What is the general solution for (sin(75))/(101)=(sin(b))/(65) ?

The general solution for (sin(75))/(101)=(sin(b))/(65) is b=0.67082…+360n,b=180-0.67082…+360n

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(sin(75))/(101)=(sin(b))/(65)

\frac{sin ( 7 5 ^{\circ} )}{101} = \frac{sin ( b )}{65}

b = 0.67082 \dots + 36 0^{\circ} n, b = 18 0^{\circ} - 0.67082 \dots + 36 0^{\circ} n

Solution steps

\frac{sin ( 7 5 ^{\circ} )}{101} = \frac{sin ( b )}{65}

sin (7 5^{\circ}) = \frac{6 + 2}{4}

\frac{\frac{6 + 2}{4}}{101} = \frac{sin ( b )}{65}

Switch sides

\frac{sin ( b )}{65} = \frac{\frac{6 + 2}{4}}{101}

Apply the fraction rule:

\frac{\frac{b}{c}}{a} = \frac{b}{c \cdot a}

\frac{sin ( b )}{65} = \frac{6 + 2}{4 \cdot 101}

Multiply the numbers:

4 \cdot 101 = 404

\frac{sin ( b )}{65} = \frac{6 + 2}{404}

Multiply both sides by

65

sin (b) = \frac{65 ( 6 + 2 )}{404}

Apply trig inverse properties

b = arcsin (\frac{65 ( 6 + 2 )}{404}) + 36 0^{\circ} n, b = 18 0^{\circ} - arcsin (\frac{65 ( 6 + 2 )}{404}) + 36 0^{\circ} n

Show solutions in decimal form

b = 0.67082 \dots + 36 0^{\circ} n, b = 18 0^{\circ} - 0.67082 \dots + 36 0^{\circ} n

3 tan (A) + 5 = 0

tan (2 x) = - \frac{( \frac{30 + 20}{2} )}{80}

15 sin^{2} (x) - 22 sin (x) + 8 = 0

1.333 \cdot sin (4 8^{\circ}) = sin (x)

x, y = cos (x^{2})

What is the general solution for (sin(75))/(101)=(sin(b))/(65) ?
The general solution for (sin(75))/(101)=(sin(b))/(65) is b=0.67082…+360n,b=180-0.67082…+360n