Answer:
[tex]\large\boxed{A=83^o}[/tex]
Step-by-step explanation:
[tex]\text{Law of cosine:}\\\\BC^2=AC^2+AB^2-2(AC)(AB)\cos\angle A\\---------------------\\\text{We have:}\\\\BC=17\ ft,\ AC=8\ ft,\ AB=16\ ft\\\\\text{Substitute:}\\\\17^2=8^2+16^2-2(8)(16)\cos\angle A\\\\289=64+256-256\cos\ma\angle a\\\\289=320-256\cos\angle A\qquad\text{subtract 320 from both sides}\\\\-31=-256\cos\angle A\qquad\text{divide both sides by (-256)}\\\\\cos\angle A=\dfrac{31}{256}\\\\\cos\angle A\approx0.1211\to m\angle A\approx83.0445^o\to m\angle\approx83^o[/tex]
