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Home/Practice questions/General Aptitude Test
General Aptitude TestQuantitative ReasoningTrigonometry

If cos(x - y) = (√3)/2 and sin (x + y) = 1, where x and y are positiv…

If cos(x - y) = $\frac{\sqrt{3}}{2}$ and sin (x + y) = 1, where x and y are positive acute angles and x ≥ y, then x and y are : (0o , (x +y) ≤ 90o)

Answer & explanation+

Correct answer: option 4

cos (x-y) = \(\frac{ √3 }{2}\)

{ we know, cos 30º = \(\frac{ √3 }{2}\) }

So, ( x - y ) = 30º    ----(1)

& sin ( x + y ) = 1

{ we know, sin 90º= 1 }

So, ( x + y ) = 90º         ----(2)

On adding 1 nad 2 .

2x = 90º + 30º

x = 60º

Now, x + y = 90º

60º  y = 90º

y = 30º

So, x = 60º and y = 30º

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