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)

Solution

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

Solution

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

Radians

b = 0.67082 \dots + 2 πn, b = π - 0.67082 \dots + 2 πn

Solution steps

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

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

sin (7 5^{\circ})

Rewrite using trig identities:

sin (4 5^{\circ}) cos (3 0^{\circ}) + cos (4 5^{\circ}) sin (3 0^{\circ})

sin (7 5^{\circ})

Write

sin (7 5^{\circ})

sin (4 5^{\circ} + 3 0^{\circ})

= sin (4 5^{\circ} + 3 0^{\circ})

Use the Angle Sum identity:

sin (s + t) = sin (s) cos (t) + cos (s) sin (t)

= sin (4 5^{\circ}) cos (3 0^{\circ}) + cos (4 5^{\circ}) sin (3 0^{\circ})

= sin (4 5^{\circ}) cos (3 0^{\circ}) + cos (4 5^{\circ}) sin (3 0^{\circ})

Use the following trivial identity:

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

sin (4 5^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{2}{2}

Use the following trivial identity:

cos (3 0^{\circ}) = \frac{3}{2}

cos (3 0^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{3}{2}

Use the following trivial identity:

cos (4 5^{\circ}) = \frac{2}{2}

cos (4 5^{\circ})

cos (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} cos (x) 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} cos (x) - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2} 0 \frac{1}{2} \frac{2}{2} \frac{3}{2}

= \frac{2}{2}

Use the following trivial identity:

sin (3 0^{\circ}) = \frac{1}{2}

sin (3 0^{\circ})

sin (x)

periodicity table with

36 0^{\circ} n

cycle:

x 0 3 0^{\circ} 4 5^{\circ} 6 0^{\circ} 9 0^{\circ} 12 0^{\circ} 13 5^{\circ} 15 0^{\circ} sin (x) 0 \frac{1}{2} \frac{2}{2} \frac{3}{2} 1 \frac{3}{2} \frac{2}{2} \frac{1}{2} x 18 0^{\circ} 21 0^{\circ} 22 5^{\circ} 24 0^{\circ} 27 0^{\circ} 30 0^{\circ} 31 5^{\circ} 33 0^{\circ} sin (x) 0 - \frac{1}{2} - \frac{2}{2} - \frac{3}{2} - 1 - \frac{3}{2} - \frac{2}{2} - \frac{1}{2}

= \frac{1}{2}

= \frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2}

Simplify

\frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2} : \frac{6 + 2}{4}

\frac{2}{2} \cdot \frac{3}{2} + \frac{2}{2} \cdot \frac{1}{2}

\frac{2}{2} \cdot \frac{3}{2} = \frac{6}{4}

\frac{2}{2} \cdot \frac{3}{2}

Multiply fractions:

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

= \frac{2 3}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 3}{4}

Simplify

23 : 6

23

Apply radical rule:

a b = a \cdot b

23 = 2 \cdot 3

= 2 \cdot 3

Multiply the numbers:

2 \cdot 3 = 6

= 6

= \frac{6}{4}

\frac{2}{2} \cdot \frac{1}{2} = \frac{2}{4}

\frac{2}{2} \cdot \frac{1}{2}

Multiply fractions:

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

= \frac{2 \cdot 1}{2 \cdot 2}

Multiply:

2 \cdot 1 = 2

= \frac{2}{2 \cdot 2}

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2}{4}

= \frac{6}{4} + \frac{2}{4}

Apply rule

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

= \frac{6 + 2}{4}

= \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

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

Multiply both sides by

65

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

Simplify

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

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

Apply trig inverse properties

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

General solutions for

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

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

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

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

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Frequently Asked Questions (FAQ)

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