Law of sines

Example 1

Given: side a = 20, side c = 24, and angle γ = 40°. Angle α is desired.

Using the law of sines, we conclude that

$${\displaystyle {\frac {\sin \alpha }{20}}={\frac {\sin(40^{\circ })}{24}}.}$$

$${\displaystyle \alpha =\arcsin \left({\frac {20\sin(40^{\circ })}{24}}\right)\approx 32.39^{\circ }.}$$

Note that the potential solution α = 147.61° is excluded because that would necessarily give α + β + γ > 180°.

Example 2

If the lengths of two sides of the triangle a and b are equal to x, the third side has length c, and the angles opposite the sides of lengths a, b, and c are α, β, and γ respectively then

$${\displaystyle {\begin{aligned}&\alpha =\beta ={\frac {180^{\circ }-\gamma }{2}}=90^{\circ }-{\frac {\gamma }{2}}\\[6pt]&\sin \alpha =\sin \beta =\sin \left(90^{\circ }-{\frac {\gamma }{2}}\right)=\cos \left({\frac {\gamma }{2}}\right)\\[6pt]&{\frac {c}{\sin \gamma }}={\frac {a}{\sin \alpha }}={\frac {x}{\cos \left({\frac {\gamma }{2}}\right)}}\\[6pt]&{\frac {c\cos \left({\frac {\gamma }{2}}\right)}{\sin \gamma }}=x\end{aligned}}}$$

As shown in the figure, let there be a circle with inscribed ${\displaystyle \triangle ABC}$ and another inscribed ${\displaystyle \triangle ADB}$ that passes through the circle's center O. The ${\displaystyle \angle AOD}$ has a central angle of ${\displaystyle 180^{\circ }}$ and thus ${\displaystyle \angle ABD=90^{\circ }}$, by Thales's theorem. Since ${\displaystyle \triangle ABD}$ is a right triangle,

$${\displaystyle \sin {\delta }={\frac {\text{opposite}}{\text{hypotenuse}}}={\frac {c}{2R}},}$$

where ${\textstyle R={\frac {d}{2}}}$ is the radius of the circumscribing circle of the triangle.[9] Angles ${\displaystyle {\gamma }}$ and ${\displaystyle {\delta }}$ have the same central angle thus they are the same, by the inscribed angle theorem: ${\displaystyle {\gamma }={\delta }}$. Therefore,

$${\displaystyle \sin {\delta }=\sin {\gamma }={\frac {c}{2R}}.}$$

Rearranging yields

$${\displaystyle 2R={\frac {c}{\sin {\gamma }}}.}$$

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

The area of a triangle is given by ${\textstyle T={\frac {1}{2}}ab\sin \theta }$, where ${\displaystyle \theta }$ is the angle enclosed by the sides of lengths a and b. Substituting the sine law into this equation gives

$${\displaystyle T={\frac {1}{2}}ab\cdot {\frac {c}{2R}}.}$$

Taking ${\displaystyle R}$ as the circumscribing radius,[10]

It can also be shown that this equality implies

$${\displaystyle {\begin{aligned}{\frac {abc}{2T}}&={\frac {abc}{2{\sqrt {s(s-a)(s-b)(s-c)}}}}\\[6pt]&={\frac {2abc}{\sqrt {{(a^{2}+b^{2}+c^{2})}^{2}-2(a^{4}+b^{4}+c^{4})}}},\end{aligned}}}$$

where T is the area of the triangle and s is the semiperimeter ${\textstyle s={\frac {1}{2}}\left(a+b+c\right).}$

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 ${\textstyle S={\frac {1}{2}}\left(\sin A+\sin B+\sin C\right)}$, we have[11]

where ${\displaystyle R}$ is the radius of the circumcircle: ${\displaystyle 2R={\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}}$.

