Discussion
In \(\Delta ABC\), \(\angle C = 90^{\circ}\), \(AC = 15\), \(BC = 8\), find sec A
Answer: B
sec A is 17/15.
With the right angle at C, the hypotenuse is AB:
\[ AB = \sqrt{15^{2} + 8^{2}} = \sqrt{289} = 17 \]
For angle A, the adjacent side is AC = 15, so \(\cos A = 15/17\) and \(\sec A = 17/15\). 15/17 is cos A, the reciprocal of the answer, and 17/5 and 5/17 do not use the sides of this triangle at all.
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