Question 392548
All you need to know is the {{{value }}}of the trigonometry functions at the {{{special }}}{{{angles}}}. 

For this problem, you need to know the following:


{{{sin (30) deg = 1/2}}}


{{{tan (45) deg = 1}}}


{{{sec (45) = sqrt(2)}}}


{{{cos (30) =sqrt(3)/2}}}


{{{sin (60) = sqrt(3)/2}}}


so, you will have:


{{{12 sin 30 deg - 6 tan 45 deg +sec 45 deg/52}}}


{{{12 (1/2)  - 6 *1 + sqrt(2)/52}}}


{{{6  - 6  + sqrt(2)/52}}}


= {{{sqrt(2)/52}}}


= {{{(1.41)/52}}}


= {{{0.03}}}