Sine Law - Demonstration and Examples

The sine law or sine theorem is a relationship applicable to any triangle (unlike the Pythagoras theorem which needs to be a right triangle), which relates the lengths of its sides to the sines of their respective opposite angles.

What is the law of sine?

Sine law

When observing the upper figure we can see that its vertices are A, B, C; its sides a, b, c and its angles α, β, γ. The law of sine states that the ratios of the relationship of each side between its opposite angle must be equal, that is:

\ LARGE \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta} = \ frac {c} {\ sin \ gamma}}

This relationship can only be applied in pairs, it does not apply in a group of three equalities:

  • \ large \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta}}
  • \ large \ mathbf {\ frac {b} {\ sin \ beta} = \ frac {c} {\ sin \ gamma}}
  • \ large \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {c} {\ sin \ gamma}}

For any triangle, using these relationships and depending on the data we have, we can find the values of angles or sides of that triangle.

Law of Sines or Sine formula

Yes in a triangle ABC, the measures of the sides opposite the angles AB y C are respectively abc, so:

\ LARGE \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta} = \ frac {c} {\ sin \ gamma}}

Example of the law of sines

If in a triangle ABC, ∠A = 90 °, ∠B = 30 °; If the side opposite ∠A is 60 cm, find the value of the side opposite ∠B and ∠C.

From the law of tangents we have:

\ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta} = \ frac {c} {\ sin \ gamma}}

B calculation

We take the first two reasons because we know A, B already:

\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta}

\ frac {60cm} {\ sin 90 ^ {\ circ}} = \ frac {b} {\ sin 30}

\ frac {60cm} {1} = \ frac {b} {\ frac {1} {2}}

60 cm = 2b

 Clearing b:

b = \ left (\ frac {60} {2} \ right) cm

\ mathbf {b = 30cm}

Calculation of c

To calculate this side we need to know the angle C; remembered that the sum of the interior angles of any triangle must be equal to 180 °, we have:

A + B + C = 180 ^ {\ circ}

\ begin {center} C = 180 ^ {\ circ} -AB = 180 ^ {\ circ} -90 ^ {\ circ} -30 ^ {\ circ} \\ \ medskip C = 60 ^ {\ circ} \ end {center}Now, in the law of sines we take two reasons where one of them contains a (c / sin C), therefore:

 

\ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {c} {\ sin \ gamma}}\ frac {60cm} {\ sin 90 ^ {\ circ}} = \ frac {c} {\ sin 60 ^ {\ circ}}

\ frac {60cm} {1} = \ frac {c} {\ frac {\ sqrt {3}} {2}}

60cm = \ frac {\ sqrt {3}} {2} c

 Clearing c:

c = \ frac {\ left (60 \ cdot 2 \ right)} {\ sqrt {3}} cm

c = 69.28cm

Demonstration

Consider a triangle whose lengths of its sides are a, b and c and their opposite angles α, β, γ.

sine theorem - proof

In order to prove the theorem, we must divide this triangle into two right triangles.

sine theorem - proof2

Let's see that the line that divides them is "h" and "AC" is the base of triangle ABC.

For the triangle on the left, we have:

\ begin {align *} & \ sin \ alpha = \ frac {h} {b} \\\ medskip & h = {b} \ cdot {\ sin \ alpha} \ hspace {2em} (1) \ end {align * }

Repeating the procedure in the one on the right:

\ begin {align *} & \ sin \ beta = \ frac {h} {a} \\\ medskip & h = {a} \ cdot {\ sin \ beta} \ hspace {2em} (2) \ end {align * }

Equating equation (1) and with (2), we have:

b \ sin \ alpha = a \ sin \ beta

or

\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta} \ hspace {2em} (3)

Similarly, we can find that

\ frac {a} {\ sin \ alpha} = \ frac {c} {\ sin \ gamma} \ hspace {2em} (4)

From (3) and (4)

\ LARGE \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta} = \ frac {c} {\ sin \ gamma}}

Which is what we wanted to demonstrate.

Exercises

  • In the attached figure we know: α = 30 °; β = 53 ° and = 75cm. Find b and c.

