Law of Sines

From Math Images

(Difference between revisions)
Jump to: navigation, search
Current revision (12:14, 20 June 2012) (edit) (undo)
 
Line 267: Line 267:
</div>
</div>
-
________________________________________________________________
+
<hr>
If the above problem asked to find the radius of the circumcircle of <math>\vartriangle ABC</math>, the law of sines could help to find the diameter.
If the above problem asked to find the radius of the circumcircle of <math>\vartriangle ABC</math>, the law of sines could help to find the diameter.

Current revision


Law of Sines
Field: Geometry
Image Created By: Richard Scott

Law of Sines

The law of sines is a tool commonly used to help solve arbitrary triangles. It is a formula that relates the sine of a given angle to its opposite side length.


Contents

Basic Description

In any triangle, there is a relationship between the measures of the angles and the lengths of the sides: the largest angle is opposite the longest side, the second-largest angle is opposite the second-longest side, and the smallest angle is opposite the shortest side.

The law of sines is an equation that more precisely expresses this relationship between the angles of a triangle and the length of their opposite sides. The law of sines states that the ratio between a length of one side of a triangle and the sine of its opposite angle is equal for all three sides. Specifically:

Given a triangle with side lengths a, b, c and opposite angles A, B, C,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}


The law of sines is used to find all of the lengths of the sides and the angle measures for an arbitrary triangle given only some of this information. This process is called solving a triangle. To use the law of sines in solving triangles, at least three elements of a triangle must be known. Whenever a side length and two angles are given, the law of sines can be used to solve the triangle.

In some cases, the law of sines can provide multiple solutions to a triangle. If two adjacent side lengths are given with one of the opposite angles, the law of sines cannot definitively determine the triangle, but instead offers zero, one, or two possible solutions in what is known as the ambiguous case.

The law of sines does not help with solving a triangle in several cases. With two known side lengths and the measure of the angle between, there is no way to use the law of sines to solve the triangle because no pair of opposite angle measure and side length is provided. The law of sines by itself is also not able to provide solutions when three side lengths are provided. Instead, the law of cosines is often used for solving triangles in these cases.

A More Mathematical Explanation

Two Derivations

There are at least two different ways to derive the law of sines: using the area [...]

Two Derivations

There are at least two different ways to derive the law of sines: using the area formula and using the definition of sine.

Using Area

The formula for area of a triangle uses the lengths of the base and height. By using these lengths and the angle measures of a triangle, we can derive the law of sines.

A triangle can be oriented so that any one side can be used as the base. Depending on which side is chosen as the base of the triangle, the height may be different. Let h_{a} be the height when the side of length a is the base. When a is the base, h_{a} is the distance from a vertex to the opposite side, such that h_{a} is perpendicular to the side. When b is oriented as the base of the triangle, h_{b} runs perpendicular to side b and is the distance from side b to the vertex B.

First, we must determine the height of the triangle for each orientation of the base.

When b is oriented as the base,

Image:Heightb.jpg


\sin A = \frac{h_{b}}{c}


h_{b} = c \sin A

When a is oriented as the base,

Image:Heighta.jpg


\sin B = \frac{h_{a}}{c}


h_{a} = c \sin B


In any triangle,


\text{Area} = \frac{\text{base} \times \text{height}}{2}


Since the area of the triangle is the same no matter how the triangle is oriented, the area of the triangle with b as the base is the same as the area of the triangle with a as the base.

\text{Area}_{\text{base} = b} = \text{Area}_{\text{base} = a}


Substituting the formula for the area of a triangle,

\frac{b h_{b}}{2} = \frac{a h_{a}}{2}


Both h_{b} and  h_{a} can be written in terms of side lengths a, b, c and angles A, B, C as shown in the "More on Height" section. Therefore, we can substitute {(c\sin A)} for h_{b} and {(c\sin B)} for  h_{a} , giving us

\frac{b(c\sin A)}{2} = \frac{a(c\sin B)}{2}


Multiplying both sides by 2 and dividing by c gives us

b \sin A = a \sin B


Then, rearranging once more gives us our equation in its most common form,

\frac{a}{\sin A} = \frac{b}{\sin B}


Since we can orient the base differently and go through the same process with other variables, we know that \frac{b}{\sin B} = \frac{c}{\sin C}, so

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

which is the law of sines.


