Question

If SinA =3/4, calculate CosA and tanA.​

Answer 1

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$given \: \sin(a) = \frac{3}{4}$

$= > \frac{BC}{DC} = \frac{3}{4}$

$= > BC= 3K \: and \: AC=4k$

where k is the constant of proportionality.

By Pythagoras theorem, we have

$= > AB^{2} =AC^{2} -BC^{2}$

$= > {(4k)}^{2} - {(3k)}^{2} = {7k}^{2}$

$= > AB = \sqrt{7} k$

$so \: cosA \: = \frac{AB}{AC} = \frac{ \sqrt{7}k }{4k} = \frac{ \sqrt{7} }{4}$

$and \: tanA \: = \frac{AB}{AC} = \frac{3k }{ \sqrt{7} k} = \frac{3}{ \sqrt{7} }$

$\pink{i \: hope \: it \: helpfull \: for \: you}$

Answer 2

Answer:

hope this helps

really please