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zhenek [66]
3 years ago
7

What is the domain of the function {(-1,1) (0,0) (1,2)}

Mathematics
1 answer:
kipiarov [429]3 years ago
7 0
The domain is {-1, 0, 1}
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Find the values of the sine, cosine, and tangent for ZA.<br><br> (TOP OF TRIANGLE IS (A))
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\bigstar\:{\underline{\sf{In\:right\:angled\:triangle\:ABC\::}}}\\\\

  • AC = 7 m
  • BC = 4 m

⠀⠀⠀

\bf{\dag}\:{\underline{\frak{By\:using\:Pythagoras\: Theorem,}}}\\\\

\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\\\\

⠀⠀⠀⠀━━━━━━━━━━━━━━━━━━━━━

☆ Now Let's find value of sin A, cos A and tan A,

⠀⠀⠀

  • sin A = Perpendicular/Hypotenus = \sf \dfrac{4}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{4 \sqrt{65}}{65}}

⠀⠀⠀

  • cos A = Base/Hypotenus = \sf \dfrac{7}{\sqrt{65}} \times \dfrac{\sqrt{65}}{\sqrt{65}} = \pink{\dfrac{7 \sqrt{65}}{65}}

⠀⠀⠀

  • tan A = Perpendicular/Base = {\sf{\pink{\dfrac{4}{7}}}}

⠀⠀⠀

\therefore\:{\underline{\sf{Hence,\: {\pmb{Option\:A)}}\:{\sf{is\:correct}}.}}}

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