Question

if sin (x + y) =1 and cos (x-y) = √3/2 , where 0°<_x+y,<90°, x>y, then find the values of x and y.

Answer 1

Hi Mate!!

Sin ( x + y ) = Sin ( 90 ) and Cos ( x - y) = Cos ( 30 )

bcoz. Sin ( 90 ) = 1. and Cos ( 30 ) = √3 / 2


=>. x + y = 90....... equation 1

and

x - y = 30..... equation 2

Adding both the equations

x = 60 and. y = 30

Answer 2

Answer:

x=60°

y=30°

Step-by-step explanation:

sin (x+y) =90° [sin90° = 1)]

x+y =90°.................(i)

again

cos(x+y) = √3/2

cos(x+y) = cos30° [ cos30°=√3/2]

x-y =30°.................................................(ii)

adding (i) and (ii), we get

x+y+x-y= 90°+30°

x+x=120°

2x=120°

x=120°/2

x=60°

now,

since,x+y=90°

y=90°-60°

y= 30°

hence, the required answer...