Solve for B sin(B)=(53(sin(29.2)))/27

$\sin (B) = \frac{53 (\sin (29.2))}{27}$

Take the inverse sine of both sides of the equation to extract $B$ from inside the sine.

$B = \arcsin (0.95765044)$

Evaluate $\arcsin (0.95765044)$ .

$B = 73.26572916$

The sine function is positive in the first and second quadrants. To find the second solution, subtract the reference angle from $180$ to find the solution in the second quadrant.

$B = 180 - 73.26572916$

Subtract $73.26572916$ from $180$ .

$B = 106.73427083$

Find the period.

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The period of the function can be calculated using $\frac{360}{| b |}$ .

$\frac{360}{| b |}$

Replace $b$ with $1$ in the formula for period.

$\frac{360}{| 1 |}$

Solve the equation.

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The absolute value is the distance between a number and zero. The distance between $0$ and $1$ is $1$ .

$\frac{360}{1}$

Divide $360$ by $1$ .

$360$

The period of the $\sin (B)$ function is $360$ so values will repeat every $360$ degrees in both directions.

$B = 73.26572916 + 360 n, 106.73427083 + 360 n$ , for any integer $n$

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