The law of sines consists of two rules that relate side lengths and angle measures in a triangle using trigonometric concepts. As the name suggests, the law of sines involves the sine function.
Law of Sines
The law of sines (also called the sine law or sine rule) is a theorem in the form of an equation that relates the sine values of angles in a triangle to the lengths of their opposite sides in the form of a ratio.
In triangle $ABC,$ the following holds:
$$\boxed{\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}}.$$
A formal proof of the law of sines can be seen in the following panel.
Construct triangle $ABC$ as follows.
Draw the altitude of the triangle from point $C$ to side $AB$ so that line $CD$ is formed. In right triangle $ACD,$ we have
$$\sin A = \dfrac{CD}{b} \Rightarrow CD = b \sin A.$$Meanwhile, in right triangle $BDC,$ we have
$$\sin B = \dfrac{CD}{a} \Rightarrow CD = a \sin B.$$From the two equations above, we obtain
$$b \sin A = a \sin B \Rightarrow \dfrac{a}{\sin A} = \dfrac{b}{\sin B}.$$Using another altitude and a similar process, we will obtain
$$\boxed{\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}}.$$Therefore, the law of sines is proven. $\blacksquare$
To gain a deeper understanding of this topic, the following problems and their solutions are provided. Hope this is helpful.
Multiple Choice Section
Given $\triangle ABC$ with side length $a = 4~\text{cm},$ $\angle A = 120^{\circ},$ and $\angle B = 30^{\circ}.$ The length of side $c = \cdots~\text{cm}.$
A. $2\sqrt2$ D. $\dfrac34\sqrt2$
B. $\dfrac43\sqrt3$ E. $\sqrt3$
C. $\dfrac34\sqrt3$
Consider the following sketch.
Since the sum of the interior angles of a triangle is always $180^{\circ}$, then $\angle C = (180-120-30)^{\circ} = 30^{\circ}.$
Next, by using the law of sines, we obtain
$\begin{aligned} \dfrac{a}{\sin A} & = \dfrac{c}{\sin C} \\ \dfrac{4}{\sin 120^{\circ}} & = \dfrac{c}{\sin 30^{\circ}} \\ \dfrac{4}{\cancel{\frac12}\sqrt3} & = \dfrac{c}{\cancel{\frac12}} \\ c & = \dfrac{4}{\sqrt3} = \dfrac43\sqrt3~\text{cm}. \end{aligned}$
Therefore, the length of side $\boxed{c = \dfrac43\sqrt3~\text{cm}}.$
(Answer B)
In $\triangle JKL$, it is known that $\sin L = \dfrac13$, $\sin J = \dfrac35$, and $JK = 5$ cm. The length of $KL$ is $\cdots~\text{cm}$.
A. $5$ C. $9$ E. $15$
B. $7$ D. $12$
In $\triangle JKL$, the side opposite $\angle L$ is $JK$, while the side opposite $\angle J$ is $KL.$ By using the law of sines, we obtain
$$\begin{aligned} \dfrac{JK}{\sin L} & = \dfrac{KL}{\sin J} \\ \dfrac{5}{\frac13} & = \dfrac{KL}{\frac35} \\ 15 & = \dfrac{KL}{\frac35} \\ KL & = 15 \cdot \dfrac35 = 9. \end{aligned}$$Therefore, the length of $KL$ is $\boxed{9~\text{cm}}.$
(Answer C)
Given triangle $PQR$ with $\angle P=60^\circ,$ side length $p = 10$ cm, and side length $q = 12$ cm. The value of $\sin Q$ is $\cdots \cdot$
A. $\dfrac15\sqrt3$ D. $\dfrac45\sqrt3$
B. $\dfrac25\sqrt3$ E. $\dfrac13\sqrt3$
C. $\dfrac35\sqrt3$
The following information is known.
- $\angle P = 60^\circ.$
- $p = 10$ cm.
- $q = 12$ cm.
