What is the value of $y$ $y$ in $\frac{sin 99}{5 y - 18} = \frac{sin 49}{3 y + 2.7}$ $sin99/(5y-18) = sin49/(3y+2.7)$ ?

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Answer 1 · 2017-12-17T11:42:35

$\Rightarrow y = - \frac{18 sin 49 + 2.7 sin 99}{3 sin 99 - 5 sin 49}$ $=> y = - (18sin49 + 2.7sin99)/(3sin99 - 5sin49 )$

Multiplying by $5 y - 18$ $5y - 18$ :

$sin 99 = \frac{sin 49 (5 y - 18)}{3 y + 2.7}$ $sin99 = (sin49 ( 5y - 18))/(3y+2.7)$

Multiplying by $3 y + 2.7$ $3y + 2.7$

$\Rightarrow sin 99 (3 y + 2.7) = sin 49 (5 y - 18)$ $=> sin99 (3y + 2.7 ) = sin49( 5y-18)$

Expanding:

$\Rightarrow 3 sin 99 y + 2.7 sin 99 = 5 sin 49 y - 18 sin 49$ $=> 3sin99 y + 2.7sin99 = 5sin49 y - 18sin49$

Rearanging:

$\Rightarrow y (3 sin 99 - 5 sin 49) = - 18 sin 49 - 2.7 sin 99$ $=> y ( 3sin99 -5sin49) = -18sin49 - 2.7sin99$

$\Rightarrow y = - \frac{18 sin 49 + 2.7 sin 99}{3 sin 99 - 5 sin 49}$ $=> y = - (18sin49 + 2.7sin99)/(3sin99 - 5sin49 )$

What is the value of y y in sin995y−18=sin493y+2.7sin99/(5y-18) = sin49/(3y+2.7) ?