Question

How do you evaluate sec 23 deg. 2'?

Answer 1

Evaluate `sec 23^@2`'

Ans: 1.09

Explanation:

Call `x = 23^@2' = 23.033`
`sec x = 1/(cos x)` . Find cos x
`cos x = cos (23.033) = 0.92`
`sec x = 1/0.92 = 1.09`