Question

If 3 tan A=4then find sin A
If 3tan A=4find sinA

Answer 1

$\huge\mathfrak{\purple{Answer:-}}$

$\sf\bullet 3 tan A=4\\ \\ \sf\bullet tanA=\dfrac{4}{3}$

Means

$\boxed{\sf{tan\theta=\dfrac{Perpendicular}{Base}}}$

$\sf\bullet P= 4\:\:\: and\:\:B=3$

Now by Pythagoras

$\sf{\purple{H{}^{2}=B{}^{2}+P{}^{2}}}\\ \\ \sf\implies H{}^{2}= (3){}^{2}+(4){}^{2}\\ \\ \sf\implies H{}^{2}= 9+16\\ \\ \sf\implies H{}^{2}=25\\ \\ \sf\implies H=\sqrt{25}\\ \\ \sf\implies H=5$

Now we have to find the Value of sinA

$\boxed{\bigstar{\sf{sin\theta=\dfrac{Perpendicular}{Height}}}}$

$\sf\implies sinA=\dfrac{P}{H}\\ \\ \sf\bullet P=4\:\:\:and\:\:H=5\\ \\ \sf\implies sinA=\dfrac{4}{5}$

Answer 2

$\large{\underline{\mathrm{\blue{Solution-}}}}$

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It is given that,

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$\sf{3TanA\:=\:4}$

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$\therefore$ $\sf{TanA\:=\:\dfrac{4}{3}}$

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Also we know that,

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$\sf{TanA\:=\:\dfrac{P}{B}}$

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: $\implies$ $\sf{\dfrac{4}{3}\:=\:\dfrac{P}{B}}$

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Let perpendicular be 4k and base be 3k.

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Using Pythagoras theorem,

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$\sf{H^2\:=\:P^2\:+\:B^2}$

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: $\implies$ $\sf{H^2\:=\:(4k)^2\:+\:(3k)^2}$

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: $\implies$ $\sf{H^2\:=16k^2\:+\:9k^2}$

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: $\implies$ $\sf{H^2\:=\:25k^2}$

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: $\implies$ $\sf{H\:=5k}$

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$\rule{200}2$

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Now, we know that,

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$\bold{\sf{SinA\:=\:\dfrac{Perpendicular}{Hypotenuse}}}$

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: $\implies$ $\sf{SinA\:=\:{\cancel{\dfrac{4k}{5k}}}}$

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$\huge{\therefore}$ $\huge{\boxed{\red{\sf{SinA\:=\dfrac{4}{5}}}}}$