Consider a unit sphere with three unit vectors OA, OB and OC drawn from the origin to the vertices of the triangle. Thus the angles α, β, and γ are the angles a, b, and c, respectively. The arc BC subtends an angle of magnitude a at the centre. Introduce a Cartesian basis with OA along the z-axis and OB in the xz-plane making an angle c with the z-axis. The vector OC projects to ON in the xy-plane and the angle between ON and the x-axis is A. Therefore, the three vectors have components:

$${\displaystyle \mathbf {OA} ={\begin{pmatrix}0\\0\\1\end{pmatrix}},\quad \mathbf {OB} ={\begin{pmatrix}\sin c\\0\\\cos c\end{pmatrix}},\quad \mathbf {OC} ={\begin{pmatrix}\sin b\cos A\\\sin b\sin A\\\cos b\end{pmatrix}}.}$$

The scalar triple product, OA ⋅ (OB × OC) is the volume of the parallelepiped formed by the position vectors of the vertices of the spherical triangle OA, OB and OC. This volume is invariant to the specific coordinate system used to represent OA, OB and OC. The value of the scalar triple product OA ⋅ (OB × OC) is the 3 × 3 determinant with OA, OB and OC as its rows. With the z-axis along OA the square of this determinant is

$${\displaystyle {\begin{aligned}{\bigl (}\mathbf {OA} \cdot (\mathbf {OB} \times \mathbf {OC} ){\bigr )}^{2}&=\left(\det {\begin{pmatrix}\mathbf {OA} &\mathbf {OB} &\mathbf {OC} \end{pmatrix}}\right)^{2}\\[4pt]&={\begin{vmatrix}0&0&1\\\sin c&0&\cos c\\\sin b\cos A&\sin b\sin A&\cos b\end{vmatrix}}^{2}=\left(\sin b\sin c\sin A\right)^{2}.\end{aligned}}}$$

Repeating this calculation with the z-axis along OB gives (sin c sin a sin B)², while with the z-axis along OC it is (sin a sin b sin C)². Equating these expressions and dividing throughout by (sin a sin b sin c)² gives

$${\displaystyle {\frac {\sin ^{2}A}{\sin ^{2}a}}={\frac {\sin ^{2}B}{\sin ^{2}b}}={\frac {\sin ^{2}C}{\sin ^{2}c}}={\frac {V^{2}}{\sin ^{2}(a)\sin ^{2}(b)\sin ^{2}(c)}},}$$

where V 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

$${\displaystyle \lim _{a\to 0}{\frac {\sin a}{a}}=1}$$

and the same for sin b and sin c.

Consider a unit sphere with:

$${\displaystyle OA=OB=OC=1}$$

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

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

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

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

By basic trigonometry, we have:

$${\displaystyle {\begin{aligned}AD&=\sin c\\AE&=\sin b\end{aligned}}}$$

But ${\displaystyle AA'=AD\sin B=AE\sin C}$

Combining them we have:

$${\displaystyle {\begin{aligned}\sin c\sin B&=\sin b\sin C\\\Rightarrow {\frac {\sin B}{\sin b}}&={\frac {\sin C}{\sin c}}\end{aligned}}}$$

By applying similar reasoning, we obtain the spherical law of sine:

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

A purely algebraic proof can be constructed from the spherical law of cosines. From the identity ${\displaystyle \sin ^{2}A=1-\cos ^{2}A}$ and the explicit expression for ${\displaystyle \cos A}$ from the spherical law of cosines

$${\displaystyle {\begin{aligned}\sin ^{2}\!A&=1-\left({\frac {\cos a-\cos b\,\cos c}{\sin b\,\sin c}}\right)^{2}\\&={\frac {\left(1-\cos ^{2}\!b\right)\left(1-\cos ^{2}\!c\right)-\left(\cos a-\cos b\,\cos c\right)^{2}}{\sin ^{2}\!b\,\sin ^{2}\!c}}\\[8pt]{\frac {\sin A}{\sin a}}&={\frac {\left[1-\cos ^{2}\!a-\cos ^{2}\!b-\cos ^{2}\!c+2\cos a\cos b\cos c\right]^{1/2}}{\sin a\sin b\sin c}}.\end{aligned}}}$$

Since the right hand side is invariant under a cyclic permutation of ${\displaystyle a,\;b,\;c}$ the spherical sine rule follows immediately.

The figure used in the Geometric proof above is used by and also provided in Banerjee[12] (see Figure 3 in this paper) to derive the sine law using elementary linear algebra and projection matrices.