Question

Is sinA=3/5 and A is acute, then find sin2A

Answer 1

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Answer 2

$\Large{\textbf{\underline{\underline{According\:to\:the\:Question}}}}$

Here we have,

$\tt{\rightarrow SinA=\dfrac{3}{5}}$

That's mean

$\tt{\rightarrow\dfrac{P}{H}=\dfrac{3}{5}}$

Here we have

P = 3

H = 5

Now using Pythagoras Theorem

(3)² + (B)² = 5²

25 - 9 = B²

16 = B²

√16 = B

Hence we get,

B = 4 .

Hence,

We get the values,

$\tt{\rightarrow SinA=\dfrac{P}{H}=\dfrac{3}{5}}$

$\tt{\rightarrow CosA=\dfrac{B}{H}=\dfrac{4}{5}}$

Now,

Sin2A

2 × SinA × CosA

$\tt{\rightarrow 2\times\dfrac{3}{5}\times\dfrac{4}{5}}$

$\tt{\rightarrow\dfrac{24}{25}}$

Therefore,

$\tt{\rightarrow Sin2A=\dfrac{24}{25}}$