Law Of Sines (example 02)

In trigonometry, the law of sines, sine law, sine formula, or sine rule is an equation relating the lengths of the sides of any triangle to the sines of its angles. According to the law,
$\frac \,=\, \frac \,=\, \frac \,=\, 2R,$
where , and are the lengths of the sides of a triangle, and , and are the opposite angles (see figure 2), while is the radius of the triangle's circumcircle. When the last part of the equation is not used, the law is sometimes stated using the reciprocals;
$\frac \,=\, \frac \,=\, \frac.$
The law of sines can be used to compute the remaining sides of a triangle when two angles and a side are known—a technique known as triangulation. It can also be used when two sides and one of the non-enclosed angles are known. In some such cases, the triangle is not uniquely determined by this data (called the ''ambiguous case'') and the technique gives two possible values for the enclosed angle.

The law of sines is one of two trigonometric equations commonly applied to find lengths and angles in scalene triangles, with the other being the law of cosines.

The law of sines can be generalized to higher dimensions on surfaces with constant curvature.

History

According to Ubiratàn D'Ambrosio and Helaine Selin, the spherical law of sines was discovered in the 10th century. It is variously attributed to Abu-Mahmud Khojandi, Abu al-Wafa' Buzjani, Nasir al-Din al-Tusi and Abu Nasr Mansur.Sesiano just lists al-Wafa as a contributor. Sesiano, Jacques (2000) "Islamic mathematics" pp. 137–157, in

Ibn Muʿādh al-Jayyānī's ''The book of unknown arcs of a sphere'' in the 11th century contains the general law of sines. The plane law of sines was later stated in the 13th century by Nasīr al-Dīn al-Tūsī. In his ''On the Sector Figure'', he stated the law of sines for plane and spherical triangles, and provided proofs for this law.

According to Glen Van Brummelen, "The Law of Sines is really Regiomontanus's foundation for his solutions of right-angled triangles in Book IV, and these solutions are in turn the bases for his solutions of general triangles." Regiomontanus was a 15th-century German mathematician.

Proof

The area of any triangle can be written as one half of its base times its height. Selecting one side of the triangle as the base, the height of the triangle relative to that base is computed as the length of another side times the sine of the angle between the chosen side and the base. Thus depending on the selection of the base, the area of the triangle can be written as any of:
$T = \frac b \left(c \sin\right) = \frac c \left(a \sin\right) = \frac a \left(b \sin\right).$
Multiplying these by gives
$\frac = \frac = \frac = \frac\,.$

The ambiguous case of triangle solution

When using the law of sines to find a side of a triangle, an ambiguous case occurs when two separate triangles can be constructed from the data provided (i.e., there are two different possible solutions to the triangle). In the case shown below they are triangles and .

:

Given a general triangle, the following conditions would need to be fulfilled for the case to be ambiguous:
* The only information known about the triangle is the angle and the sides and .
* The angle is acute (i.e., < 90°).
* The side is shorter than the side (i.e., ).
* The side is longer than the altitude from angle , where (i.e., ).

If all the above conditions are true, then each of angles and produces a valid triangle, meaning that both of the following are true:
$' = \arcsin\frac \quad \text \quad = \pi - \arcsin\frac.$

From there we can find the corresponding and or and if required, where is the side bounded by vertices and and is bounded by and .

Examples

The following are examples of how to solve a problem using the law of sines.

Example 1

Given: side , side , and angle . Angle is desired.

Using the law of sines, we conclude that
$\frac = \frac.$
$\alpha = \arcsin\left( \frac \right) \approx 32.39^\circ.$

Note that the potential solution is excluded because that would necessarily give .

Example 2

If the lengths of two sides of the triangle and are equal to , the third side has length , and the angles opposite the sides of lengths , , and are , , and respectively then
\begin
& \alpha = \beta = \frac= 90^\circ-\frac \\ pt& \sin \alpha = \sin \beta = \sin \left(90^\circ-\frac\right) = \cos \left(\frac\right) \\ pt& \frac=\frac=\frac \\ pt& \frac = x
\end

Relation to the circumcircle

In the identity
$\frac = \frac = \frac,$
the common value of the three fractions is actually the diameter of the triangle's circumcircle. This result dates back to Ptolemy.

Proof

As shown in the figure, let there be a circle with inscribed $\triangle ABC$ and another inscribed $\triangle ADB$ that passes through the circle's center O. The $\angle AOD$ has a central angle of $180^\circ$ and thus $\angle ABD = 90^\circ$. Since $\triangle ABD$ is a right triangle,
$\sin= \frac= \frac,$
where $R= \frac$ is the radius of the circumscribing circle of the triangle. Angles $$ and $$ have the same central angle thus they are the same: $=$. Therefore,
$\sin = \sin = \frac.$

Rearranging yields
$2R = \frac.$

Repeating the process of creating $\triangle ADB$ with other points gives

Relationship to the area of the triangle

The area of a triangle is given by $T = \fracab \sin \theta$, where $\theta$ is the angle enclosed by the sides of lengths and . Substituting the sine law into this equation gives
$T=\fracab \cdot \frac .$

Taking $R$ as the circumscribing radius,

It can also be shown that this equality implies
\begin
\frac
& = \frac \\ pt& = \frac ,
\end
where is the area of the triangle and is the semiperimeter $s = \frac.$

The second equality above readily simplifies to Heron's formula for the area.

The sine rule can also be used in deriving the following formula for the triangle's area: Denoting the semi-sum of the angles' sines as $S =\frac$, we have

where $R$ is the radius of the circumcircle: $2R = \frac = \frac = \frac$.

