Question

Given that sin(5y-28)=cos(3y-50). Find the value of y where 0°≤y≤90º

Answer 1

$Given \: sin(5y-28) = cos(3y-50)$

$\implies sin(5y-28) = sin[ 90 -(3y-50)]$

$\boxed{\pink{ \because sin (90 - \theta ) = cos \theta }}$

$\implies 5y - 28 = 90 - (3y-50)$

$\implies 5y - 28 = 90 - 3y+50$

$\implies 5y - 28 = 140 - 3y$

$\implies 5y + 3y = 140 + 28$

$\implies 8y = 168$

$\implies y = \frac{168}{8}$

$\implies y = 21$

Therefore.,

$\red{ Value \:of \: y }\green{ = 21 }$

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