- Source: Law of sines
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,
a
sin
α
=
b
sin
β
=
c
sin
γ
=
2
R
,
{\displaystyle {\frac {a}{\sin {\alpha }}}\,=\,{\frac {b}{\sin {\beta }}}\,=\,{\frac {c}{\sin {\gamma }}}\,=\,2R,}
where a, b, and c are the lengths of the sides of a triangle, and α, β, and γ are the opposite angles (see figure 2), while R 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;
sin
α
a
=
sin
β
b
=
sin
γ
c
.
{\displaystyle {\frac {\sin {\alpha }}{a}}\,=\,{\frac {\sin {\beta }}{b}}\,=\,{\frac {\sin {\gamma }}{c}}.}
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
The Persian mathematician Nasir al-Din al-Tusi was the first to state and prove the law of sines, which applies to all plane triangles. He stated:In any plane triangle, the ratio of the sides is equal to the ratio of the sines of the angles opposite to those sides. That is, in triangle ABC, we have AB : AC = Sin(∠ACB) : Sin(∠ABC)By employing the law of sines, al-Tusi could solve triangles where either two angles and a side were known or two sides and an angle opposite one of them were given. For triangles with two sides and the included angle, he divided them into right triangles that he could then solve. When three sides were given, he dropped a perpendicular line and then used Proposition II-13 of Euclid's Elements. Al-Tusi established the important result that if the sum or difference of two arcs is provided along with the ratio of their sines, then the arcs can be calculated.
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.
Ibn Muʿādh al-Jayyānī's The book of unknown arcs of a sphere in the 11th century contains the spherical 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
With the side of length a as the base, the triangle's altitude can be computed as b sin γ or as c sin β. Equating these two expressions gives
sin
β
b
=
sin
γ
c
,
{\displaystyle {\frac {\sin \beta }{b}}={\frac {\sin \gamma }{c}}\,,}
and similar equations arise by choosing the side of length b or the side of length c as the base of the triangle.
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 ABC and ABC′.
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 α and the sides a and c.
The angle α is acute (i.e., α < 90°).
The side a is shorter than the side c (i.e., a < c).
The side a is longer than the altitude h from angle β, where h = c sin α (i.e., a > h).
If all the above conditions are true, then each of angles β and β′ produces a valid triangle, meaning that both of the following are true:
γ
′
=
arcsin
c
sin
α
a
or
γ
=
π
−
arcsin
c
sin
α
a
.
{\displaystyle {\gamma }'=\arcsin {\frac {c\sin {\alpha }}{a}}\quad {\text{or}}\quad {\gamma }=\pi -\arcsin {\frac {c\sin {\alpha }}{a}}.}
From there we can find the corresponding β and b or β′ and b′ if required, where b is the side bounded by vertices A and C and b′ is bounded by A and C′.
Examples
The following are examples of how to solve a problem using the 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
sin
α
20
=
sin
(
40
∘
)
24
.
{\displaystyle {\frac {\sin \alpha }{20}}={\frac {\sin(40^{\circ })}{24}}.}
α
=
arcsin
(
20
sin
(
40
∘
)
24
)
≈
32.39
∘
.
{\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
α
=
β
=
180
∘
−
γ
2
=
90
∘
−
γ
2
sin
α
=
sin
β
=
sin
(
90
∘
−
γ
2
)
=
cos
(
γ
2
)
c
sin
γ
=
a
sin
α
=
x
cos
(
γ
2
)
c
cos
(
γ
2
)
sin
γ
=
x
{\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}}}
Relation to the circumcircle
In the identity
a
sin
α
=
b
sin
β
=
c
sin
γ
,
{\displaystyle {\frac {a}{\sin {\alpha }}}={\frac {b}{\sin {\beta }}}={\frac {c}{\sin {\gamma }}},}
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
△
A
B
C
{\displaystyle \triangle ABC}
and another inscribed
△
A
D
B
{\displaystyle \triangle ADB}
that passes through the circle's center O. The
∠
A
O
D
{\displaystyle \angle AOD}
has a central angle of
180
∘
{\displaystyle 180^{\circ }}
and thus
∠
A
B
D
=
90
∘
{\displaystyle \angle ABD=90^{\circ }}
, by Thales's theorem. Since
△
A
B
D
{\displaystyle \triangle ABD}
is a right triangle,
sin
δ
=
opposite
hypotenuse
=
c
2
R
,
{\displaystyle \sin {\delta }={\frac {\text{opposite}}{\text{hypotenuse}}}={\frac {c}{2R}},}
where
R
=
d
2
{\textstyle R={\frac {d}{2}}}
is the radius of the circumscribing circle of the triangle. Angles
γ
{\displaystyle {\gamma }}
and
δ
{\displaystyle {\delta }}
lie on the same circle and subtend the same chord c; thus, by the inscribed angle theorem,
γ
=
δ
{\displaystyle {\gamma }={\delta }}
. Therefore,
sin
δ
=
sin
γ
=
c
2
R
.
