• Choice A

    With the right angle at BB, BC‾\overline{BC} is opposite ∠A\angle A and AC‾\overline{AC} is adjacent to ∠A\angle A. tan⁡70∘=BCAC=9AC\tan 70^\circ = \frac{BC}{AC} = \frac{9}{AC} AC=9tan⁡70∘AC = \frac{9}{\tan 70^\circ}

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