kayleecaswell04 kayleecaswell04
  • 03-10-2018
  • Mathematics
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PLEAS I REALLY NEED HELP!! TRIGONOMETRIC RATIOS IN RIGHT TRIANGLES

PLEAS I REALLY NEED HELP TRIGONOMETRIC RATIOS IN RIGHT TRIANGLES class=
PLEAS I REALLY NEED HELP TRIGONOMETRIC RATIOS IN RIGHT TRIANGLES class=

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Аноним Аноним
  • 03-10-2018

sin x  = opposite side / hypotenuse  (the hypotenuse is the longest side ) so sin A in the given triangle = 5 / 13

cos x = adjacent side / hypotenuse.  The adjacent side to angle A is AC  so cos A = AC / 13

tan x = opposite side / adjacent side  In the diagram tan A =  5 / AC

csc x = 1 / sin x  = H / O     ( H=hypotenuse and O = opposite) side

sec x = 1 / cos x = H/ A    ( A = adjacent side)

cot x  = 1 / tanx =  A / O

In the triangle   AC^2 = 13^2 - 5^2 = 144 so AC = 12  ( By Pythagoras)

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