Question

in triangle ABC,<B=90°,sin A=4/5.find cos A?
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Answer 1

Answer:

cosA = 3/5

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Answer 2

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In triangle ABC,<B=90°,sin A=4/5.find cos A?

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$sin \: A = \frac{4}{5} = \frac{opp}{hyp} = \frac{BC}{AC}$

$\triangle \:ABC\:, \angle B \:=\: 90°$

$by \:Pythagoras \:Theorem$

$AC² = AB² + BC²$

$(5)² = AB² + (4)²$

$25 = AB² + 16$

$25 - 16 = AB²$

$9 = AB²$

$AB = \sqrt{9}$

$AB = 3cm$

$cos A = \frac{adj}{hyp} = \frac{AB}{AC} = \frac{3cm}{5cm}$

$cos A = \frac{3}{5}$

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