The spherical law of sines

The spherical law of sines deals with triangles on a sphere, whose sides are arcs of great circles.

Suppose the radius of the sphere is 1. Let , , and be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, , , and are the angles at the center of the sphere subtended by those arcs, in radians. Let , , and be the angles opposite those respective sides. These are dihedral angles between the planes of the three great circles.

Then the spherical law of sines says:
$\frac = \frac = \frac.$

Vector proof

Consider a unit sphere with three unit vectors , and drawn from the origin to the vertices of the triangle. Thus the angles , , and are the angles , , and , respectively. The arc subtends an angle of magnitude at the centre. Introduce a Cartesian basis with along the -axis and in the -plane making an angle with the -axis. The vector projects to in the -plane and the angle between and the -axis is . Therefore, the three vectors have components:
$\mathbf = \begin0 \\ 0 \\ 1\end, \quad \mathbf = \begin\sin c \\ 0 \\ \cos c\end, \quad \mathbf = \begin\sin b\cos A \\ \sin b\sin A \\ \cos b\end.$

The scalar triple product, is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle , and . This volume is invariant to the specific coordinate system used to represent , and . The value of the scalar triple product is the determinant with , and as its rows. With the -axis along the square of this determinant is
\begin
\bigl(\mathbf \cdot (\mathbf \times \mathbf)\bigr)^2
& = \left(\det \begin\mathbf & \mathbf & \mathbf\end\right)^2 \\ pt& = \begin
0 & 0 & 1 \\
\sin c & 0 & \cos c \\
\sin b \cos A & \sin b \sin A & \cos b
\end ^2
= \left(\sin b \sin c \sin A\right)^2.
\end
Repeating this calculation with the -axis along gives , while with the -axis along it is . Equating these expressions and dividing throughout by gives
$\frac = \frac = \frac = \frac,$
where is the volume of the parallelepiped formed by the position vector of the vertices of the spherical triangle. Consequently, the result follows.

It is easy to see how for small spherical triangles, when the radius of the sphere is much greater than the sides of the triangle, this formula becomes the planar formula at the limit, since
$\lim_ \frac = 1$
and the same for and .

Geometric proof

Consider a unit sphere with:
$OA = OB = OC = 1$

Construct point $D$ and point $E$ such that $\angle ADO = \angle AEO = 90^\circ$

Construct point $A'$ such that $\angle A'DO = \angle A'EO = 90^\circ$

It can therefore be seen that $\angle ADA' = B$ and $\angle AEA' = C$

Notice that $A'$ is the projection of $A$ on plane $OBC$. Therefore $\angle AA'D = \angle AA'E = 90^\circ$

By basic trigonometry, we have:
$AD = \sin c$
$AE = \sin b$

But $AA' = AD \sin B = AE \sin C$

Combining them we have:
$\sin c \sin B = \sin b \sin C$
$\frac =\frac$

By applying similar reasoning, we obtain the spherical law of sine:
$\frac =\frac =\frac$

Other proofs

A purely algebraic proof can be constructed from the spherical law of cosines. From the identity $\sin^2 A = 1 - \cos^2 A$ and the explicit expression for $\cos A$ from the spherical law of cosines
\begin
\sin^2\!A &= 1-\left(\frac\right)^2\\
&=\frac
\\ pt \frac
&= \frac.
\end
Since the right hand side is invariant under a cyclic permutation of $a,\;b,\;c$ the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.

Hyperbolic case

In hyperbolic geometry when the curvature is −1, the law of sines becomes
$\frac = \frac = \frac \,.$

In the special case when is a right angle, one gets
$\sin C = \frac$

which is the analog of the formula in Euclidean geometry expressing the sine of an angle as the opposite side divided by the hypotenuse.

The case of surfaces of constant curvature

Define a generalized sine function, depending also on a real parameter :
$\sin_K x = x - \frac + \frac - \frac + \cdots.$

The law of sines in constant curvature reads as
$\frac = \frac = \frac \,.$

By substituting , , and , one obtains respectively the Euclidean, spherical, and hyperbolic cases of the law of sines described above.

Let indicate the circumference of a circle of radius in a space of constant curvature . Then . Therefore, the law of sines can also be expressed as:
$\frac = \frac = \frac \,.$

This formulation was discovered by János Bolyai.

Higher dimensions

For an -dimensional simplex (i.e., triangle (), tetrahedron (), pentatope (), etc.) in -dimensional Euclidean space, the absolute value of the polar sine () of the normal vectors of the facets that meet at a vertex, divided by the hyperarea of the facet opposite the vertex is independent of the choice of the vertex. Writing for the hypervolume of the -dimensional simplex and for the product of the hyperareas of its -dimensional facets, the common ratio is
$\frac.$

For example, a tetrahedron has four triangular facets. The absolute value of the polar sine of the normal vectors to the three facets that share a vertex, divided by the area of the fourth facet will not depend upon the choice of the vertex:
\begin
& \frac =
\frac =
\frac =
\frac \\ pt= & \frac\,.
\end

See also

* Gersonides
* Half-side formula for solving spherical triangles
* Law of cosines
* Law of tangents
* Law of cotangents
* Mollweide's formula for checking solutions of triangles
* Solution of triangles
* Surveying

References

External links

*

The Law of Sines
at cut-the-knot

Degree of Curvature

{{DEFAULTSORT:Law Of Sines
Trigonometry
Angle
Articles containing proofs
Theorems about triangles