What is the general solution for (sin(C))/(18)=(sin(63))/(17) ?

The general solution for (sin(C))/(18)=(sin(63))/(17) is C=1.23279…+360n,C=180-1.23279…+360n

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(sin(C))/(18)=(sin(63))/(17)

\frac{sin ( C )}{18} = \frac{sin ( 6 3 ^{\circ} )}{17}

C = 1.23279 \dots + 36 0^{\circ} n, C = 18 0^{\circ} - 1.23279 \dots + 36 0^{\circ} n

Solution steps

\frac{sin ( C )}{18} = \frac{sin ( 6 3 ^{\circ} )}{17}

sin (6 3^{\circ}) = \frac{2 4 + 2 5 - 5}{4}

\frac{sin ( C )}{18} = \frac{\frac{2 4 + 2 5 - 5}{4}}{17}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{sin ( C )}{18} = \frac{2 4 + 2 5 - 5}{4 \cdot 17}

Multiply the numbers:

4 \cdot 17 = 68

\frac{sin ( C )}{18} = \frac{2 4 + 2 5 - 5}{68}

Multiply both sides by

18

sin (C) = \frac{9 2 2 5 - 5 + 4}{34}

Apply trig inverse properties

C = arcsin \frac{9 2 2 5 - 5 + 4}{34} + 36 0^{\circ} n, C = 18 0^{\circ} - arcsin \frac{9 2 2 5 - 5 + 4}{34} + 36 0^{\circ} n

Show solutions in decimal form

C = 1.23279 \dots + 36 0^{\circ} n, C = 18 0^{\circ} - 1.23279 \dots + 36 0^{\circ} n

cos (\frac{1}{2} x) = - 1

\frac{- ( 3 \cdot cos ( Q ^{1} ) )}{2} = 0

tan^{2} (x) + 3 tan (x) = 0, 0^{\circ} \leq x < 36 0^{\circ}

\frac{1}{2} (6.50941) + cos (θ) (10.40478) = 12

- 8 = - 4 + 4 tan (θ)

What is the general solution for (sin(C))/(18)=(sin(63))/(17) ?
The general solution for (sin(C))/(18)=(sin(63))/(17) is C=1.23279…+360n,C=180-1.23279…+360n