What is the value of 40*(sin(135))/(sin(105)) ?

The value of 40*(sin(135))/(sin(105)) is 40(sqrt(3)-1)

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40*(sin(135))/(sin(105))

Solution

40 \cdot \frac{sin ( 13 5 ^{\circ} )}{sin ( 10 5 ^{\circ} )}

Solution

40 (3 - 1)

Decimal Notation

29.28203 \dots

Solution steps

40 \cdot \frac{sin ( 13 5 ^{\circ} )}{sin ( 10 5 ^{\circ} )}

Use the following trivial identity:

sin (13 5^{\circ}) = \frac{2}{2}

sin (13 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}

Rewrite using trig identities:

sin (10 5^{\circ}) = \frac{2 ( 3 + 1 )}{4}

sin (10 5^{\circ})

Rewrite using trig identities:

sin (6 0^{\circ}) cos (4 5^{\circ}) + cos (6 0^{\circ}) sin (4 5^{\circ})

sin (10 5^{\circ})

Write

sin (10 5^{\circ})

sin (6 0^{\circ} + 4 5^{\circ})

= sin (6 0^{\circ} + 4 5^{\circ})

Use the Angle Sum identity:

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

= sin (6 0^{\circ}) cos (4 5^{\circ}) + cos (6 0^{\circ}) sin (4 5^{\circ})

= sin (6 0^{\circ}) cos (4 5^{\circ}) + cos (6 0^{\circ}) sin (4 5^{\circ})

Use the following trivial identity:

sin (6 0^{\circ}) = \frac{3}{2}

sin (6 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{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:

cos (6 0^{\circ}) = \frac{1}{2}

cos (6 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{1}{2}

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}

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

Simplify

\frac{3}{2} \cdot \frac{2}{2} + \frac{1}{2} \cdot \frac{2}{2} : \frac{2 ( 3 + 1 )}{4}

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

Factor out common term

\frac{2}{2}

= \frac{2}{2} (\frac{3}{2} + \frac{1}{2})

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

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

Apply rule

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

= \frac{3 + 1}{2}

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

Multiply fractions:

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

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

Multiply the numbers:

2 \cdot 2 = 4

= \frac{2 ( 1 + 3 )}{4}

= \frac{2 ( 3 + 1 )}{4}

= 40 \cdot \frac{\frac{2}{2}}{\frac{2 ( 3 + 1 )}{4}}

Simplify

40 \cdot \frac{\frac{2}{2}}{\frac{2 ( 3 + 1 )}{4}} : 40 (3 - 1)

40 \cdot \frac{\frac{2}{2}}{\frac{2 ( 3 + 1 )}{4}}

\frac{\frac{2}{2}}{\frac{2 ( 3 + 1 )}{4}} = \frac{2}{3 + 1}

\frac{\frac{2}{2}}{\frac{2 ( 3 + 1 )}{4}}

Divide fractions:

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

= \frac{2 \cdot 4}{2 2 ( 3 + 1 )}

Cancel the common factor:

2

= \frac{4}{2 ( 3 + 1 )}

Divide the numbers:

\frac{4}{2} = 2

= \frac{2}{( 3 + 1 )}

Remove parentheses:

(a) = a

= \frac{2}{3 + 1}

= 40 \cdot \frac{2}{1 + 3}

Multiply fractions:

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

= \frac{2 \cdot 40}{3 + 1}

Multiply the numbers:

2 \cdot 40 = 80

= \frac{80}{3 + 1}

Rationalize

\frac{80}{3 + 1} : 40 (3 - 1)

\frac{80}{3 + 1}

Multiply by the conjugate

\frac{3 - 1}{3 - 1}

= \frac{80 ( 3 - 1 )}{( 3 + 1 ) ( 3 - 1 )}

(3 + 1) (3 - 1) = 2

(3 + 1) (3 - 1)

Apply Difference of Two Squares Formula:

(a + b) (a - b) = a^{2} - b^{2}

a = 3, b = 1

= (3)^{2} - 1^{2}

Simplify

(3)^{2} - 1^{2} : 2

(3)^{2} - 1^{2}

Apply rule

1^{a} = 1

1^{2} = 1

= (3)^{2} - 1

(3)^{2} = 3

(3)^{2}

Apply radical rule:

a = a^{\frac{1}{2}}

= (3^{\frac{1}{2}})^{2}

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

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

\frac{1}{2} \cdot 2

Multiply fractions:

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

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

Cancel the common factor:

2

= 1

= 3

= 3 - 1

Subtract the numbers:

3 - 1 = 2

= 2

= 2

= \frac{80 ( 3 - 1 )}{2}

Divide the numbers:

\frac{80}{2} = 40

= 40 (3 - 1)

= 40 (3 - 1)

= 40 (3 - 1)

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Frequently Asked Questions (FAQ)

What is the value of 40*(sin(135))/(sin(105)) ?
The value of 40*(sin(135))/(sin(105)) is 40(sqrt(3)-1)