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Given cot (x) = 3/4, ( let theta be x)
so,
cos (x) / sin (x) = 3/4
cos (x) = sin (90 - x)
sin (x) = cos (90 - x)
so,
sin (90 - x) / cos (90 - x) = 3/4
but Given that, x - a = 90
so,
-a = 90 - x
on substituting in the main equation,
we get,
sin (-a) / cos (-a) = 3/4
but,
sin (-a) = -sin(a) and cos (-a) = cos (a)
so,
-tan (a) = 3/4
a = tan^-1 (-3/4)
a = -36.86°
Hope this helps
- WonderGirl
so,
cos (x) / sin (x) = 3/4
cos (x) = sin (90 - x)
sin (x) = cos (90 - x)
so,
sin (90 - x) / cos (90 - x) = 3/4
but Given that, x - a = 90
so,
-a = 90 - x
on substituting in the main equation,
we get,
sin (-a) / cos (-a) = 3/4
but,
sin (-a) = -sin(a) and cos (-a) = cos (a)
so,
-tan (a) = 3/4
a = tan^-1 (-3/4)
a = -36.86°
Hope this helps
- WonderGirl
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