Question

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Q . In a right triangle ABC, tan(A) = 3/4. Find sin(A) and cos(A).


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Answer 1

Step-by-step explanation:

$heloo... \\ \tan(a) = \frac{3}{4} \\ so \: \: we \: \: already \: know \: 3 \: 4 \: 5 \: triangle \: .. \\ \\ \: where \: 3 \: is \: opposite \: side \\ 4 \: is \: adjacent \: one \: \\and \: 5 \: is \: hypotenuse. \\ \\ so \: we \: know \: \tan(a ) = \frac{opp}{adj} = \frac{3}{4} \\ so \\ \sin(a) = \frac{opp}{hyp} = \frac{3}{5} .. \\ and \\ \cos(a) = \frac{adj}{hyp} = \frac{4}{5} \\ \\ hope \: it \: helps \: uh$

Answer 2

Given

tanA = 3/4

on sqiring LHS and RHS

tan^2A = 9/16

sec^2A - 1 = 9/16

sec^2A = 9/16 + 1

sec^2A = 25/16

secA = 5/4 and - 5/4

cos A = 4/5 and - 4/5

since secA = 1/cosA

Now

CosA = 4/5

on sqiring both side

cos^2A = 16/25

1 - sin^2A = 16/25

sin^2A = 1- 16/25

sin^2A = 9/25

sinA = 3/5 and - 3/5