Sine law - exercise

The triangle is oblique and we know an angle and its opposite side, therefore we apply the sine law using the proportions that best fit the given data:

B calculation

\ begin {align *} & \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {b} {\ sin \ beta}} \\\ medskip & b = \ frac {a \ sin \ beta} { \ sin \ alpha} \ end {align *}

For α = 30 °; β = 53 ° and = 75cm, we have:

\ begin {align *} & b = \ frac {75 \ sin 53 ^ {\ circ}} {\ sin 30 ^ {\ circ}} cm \\\ medskip & b = \ frac {59,89766} {\ frac {1 } {2}} cm \\\ medskip & b = 119.79 cm \ end {align *}

Calculation of c

To calculate c we need to know the angle γ, which we calculate applying the property that in any triangle the sum of its internal angles is 180 °, therefore:

\ begin {align *} & \ alpha + \ beta + \ gamma = 180 ^ {\ circ} \\\ medskip & \ gamma = 180 ^ {\ circ} - \ alpha - \ beta \ end {align *}

For α = 30 ° and β = 53 °, we have:

\ gamma = 180 ^ {\ circ} -30 ^ {\ circ} -53 ^ {\ circ} = 97 ^ {\ circ}

Now:

\ begin {align *} & \ mathbf {\ frac {a} {\ sin \ alpha} = \ frac {c} {\ sin \ gamma}} \\\ medskip & c = \ frac {a \ sin \ gamma} { sin \ alpha} \ end {align *}

For a = 75cm; α = 30 ° and γ = 97 °:

\ begin {align *} & c = \ frac {75 \ sin 97 ^ {\ circ}} {sin 30 ^ {\ circ}} cm = \\\ medskip & c = \ frac {75 \ sin 97 ^ {\ circ} } {sin 30 ^ {\ circ}} cm = \\\ medskip & c = 148,88cm \ end {align *}

  • Calculate distance dc of the following figure using the law of sine:

Sine theorem - exercise

We draw the two figures separately:

sine law - exercise 2

From here we observe that:

  • Both triangles share the side bd, which is the hypotenuse of the first triangle.
  • If we calculate this side, in the second triangle we will be able to know an angle (30 °) and its opposite side (bd).
  • In addition, we can calculate the angle γ, since this is adjacent that of 60 °, which means that they are complementary and add up to 180 °.
  • Knowing the value of this angle, we can calculate α applying the property that in any triangle the sum of its internal angles is 180 °. This angle is the opposite of the dc side, which is the one we want to calculate.
  • In this way, in the second triangle we can apply the sine law using the proportions that best fit the data to calculate the distance dc.

Side calculation bd

From the previous figure, we see that the side bc is the hypotenuse of the first triangle, therefore we use the sine trigonometric relation to calculate it:

\ begin {align *} & \ sin 60 ^ {\ circ} = \ frac {ab} {bd} \\\ medskip & bd = \ frac {ab} {\ sin 60 ^ \ circ} = \ frac {300cm} { \ frac {\ sqrt {3}} {2}} = \\\ medskip & bd = \ frac {2 \ cdot 300cm} {\ sqrt {3}} = \ frac {600cm} {\ sqrt {3}} \ end {align *}

Now we rationalize the result:

\ begin {align *} & bd = \ frac {600m} {\ sqrt {3}} \ frac {\ sqrt {3}} {\ sqrt {3}} = \ frac {600 \ cdot \ sqrt {3}} { 3} = \\\ medskip & bd = 200 \ sqrt {3} cm \ end {align *}

Angle calculation γ (gamma)

From the figure we observe that γ and the angle of 60 ° are adjacent. Since two adjacent angles are complementary, they both add up to 180 °. Thus:

\ begin {align *} & \ gamma +60 ^ {\ circ} = 180 ^ {\ circ} \\\ medskip & \ gamma = 180 ^ {\ circ} -60 ^ {\ circ} \\\ medskip & \ gamma = 120 ^ {\ circ} \ end {align *}

Calculation of angle α (alpha)

From the second triangle we know its angles β and γ; knowing that in any triangle the sum of its internal angles is worth 180 ° we have:

\ begin {align *} & \ alpha + \ beta + \ gamma = 180 ^ {\ circ} \\\ medskip & \ alpha = 180 ^ {\ circ} - \ beta- \ gamma \ end {align *}

For β = 30 ° and γ = 120 °, we have:

\ begin {align *} & \ alpha = 180 ^ {\ circ} -30 ^ {\ circ} -120 ^ {\ circ} \\\ medskip & \ alpha = 30 ^ {\ circ} \ end {align *}

DC side calculation

With the data we have calculated so far, our second triangle would be:

We now apply the sine theorem, taking into account that a = bd and b = dc:

\ begin {align *} & {\ frac {bd} {\ sin \ alpha} = \ frac {dc} {\ sin \ beta}} \\\ medskip & dc = \ frac {bd \ cdot \ sin \ beta} { \ sin \ alpha} \ end {align *}

For α = 30 °; β = 30 ° and bd = 200√3 cm, we have:

\ begin {align *} & dc = \ frac {200 \ sqrt {3} \ cdot \ sin30 ^ \ circ} {\ sin30 ^ \ circ} \\\ medskip & dc = 200 \ sqrt {3} \ end {align *}

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