Question

answer with explanation :|​

Answer 1

Step-by-step explanation:

$ΔABC be a right angled at B</p><p>Let ∠ACB=θ</p><p>Given that, sin θ = 3/5</p><p>AB/AC = 3/5</p><p>Let AB = 3x</p><p>then AC = 5x</p><p>In right angled ΔABC,</p><p>By Pythagoras theorem,</p><p>We get</p><p>(5x)2=(3x)2+BC2</p><p>BC2=(5x)2−(3x)2</p><p>BC2=(2x)2</p><p>BC=4x</p><p>(i) cos θ = Base/ Hypotenuse</p><p>= BC / AC</p><p>= 4x /5x</p><p>= 4/5</p><p>(ii) tan θ = perpendicular/Base</p><p>= AB/BC</p><p>= 3x/4x</p><p>=3/4</p><p></p><p>$

Hope it helps !

Answer 2

$\Huge\bf\maltese{\underline{\green{Answer°᭄}}}\maltese$

$\implies$$\large\bf{\underline{\red{VERIFIED✔}}}$

$\begin{gathered} \sin( \alpha ) = \frac{perpendicular}{hypotenuse} \\ = \frac{AC}{AB} = \frac{3}{5} \\ hence \: bc = 4 \: by \: pythagoras \\ hence \: \cos( \alpha ) = \frac{base}{hypotenuse} \\ = \frac{4}{5} \\ tan \alpha = \frac{perpendicular}{base} = \frac{3}{4} \end{gathered} \\ sin(α)=hypotenuseperpendicular=ABAC=53 \\ hence \: bc=4 \: by \: pythagoras \\ hence \: cos(α)=hypotenusebase=54 \\ tanα=baseperpendicular=43$

$\boxed{I \:Hope\: it's \:Helpful}$

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