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schepotkina [342]
3 years ago
6

If 0°<0<90° and sin0=12/13, find cos0 using trigonometric identities.

Mathematics
1 answer:
Aleks04 [339]3 years ago
3 0

here's the solution,

we know :

\sin( \theta)  =  \dfrac{perpendicular}{hypotenuse}

So,

\dfrac{p}{h}  =  \dfrac{12}{13}

so.. let the perpendicular be 12x and hypotenuse be 13x

now,

by applying pythagoras theorem,

  • b {}^{2}  = h {}^{2}  - p {}^{2}

where,

  • b = base
  • h = hypotenuse
  • p = perpendicular

So,

  • b {}^{2}  =  ({13x})^{2}  -  ({12x})^{2}
  • b  {}^{2} = 169x {}^{2}  - 144x {}^{2}
  • {b}^{2}  = 25x {}^{2}
  • b  = 5x

so,

  • \cos( \theta)  =  \dfrac{b}{h}

  • \cos( \theta)  =   \dfrac{5x}{13x}

  • \cos( \theta)  =  \dfrac{5}{13}

hope it helps !!

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