{\displaystyle \sin {\delta }=\sin {\gamma }={\frac {c}{2R}}.}
Rearranging yields
2
R
=
c
sin
γ
.
{\displaystyle 2R={\frac {c}{\sin {\gamma }}}.}
Repeating the process of creating
△
A
D
B
{\displaystyle \triangle ADB}
with other points gives
= Relationship to the area of the triangle
=The area of a triangle is given by
T
=
1
2
a
b
sin
θ
{\textstyle T={\frac {1}{2}}ab\sin \theta }
, where
θ
{\displaystyle \theta }
is the angle enclosed by the sides of lengths a and b. Substituting the sine law into this equation gives
T
=
1
2
a
b
⋅
c
2
R
.
{\displaystyle T={\frac {1}{2}}ab\cdot {\frac {c}{2R}}.}
Taking
R
{\displaystyle R}
as the circumscribing radius,
It can also be shown that this equality implies
a
b
c
2
T
=
a
b
c
2
s
(
s
−
a
)
(
s
−
b
)
(
s
−
c
)
=
2
a
b
c
(
a
2
+
b
2
+
c
2
)
2
−
2
(
a
4
+
b
4
+
c
4
)
,
{\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
s
=
1
2
(
a
+
b
+
c
)
.
{\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
S
=
1
2
(
sin
A
+
sin
B
+
sin
C
)
{\textstyle S={\frac {1}{2}}\left(\sin A+\sin B+\sin C\right)}
, we have
where
R
{\displaystyle R}
is the radius of the circumcircle:
2
R
=
a
sin
A
=
b
sin
B
=
c
sin
C
{\displaystyle 2R={\frac {a}{\sin A}}={\frac {b}{\sin B}}={\frac {c}{\sin C}}}
.
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 a, b, and c be the lengths of the great-arcs that are the sides of the triangle. Because it is a unit sphere, a, b, and c are the angles at the center of the sphere subtended by those arcs, in radians. Let A, B, and C 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:
sin
A
sin
a
=
sin
B
sin
b
=
sin
C
sin
c
.
{\displaystyle {\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}.}
= Vector proof
=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:
O
A
=
(
0
0
1
)
,
O
B
=
(
sin
c
0
cos
c
)
,
O
C
=
(
sin
b
cos
A
sin
b
sin
A
cos
b
)
.
{\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
(
O
A
⋅
(
O
B
×
O
C
)
)
2
=
(
det
(
O
A
O
B
O
C
)
)
2
=
|
0
0
1
sin
c
0
cos
c
sin
b
cos
A
sin
b
sin
A
cos
b
|
2
=
(
sin
b
sin
c
sin
A
)
2
.
{\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
sin
2
A
sin
2
a
=
sin
2
B
sin
2
b
=
sin
2
C
sin
2
c
=
V
2
sin
2
(
a
)
sin
2
(
b
)
sin
2
(
c
)
,
{\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
lim
a
→
0
sin
a
a
=
1
{\displaystyle \lim _{a\to 0}{\frac {\sin a}{a}}=1}
and the same for sin b and sin c.
