) = 2/(3√3)

sec(0) = 1/cos(0) = 1/(2/√31) = √31/2

I hope this helps

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