[tex]\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\[/tex]
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[tex]\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\[/tex]
[tex]\star\:{\underline{\boxed{\frak{\purple{(Hypotenus)^2 = (Perpendicular)^2 + (Base)^2}}}}}\\\\\\ :\implies\sf (AB)^2 = (AC)^2 + (BC)^2\\\\\\ :\implies\sf (AB)^2 = (AB)^2 = (7)^2 = (4)^2\\\\\\ :\implies\sf (AB)^2 = 49 + 16\\\\\\ :\implies\sf (AB)^2 = 65\\\\\\ :\implies{\underline{\boxed{\pmb{\frak{AB = \sqrt{65}}}}}}\:\bigstar\\\\[/tex]
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☆ Now Let's find value of sin A, cos A and tan A,
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- sin A = Perpendicular/Hypotenus = [tex]\sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}[/tex]
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- cos A = Base/Hypotenus = [tex]\sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}[/tex]
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- tan A = Perpendicular/Base = [tex]{\sf{\pink{\dfrac{4}{7}}}}[/tex]
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[tex]\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}[/tex]