= Geometric proof
=Consider a unit sphere with:
O
A
=
O
B
=
O
C
=
1
{\displaystyle OA=OB=OC=1}
Construct point
D
{\displaystyle D}
and point
E
{\displaystyle E}
such that
∠
A
D
O
=
∠
A
E
O
=
90
∘
{\displaystyle \angle ADO=\angle AEO=90^{\circ }}
Construct point
A
′
{\displaystyle A'}
such that
∠
A
′
D
O
=
∠
A
′
E
O
=
90
∘
{\displaystyle \angle A'DO=\angle A'EO=90^{\circ }}
It can therefore be seen that
∠
A
D
A
′
=
B
{\displaystyle \angle ADA'=B}
and
∠
A
E
A
′
=
C
{\displaystyle \angle AEA'=C}
Notice that
A
′
{\displaystyle A'}
is the projection of
A
{\displaystyle A}
on plane
O
B
C
{\displaystyle OBC}
. Therefore
∠
A
A
′
D
=
∠
A
A
′
E
=
90
∘
{\displaystyle \angle AA'D=\angle AA'E=90^{\circ }}
By basic trigonometry, we have:
A
D
=
sin
c
A
E
=
sin
b
{\displaystyle {\begin{aligned}AD&=\sin c\\AE&=\sin b\end{aligned}}}
But
A
A
′
=
A
D
sin
B
=
A
E
sin
C
{\displaystyle AA'=AD\sin B=AE\sin C}
Combining them we have:
sin
c
sin
B
=
sin
b
sin
C
⇒
sin
B
sin
b
=
sin
C
sin
c
{\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:
sin
A
sin
a
=
sin
B
sin
b
=
sin
C
sin
c
{\displaystyle {\frac {\sin A}{\sin a}}={\frac {\sin B}{\sin b}}={\frac {\sin C}{\sin c}}}
= 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
{\displaystyle \sin ^{2}A=1-\cos ^{2}A}
and the explicit expression for
cos
A
{\displaystyle \cos A}
from the spherical law of cosines
sin
2
A
=
1
−
(
cos
a
−
cos
b
cos
c
sin
b
sin
c
)
2
=
(
1
−
cos
2
b
)
(
1
−
cos
2
c
)
−
(
cos
a
−
cos
b
cos
c
)
2
sin
2
b
sin
2
c
sin
A
sin
a
=
[
1
−
cos
2
a
−
cos
2
b
−
cos
2
c
+
2
cos
a
cos
b
cos
c
]
1
/
2
sin
a
sin
b
sin
c
.
{\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
a
,
b
,
c
{\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 (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
sin
A
sinh
a
=
sin
B
sinh
b
=
sin
C
sinh
c
.
{\displaystyle {\frac {\sin A}{\sinh a}}={\frac {\sin B}{\sinh b}}={\frac {\sin C}{\sinh c}}\,.}
In the special case when B is a right angle, one gets
sin
C
=
sinh
c
sinh
b
{\displaystyle \sin C={\frac {\sinh c}{\sinh b}}}
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
κ
{\displaystyle \kappa }
:
sin
κ
(
x
)
=
x
−
κ
3
!
x
3
+
κ
2
5
!
x
5
−
κ
3
7
!
x
7
+
⋯
=
∑
n
=
0
∞
(
−
1
)
n
κ
n
(
2
n
+
1
)
!
x
2
n
+
1
.
{\displaystyle \sin _{\kappa }(x)=x-{\frac {\kappa }{3!}}x^{3}+{\frac {\kappa ^{2}}{5!}}x^{5}-{\frac {\kappa ^{3}}{7!}}x^{7}+\cdots =\sum _{n=0}^{\infty }{\frac {(-1)^{n}\kappa ^{n}}{(2n+1)!}}x^{2n+1}.}
The law of sines in constant curvature
κ
{\displaystyle \kappa }
reads as
sin
A
sin
κ
a
=
sin
B
sin
κ
b
=
sin
C
sin
κ
c
.
{\displaystyle {\frac {\sin A}{\sin _{\kappa }a}}={\frac {\sin B}{\sin _{\kappa }b}}={\frac {\sin C}{\sin _{\kappa }c}}\,.}
By substituting
κ
=
0
{\displaystyle \kappa =0}
,
κ
=
1
{\displaystyle \kappa =1}
, and
κ
=
−
1
{\displaystyle \kappa =-1}
, one obtains respectively
sin
0
(
x
)
=
x
{\displaystyle \sin _{0}(x)=x}
,
sin
1
(
x
)
=
sin
x
{\displaystyle \sin _{1}(x)=\sin x}
, and
sin
−
1
(
x
)
=
sinh
x
{\displaystyle \sin _{-1}(x)=\sinh x}
, that is, the Euclidean, spherical, and hyperbolic cases of the law of sines described above.
