tan2A + 5 sec A = 13, where 0 < A < 90°. Solve for A (in degrees).

0 60 45 30

60

tan2A + 5 sec A = 13 ( Sec²A - 1 ) + 5 SecA = 13 Sec²A + 5SecA = 14 Sec²A + 7SecA - 2SecA - 14= 0 SecA ( SecA + 7 ) -2 ( SecA + 7 ) = 0 ( SecA - 2 ) ( SecA + 7 ) = 0 Either ( SecA - 2 = 0 ) OR ( SecA + 7 = 0  ). SecA = - 7 is not possible because 0 < A < 90° So , SecA = 2 ⇒  SecA = Sec60° A = 60°