What is the general solution for (sin(79)}{12}=\frac{sin(B))/7 ?

The general solution for (sin(79)}{12}=\frac{sin(B))/7 is B=0.60969…+360n,B=180-0.60969…+360n

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(sin(79)}{12}=\frac{sin(B))/7

\frac{sin ( 7 9 ^{\circ} )}{12} = \frac{sin ( B )}{7}

B = 0.60969 \dots + 36 0^{\circ} n, B = 18 0^{\circ} - 0.60969 \dots + 36 0^{\circ} n

Radians

B = 0.60969 \dots + 2 πn, B = π - 0.60969 \dots + 2 πn

Solution steps

\frac{sin ( 7 9 ^{\circ} )}{12} = \frac{sin ( B )}{7}

Switch sides

\frac{sin ( B )}{7} = \frac{sin ( 7 9 ^{\circ} )}{12}

Multiply both sides by

7

\frac{sin ( B )}{7} = \frac{sin ( 7 9 ^{\circ} )}{12}

Multiply both sides by

7

\frac{7 sin ( B )}{7} = \frac{7 sin ( 7 9 ^{\circ} )}{12}

Simplify

sin (B) = \frac{7 sin ( 7 9 ^{\circ} )}{12}

sin (B) = \frac{7 sin ( 7 9 ^{\circ} )}{12}

Apply trig inverse properties

sin (B) = \frac{7 sin ( 7 9 ^{\circ} )}{12}

General solutions for

sin (B) = \frac{7 sin ( 7 9 ^{\circ} )}{12}

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

B = arcsin (\frac{7 sin ( 7 9 ^{\circ} )}{12}) + 36 0^{\circ} n, B = 18 0^{\circ} - arcsin (\frac{7 sin ( 7 9 ^{\circ} )}{12}) + 36 0^{\circ} n

B = arcsin (\frac{7 sin ( 7 9 ^{\circ} )}{12}) + 36 0^{\circ} n, B = 18 0^{\circ} - arcsin (\frac{7 sin ( 7 9 ^{\circ} )}{12}) + 36 0^{\circ} n

Show solutions in decimal form

B = 0.60969 \dots + 36 0^{\circ} n, B = 18 0^{\circ} - 0.60969 \dots + 36 0^{\circ} n

\frac{( sin ^{2} ( x ) )}{( 1 - cos ( x ) )} = 1.23

- 1 = tan (2 (0)^{3} + C)

sin (x) = \frac{1}{2}, 0 < x < \frac{π}{2}

sin (2 x) = cos (8 x)

8 = sin (x)

What is the general solution for (sin(79)}{12}=\frac{sin(B))/7 ?
The general solution for (sin(79)}{12}=\frac{sin(B))/7 is B=0.60969…+360n,B=180-0.60969…+360n