What is the general solution for (sin(115))/(20)=(sin(B))/(15) ?

The general solution for (sin(115))/(20)=(sin(B))/(15) is B=0.74739…+360n,B=180-0.74739…+360n

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(sin(115))/(20)=(sin(B))/(15)

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( B )}{15}

B = 0.74739 \dots + 36 0^{\circ} n, B = 18 0^{\circ} - 0.74739 \dots + 36 0^{\circ} n

Solution steps

\frac{sin ( 11 5 ^{\circ} )}{20} = \frac{sin ( B )}{15}

Switch sides

\frac{sin ( B )}{15} = \frac{sin ( 11 5 ^{\circ} )}{20}

Multiply both sides by

10

\frac{2 sin ( B )}{3} = \frac{sin ( 11 5 ^{\circ} )}{2}

Multiply both sides by

3

2 sin (B) = \frac{3 sin ( 11 5 ^{\circ} )}{2}

Divide both sides by

2

sin (B) = \frac{3 sin ( 11 5 ^{\circ} )}{4}

Apply trig inverse properties

B = arcsin (\frac{3 sin ( 11 5 ^{\circ} )}{4}) + 36 0^{\circ} n, B = 18 0^{\circ} - arcsin (\frac{3 sin ( 11 5 ^{\circ} )}{4}) + 36 0^{\circ} n

Show solutions in decimal form

B = 0.74739 \dots + 36 0^{\circ} n, B = 18 0^{\circ} - 0.74739 \dots + 36 0^{\circ} n

3 tan (θ) + 1 = 2 tan (θ)

cos (x) = \frac{18}{25}

sin (x) = - \frac{2}{2}, - π \leq x \leq π

w, s (t) = A e^{- c t} cos (wt + b)

0 = 2 sin (t) + cos (2 t)

What is the general solution for (sin(115))/(20)=(sin(B))/(15) ?
The general solution for (sin(115))/(20)=(sin(B))/(15) is B=0.74739…+360n,B=180-0.74739…+360n