What is the general solution for (sin(42))/(22)=(sin(B))/(12) ?

The general solution for (sin(42))/(22)=(sin(B))/(12) is B=0.37361…+360n,B=180-0.37361…+360n

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(sin(42))/(22)=(sin(B))/(12)

\frac{sin ( 4 2 ^{\circ} )}{22} = \frac{sin ( B )}{12}

B = 0.37361 \dots + 36 0^{\circ} n, B = 18 0^{\circ} - 0.37361 \dots + 36 0^{\circ} n

Radians

B = 0.37361 \dots + 2 πn, B = π - 0.37361 \dots + 2 πn

Solution steps

\frac{sin ( 4 2 ^{\circ} )}{22} = \frac{sin ( B )}{12}

Switch sides

\frac{sin ( B )}{12} = \frac{sin ( 4 2 ^{\circ} )}{22}

Multiply both sides by

12

\frac{sin ( B )}{12} = \frac{sin ( 4 2 ^{\circ} )}{22}

Multiply both sides by

12

\frac{12 sin ( B )}{12} = \frac{12 sin ( 4 2 ^{\circ} )}{22}

Simplify

sin (B) = \frac{6 sin ( 4 2 ^{\circ} )}{11}

sin (B) = \frac{6 sin ( 4 2 ^{\circ} )}{11}

Apply trig inverse properties

sin (B) = \frac{6 sin ( 4 2 ^{\circ} )}{11}

General solutions for

sin (B) = \frac{6 sin ( 4 2 ^{\circ} )}{11}

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

B = arcsin (\frac{6 sin ( 4 2 ^{\circ} )}{11}) + 36 0^{\circ} n, B = 18 0^{\circ} - arcsin (\frac{6 sin ( 4 2 ^{\circ} )}{11}) + 36 0^{\circ} n

B = arcsin (\frac{6 sin ( 4 2 ^{\circ} )}{11}) + 36 0^{\circ} n, B = 18 0^{\circ} - arcsin (\frac{6 sin ( 4 2 ^{\circ} )}{11}) + 36 0^{\circ} n

Show solutions in decimal form

B = 0.37361 \dots + 36 0^{\circ} n, B = 18 0^{\circ} - 0.37361 \dots + 36 0^{\circ} n

sin (θ) = \frac{16}{14}

tan (θ) = \frac{83}{47}

x, 2 sin (x) - cos (x) sin (x) = 0

5 = 5 sin (4 θ)

cos (x) = \frac{cos ( x )}{csc ( x )}

What is the general solution for (sin(42))/(22)=(sin(B))/(12) ?
The general solution for (sin(42))/(22)=(sin(B))/(12) is B=0.37361…+360n,B=180-0.37361…+360n