What is the general solution for (sin(118))/(18.03)=(sin(b))/(17) ?

The general solution for (sin(118))/(18.03)=(sin(b))/(17) is No Solution for b\in\mathbb{R}

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(sin(118))/(18.03)=(sin(b))/(17)

\frac{sin ( 11 8 ^{\circ} )}{18.03} = \frac{sin ( b ^{\circ} )}{17}

No Solution for b \in R

Solution steps

\frac{sin ( 11 8 ^{\circ} )}{18.03} = \frac{sin ( b ^{\circ} )}{17}

Switch sides

\frac{sin ( b ^{\circ} )}{17} = \frac{sin ( 11 8 ^{\circ} )}{18.03}

Multiply both sides by

17

\frac{sin ( b ^{\circ} )}{17} = \frac{sin ( 11 8 ^{\circ} )}{18.03}

Multiply both sides by

17

\frac{17 sin ( b ^{\circ} )}{17} = \frac{17 sin ( 11 8 ^{\circ} )}{18.03}

Simplify

sin (b^{\circ}) = \frac{17 sin ( 11 8 ^{\circ} )}{18.03}

sin (b^{\circ}) = \frac{17 sin ( 11 8 ^{\circ} )}{18.03}

Apply trig inverse properties

sin (b^{\circ}) = \frac{17 sin ( 11 8 ^{\circ} )}{18.03}

General solutions for

sin (b^{\circ}) = \frac{17 sin ( 11 8 ^{\circ} )}{18.03}

sin (x) = a \Rightarrow x = arcsin (a) + 36 0^{\circ} n, x = 18 0^{\circ} - arcsin (a) + 36 0^{\circ} n

b^{\circ} = arcsin (\frac{17 sin ( 11 8 ^{\circ} )}{18.03}) + 36 0^{\circ} n, b^{\circ} = 18 0^{\circ} - arcsin (\frac{17 sin ( 11 8 ^{\circ} )}{18.03}) + 36 0^{\circ} n

b^{\circ} = arcsin (\frac{17 sin ( 11 8 ^{\circ} )}{18.03}) + 36 0^{\circ} n, b^{\circ} = 18 0^{\circ} - arcsin (\frac{17 sin ( 11 8 ^{\circ} )}{18.03}) + 36 0^{\circ} n

No Solution for b \in R

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What is the general solution for (sin(118))/(18.03)=(sin(b))/(17) ?
The general solution for (sin(118))/(18.03)=(sin(b))/(17) is No Solution for b\in\mathbb{R}