What is the general solution for (sin(θ))/(140)=(sin(110))/(622) ?

The general solution for (sin(θ))/(140)=(sin(110))/(622) is θ=0.21311…+360n,θ=180-0.21311…+360n

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(sin(θ))/(140)=(sin(110))/(622)

\frac{sin ( θ )}{140} = \frac{sin ( 11 0 ^{\circ} )}{622}

θ = 0.21311 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.21311 \dots + 36 0^{\circ} n

Radians

θ = 0.21311 \dots + 2 πn, θ = π - 0.21311 \dots + 2 πn

Solution steps

\frac{sin ( θ )}{140} = \frac{sin ( 11 0 ^{\circ} )}{622}

Multiply both sides by

10

\frac{sin ( θ )}{140} = \frac{sin ( 11 0 ^{\circ} )}{622}

Multiply both sides by

10

\frac{sin ( θ )}{140} \cdot 10 = \frac{sin ( 11 0 ^{\circ} )}{622} \cdot 10

Simplify

\frac{sin ( θ )}{140} \cdot 10 = \frac{sin ( 11 0 ^{\circ} )}{622} \cdot 10

Simplify

\frac{sin ( θ )}{140} \cdot 10 : \frac{sin ( θ )}{14}

\frac{sin ( θ )}{140} \cdot 10

Multiply fractions:

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

= \frac{sin ( θ ) \cdot 10}{140}

Cancel the common factor:

10

= \frac{sin ( θ )}{14}

Simplify

\frac{sin ( 11 0 ^{\circ} )}{622} \cdot 10 : \frac{5 sin ( 11 0 ^{\circ} )}{311}

\frac{sin ( 11 0 ^{\circ} )}{622} \cdot 10

Multiply fractions:

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

= \frac{sin ( 11 0 ^{\circ} ) \cdot 10}{622}

Cancel the common factor:

2

= \frac{5 sin ( 11 0 ^{\circ} )}{311}

\frac{sin ( θ )}{14} = \frac{5 sin ( 11 0 ^{\circ} )}{311}

\frac{sin ( θ )}{14} = \frac{5 sin ( 11 0 ^{\circ} )}{311}

\frac{sin ( θ )}{14} = \frac{5 sin ( 11 0 ^{\circ} )}{311}

Multiply both sides by

14

\frac{sin ( θ )}{14} = \frac{5 sin ( 11 0 ^{\circ} )}{311}

Multiply both sides by

14

\frac{14 sin ( θ )}{14} = \frac{14 \cdot 5 sin ( 11 0 ^{\circ} )}{311}

Simplify

\frac{14 sin ( θ )}{14} = \frac{14 \cdot 5 sin ( 11 0 ^{\circ} )}{311}

Simplify

\frac{14 sin ( θ )}{14} : sin (θ)

\frac{14 sin ( θ )}{14}

Divide the numbers:

\frac{14}{14} = 1

= sin (θ)

Simplify

\frac{14 \cdot 5 sin ( 11 0 ^{\circ} )}{311} : \frac{70 sin ( 11 0 ^{\circ} )}{311}

\frac{14 \cdot 5 sin ( 11 0 ^{\circ} )}{311}

Multiply the numbers:

14 \cdot 5 = 70

= \frac{70 sin ( 11 0 ^{\circ} )}{311}

sin (θ) = \frac{70 sin ( 11 0 ^{\circ} )}{311}

sin (θ) = \frac{70 sin ( 11 0 ^{\circ} )}{311}

sin (θ) = \frac{70 sin ( 11 0 ^{\circ} )}{311}

Apply trig inverse properties

sin (θ) = \frac{70 sin ( 11 0 ^{\circ} )}{311}

General solutions for

sin (θ) = \frac{70 sin ( 11 0 ^{\circ} )}{311}

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

θ = arcsin (\frac{70 sin ( 11 0 ^{\circ} )}{311}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{70 sin ( 11 0 ^{\circ} )}{311}) + 36 0^{\circ} n

θ = arcsin (\frac{70 sin ( 11 0 ^{\circ} )}{311}) + 36 0^{\circ} n, θ = 18 0^{\circ} - arcsin (\frac{70 sin ( 11 0 ^{\circ} )}{311}) + 36 0^{\circ} n

Show solutions in decimal form

θ = 0.21311 \dots + 36 0^{\circ} n, θ = 18 0^{\circ} - 0.21311 \dots + 36 0^{\circ} n

What is the general solution for (sin(θ))/(140)=(sin(110))/(622) ?
The general solution for (sin(θ))/(140)=(sin(110))/(622) is θ=0.21311…+360n,θ=180-0.21311…+360n