Mathematics: If 0° 0 90° and sin0=12/13, find cos0 using trigonometric identities. • 12/5 • 5/12 • 12/13 • 5/13

If 0° 0 90° and sin0=12/13, find cos0 using trigonometric - id009001213013115093, lillianmitchell580

Ответ:

kaykay9243

08.05.2021

Mathematics

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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Ответ:

ezequiel2k

14.05.2021

Mathematics

Step-by-step explanation:

$a_n=\dfrac{1}{2} *(\dfrac{-1}{5} )^{n-1}\\\\a_{n+1}=\dfrac{1}{2} *(\dfrac{-1}{5} )^{n}\\\\\\\\\dfrac{a_{n+1}}{a_n} =\dfrac{\dfrac{1}{2} *(\dfrac{-1}{5} )^{n}}{\dfrac{1}{2} *(\dfrac{-1}{5} )^{n-1}} =\dfrac{-1}{5} \\a_1=\dfrac{1}{2} *(\dfrac{-1}{5} )^{0}=\dfrac{1}{2} \\\\\\\boxed{a_1=\dfrac{1}{2}}\\\boxed{a_{n } =-\dfrac{1}{5} *a_{n-1} }\\\\\\a_1=\dfrac{1}{2}\\\\a_2=\dfrac{-1}{10}\\\\a_3=\dfrac{1}{50}\\\\a_4=\dfrac{-1}{250}\\\\$

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If 0°<0<90° and sin0=12/13, find cos0 using trigonometric identities. • 12/5
• 5/12
• 12/13
• 5/13

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If 0°<0<90° and sin0=12/13, find cos0 using trigonometric identities. • 12/5 • 5/12 • 12/13 • 5/13

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If 0°<0<90° and sin0=12/13, find cos0 using trigonometric identities. • 12/5
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• 5/13