Let
p
κ
(
r
)
{\displaystyle p_{\kappa }(r)}
indicate the circumference of a circle of radius
r
{\displaystyle r}
in a space of constant curvature
κ
{\displaystyle \kappa }
. Then
p
κ
(
r
)
=
2
π
sin
κ
(
r
)
{\displaystyle p_{\kappa }(r)=2\pi \sin _{\kappa }(r)}
. Therefore, the law of sines can also be expressed as:
sin
A
p
κ
(
a
)
=
sin
B
p
κ
(
b
)
=
sin
C
p
κ
(
c
)
.
{\displaystyle {\frac {\sin A}{p_{\kappa }(a)}}={\frac {\sin B}{p_{\kappa }(b)}}={\frac {\sin C}{p_{\kappa }(c)}}\,.}
This formulation was discovered by János Bolyai.
Higher dimensions
A tetrahedron has four triangular facets. The absolute value of the polar sine (psin) of the normal vectors to the three facets that share a vertex of the tetrahedron, divided by the area of the fourth facet will not depend upon the choice of the vertex:
|
psin
(
b
,
c
,
d
)
|
A
r
e
a
a
=
|
psin
(
a
,
c
,
d
)
|
A
r
e
a
b
=
|
psin
(
a
,
b
,
d
)
|
A
r
e
a
c
=
|
psin
(
a
,
b
,
c
)
|
A
r
e
a
d
=
(
3
V
o
l
u
m
e
t
e
t
r
a
h
e
d
r
o
n
)
2
2
A
r
e
a
a
A
r
e
a
b
A
r
e
a
c
A
r
e
a
d
.
{\displaystyle {\begin{aligned}&{\frac {\left|\operatorname {psin} (\mathbf {b} ,\mathbf {c} ,\mathbf {d} )\right|}{\mathrm {Area} _{a}}}={\frac {\left|\operatorname {psin} (\mathbf {a} ,\mathbf {c} ,\mathbf {d} )\right|}{\mathrm {Area} _{b}}}={\frac {\left|\operatorname {psin} (\mathbf {a} ,\mathbf {b} ,\mathbf {d} )\right|}{\mathrm {Area} _{c}}}={\frac {\left|\operatorname {psin} (\mathbf {a} ,\mathbf {b} ,\mathbf {c} )\right|}{\mathrm {Area} _{d}}}\\[4pt]={}&{\frac {(3~\mathrm {Volume} _{\mathrm {tetrahedron} })^{2}}{2~\mathrm {Area} _{a}\mathrm {Area} _{b}\mathrm {Area} _{c}\mathrm {Area} _{d}}}\,.\end{aligned}}}
More generally, for an n-dimensional simplex (i.e., triangle (n = 2), tetrahedron (n = 3), pentatope (n = 4), etc.) in n-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 V for the hypervolume of the n-dimensional simplex and P for the product of the hyperareas of its (n − 1)-dimensional facets, the common ratio is
|
psin
(
b
,
…
,
z
)
|
A
r
e
a
a
=
⋯
=
|
psin
(
a
,
…
,
y
)
|
A
r
e
a
z
=
(
n
V
)
n
−
1
(
n
−
1
)
!
P
.
{\displaystyle {\frac {\left|\operatorname {psin} (\mathbf {b} ,\ldots ,\mathbf {z} )\right|}{\mathrm {Area} _{a}}}=\cdots ={\frac {\left|\operatorname {psin} (\mathbf {a} ,\ldots ,\mathbf {y} )\right|}{\mathrm {Area} _{z}}}={\frac {(nV)^{n-1}}{(n-1)!P}}.}
See also
Gersonides – Medieval Jewish philosopher
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
"Sine theorem", Encyclopedia of Mathematics, EMS Press, 2001 [1994]
The Law of Sines at cut-the-knot
Degree of Curvature
Finding the Sine of 1 Degree
Generalized law of sines to higher dimensions
Kata Kunci Pencarian:
- Aturan sinus
- Optika geometris
- Logaritma
- Segitiga
- Law of sines
- Law of cosines
- Law of tangents
- Trigonometric functions
- Sines
- Trigonometry
- Tetrahedron
- Sine and cosine
- Solution of triangles
- Law of cotangents
