Question

if 3cosA=5 then find the value of sinA​

Answer 1

$\huge{\colorbox{cyan}{\underline{\underline{\bf{\pink{☆• answer \: •☆}}}}}}$

$3cos \: a = 5 \\ cos \: a = \frac{5}{3} \\ \\ using \: \: pythagoras \: theorem \: \\ {h}^{2} = {p}^{2} + {b}^{2} \\ {3}^{2} = {p}^{2} + {5}^{2} \\ 9 = {p}^{2} + 25 \\ {p}^{2} = - 9 + 25 \\ {p}^{2} = 16 \\ p = 4cm \\ \\ \sin(a) = \frac{p}{h} = \frac{4}{3}$

Answer 2

hello dear,

Answer: -

$3 \cos(a) = 5 \\ \cos(a) = \frac{5}{3} \\ ( \cos(a) = \frac{b}{h} )$

$p {}^{2} = b {}^{2} - h {}^{2} \\ p {}^{2} = (5) {}^{2} - (3) {}^{2} \\ p {}^{2} = 25 - 9 = 16 \\ p {}^{2} = 16 \\ p = \sqrt{16} = 4$

Thae valve of Sin A: -

$( \sin(a) = \frac{p}{h} ) \\ p = 4 \: and \: h = 3 \\ \frac{p}{h} = \frac{4}{3}$

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