Using the Definition of Sine

We know that, in a right triangle,

\sin A =\frac{\text{opposite}}{\text{hypotenuse}}


Letting h represent height and a, b represent the lengths of the sides opposite A, B, respectively, plug in the appropriate measures to solve for  \sin A, \sin B.

\sin A = \frac{h}{b}


Clearing the fractions,

b \sin A = h


\sin B = \frac{h}{a}


Clearing the fractions,

a \sin B = h

Image:Derivation_triangle.jpg

Set both equations for  h equal to each other to get

b \sin A = a \sin B


Divide both sides by  \sin A, \sin B for

\frac{a}{\sin A} = \frac{b}{\sin B}


Since we can go through the same process using a different angle and different variables, we know that \frac{b}{\sin B} = \frac{c}{\sin C}, so

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}


A Geometric Extension

For every triangle, there is some circle for which the vertices of the triangle lay on the circumference. This triangle is known as an inscribed triangle, and the circle is known as the circumcircle or circumscribed circle.

By the extended law of sines,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r


where r is the radius of the circumcircle.


Proof

Let there be two inscribed triangles on a circle of radius r. Let \vartriangle ABD be a triangle that has a hypotenuse that goes through the center of the circle. Let \vartriangle ABC be an oblique triangle that shares \overline{AB} with \vartriangle ABD.

Image:Geometric proof 1.jpg

For \vartriangle ABD,


\sin D = \frac{\overline{AB}}{ \ \overline{AD} \ }


Angle C is equal to angle D because they are both inscribed angles that cut the same arc. According to properties of inscribed angles, two inscribed angles that cut the same arc in circles of the came radius are equal. Since  \angle{C} and  \angle{D} are the same, so are \sin C and \sin D.

Substituting \sin C for \sin D gives us


\sin C = \frac{\overline{AB}}{ \ \overline{AD} \ }


Solving for \overline{AD} gives us


\overline{AD} = \frac{\overline{AB}}{\sin C}


Since \overline{AD} is the diameter, \overline{AD} = 2r


2r = \frac{\overline{AB}}{\sin C}


or equivalently,


2r = \frac{c}{\sin C}


since c is the length of the side opposite \angle{C}.

By the law of sines, we know that

\frac{c}{\sin C}= \frac{a}{\sin A} = \frac{b}{\sin B}

and therefore by the transitive property,

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r




Example Problem

Solve the triangle. Find all of its parts, \vartriangle ABC, given b = 10, A = 60^\circ,B = 30^\circ.

Image:Law of Sines Example_1.jpg


Solution

Since all of the angle measures in a triangle add up to 180^\circ, C = 90^\circ


\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}


Image:Example_1_Solution.jpg

\frac{a}{\sin 60^\circ} = \frac{10}{\sin 30^\circ}


Cross-multiplying gives us

a \sin 30^\circ = (10) \sin 60^\circ


Since  \sin 30^\circ= \frac{1}{2} and  \sin 60^\circ= \frac{\sqrt{3}}{2},

a \left( \frac{1}{2} \right) = (10) \left( \frac{\sqrt{3}}{2} \right)


a = 10 \sqrt{3}


\frac{c}{\sin 90^\circ} = \frac{10}{\sin 30^\circ}


Cross-multiplying gives us

c \sin 30^\circ = (10) \sin 90^\circ


Since  \sin 30^\circ= \frac{1}{2} and  \sin 90^\circ= 1,

c \left( \frac{1}{2} \right) = (10) (1)



c = 20

If the above problem asked to find the radius of the circumcircle of \vartriangle ABC, the law of sines could help to find the diameter.

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C} = 2r


Image:Example 1 circle solution.jpg

\frac{b}{\sin B} = 2r


Substituting the values for  b, B,

\frac{10}{\sin 30^\circ} = 2r


Since  \sin 30^\circ= \frac{1}{2},

 \cfrac{10}{\left( \frac{1}{2} \right)} = 2r


 20 = 2r
 10 = r




Teaching Materials

There are currently no teaching materials for this page. Add teaching materials.




References

All images were made by the page's author using Adobe Photoshop and Cinderella2.





If you are able, please consider adding to or editing this page!

Have questions about the image or the explanations on this page?
Leave a message on the discussion page by clicking the 'discussion' tab at the top of this image page.






Personal tools