What is the general solution for (sin(b))/(13.1)=(sin(112.1))/(24.1) ?

The general solution for (sin(b))/(13.1)=(sin(112.1))/(24.1) is b=0.52779…+360n,b=180-0.52779…+360n

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(sin(b))/(13.1)=(sin(112.1))/(24.1)

\frac{sin ( b )}{13.1} = \frac{sin ( 112. 1 ^{\circ} )}{24.1}

b = 0.52779 \dots + 36 0^{\circ} n, b = 18 0^{\circ} - 0.52779 \dots + 36 0^{\circ} n

Radians

b = 0.52779 \dots + 2 πn, b = π - 0.52779 \dots + 2 πn

Solution steps

\frac{sin ( b )}{13.1} = \frac{sin ( 112. 1 ^{\circ} )}{24.1}

Multiply both sides by

13.1

\frac{sin ( b )}{13.1} = \frac{sin ( 112. 1 ^{\circ} )}{24.1}

Multiply both sides by

13.1

\frac{13.1 sin ( b )}{13.1} = \frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}

Simplify

sin (b) = \frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}

sin (b) = \frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}

Apply trig inverse properties

sin (b) = \frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}

General solutions for

sin (b) = \frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}

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

b = arcsin (\frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}) + 36 0^{\circ} n, b = 18 0^{\circ} - arcsin (\frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}) + 36 0^{\circ} n

b = arcsin (\frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}) + 36 0^{\circ} n, b = 18 0^{\circ} - arcsin (\frac{13.1 sin ( 112. 1 ^{\circ} )}{24.1}) + 36 0^{\circ} n

Show solutions in decimal form

b = 0.52779 \dots + 36 0^{\circ} n, b = 18 0^{\circ} - 0.52779 \dots + 36 0^{\circ} n

tan (α) = \frac{0.006}{π \cdot 0.027}

sec^{2} (θ) tan (θ) = 2 tan (θ)

tan^{5} (x) + tan^{3} (x) = 0

0 = cos (x) (1 - cos (x)) + sin^{2} (x)

sin (2 x) = 3 cos^{2} (x)

What is the general solution for (sin(b))/(13.1)=(sin(112.1))/(24.1) ?
The general solution for (sin(b))/(13.1)=(sin(112.1))/(24.1) is b=0.52779…+360n,b=180-0.52779…+360n