The terminal ray of ∠A passes through the point (−6,3) . ∠A is drawn in standard position. What is the value of secA ?

  sec A  =  r / x

We  can  find  r  as   √ [ (-6)^2  + (3)^2  ] =  √  [36 + 9]  =  √ 45  =  √ [9 * 5 ] = 3√ 5

So

sec A  =  3√ 5 / -6  =   √ 5 / -2   =  -√ 5  /  2