By using the law of sines, we obtain
$$\begin{aligned} \dfrac{p}{\sin P} & = \dfrac{q}{\sin Q} \\ \sin Q & = q \cdot \dfrac{\sin P}{p} \\ \sin Q & = 12 \cdot \dfrac{\sin 60^\circ}{10} \\ \sin Q & = 12 \cdot \dfrac{\frac12\sqrt3}{10} \\ \sin Q & = \dfrac{6\sqrt3}{10} = \dfrac35\sqrt3. \end{aligned}$$Therefore, the value of $\sin Q$ is $\boxed{\dfrac35\sqrt3}.$
(Answer C)
In an acute triangle $ABC,$ suppose that $a=8$ cm, $b=4\sqrt2$ cm, and $\angle A=45^\circ.$ The measure of angle $B$ is $\cdots \cdot$
A. $30^\circ$ D. $90^\circ$
B. $45^\circ$ E. $150^\circ$
C. $60^\circ$
The following information is known.
- $a = 8$ cm.
- $b = 4\sqrt2$ cm.
- $\angle A = 45^\circ.$
By using the law of sines, we obtain
$$\begin{aligned} \dfrac{a}{\sin A} & = \dfrac{b}{\sin B} \\ \sin B & = b \cdot \dfrac{\sin A}{a} \\ \sin B & = 4\sqrt2 \cdot \dfrac{\sin 45^\circ}{8} \\ \sin B & = 4\sqrt2 \cdot \dfrac{\frac12\sqrt2}{8} \\ \sin B & = \dfrac12. \end{aligned}$$This means that $\angle B = 30^\circ$ or $\angle B = 150^\circ.$ However, $\triangle ABC$ is acute, which means every angle in the triangle must be less than $90^\circ,$ so the possible value of angle $B$ is $30^\circ.$
Therefore, the measure of angle $B$ is $\boxed{30^\circ}.$
(Answer A)
Observe the $\triangle ABC$ shown below.
The ratio of the lengths $BC$ and $AC$ is $\cdots \cdot$
A. $3 : 4$
B. $4 : 3$
C. $\sqrt2 : \sqrt3$
D. $\sqrt3 : 2\sqrt2$
E. $\sqrt3 : \sqrt2$
Notice that we are looking for the lengths of the sides opposite the known angles. By using the law of sines, we obtain
$$\begin{aligned} \dfrac{BC}{\sin A} & = \dfrac{AC}{\sin B} \\ \dfrac{BC}{AC} & = \dfrac{\sin A}{\sin B} \\ & = \dfrac{\sin 45^\circ}{\sin 60^\circ} \\ & = \dfrac{\cancel{\dfrac12}\sqrt2}{\cancel{\dfrac12}\sqrt3} \\ & = \dfrac{\sqrt2}{\sqrt3}. \end{aligned}$$Therefore, the ratio $BC : AC$ is $\boxed{\sqrt2 : \sqrt3}.$
(Answer C)
Observe the following figure.
The value of $\sin D$ is $\cdots \cdot$
A. $\dfrac12\sqrt6$ D. $\dfrac18\sqrt6$
B. $\dfrac14\sqrt6$ E. $\dfrac18\sqrt3$
C. $\dfrac16\sqrt6$
From the given figure, the following information is known.
- $AC = 2.$
- $BD = 4.$
- $\angle BAC = 30^\circ.$
- $\angle ABC = 45^\circ.$
- $\angle BCD = 60^\circ.$
Using the law of sines in triangle $ABC,$ we first determine the length of side $BC.$
$$\begin{aligned} \dfrac{AC}{\sin \angle ABC} & = \dfrac{BC}{\sin \angle BAC} \\ BC & = \dfrac{AC}{\sin \angle ABC} \cdot \sin \angle BAC \\ BC & = \dfrac{2}{\sin 45^\circ} \cdot \sin 30^\circ \\ BC & = \dfrac{2}{\cancel{\frac12}\sqrt2} \cdot \cancel{\dfrac12} \\ BC & = \dfrac{2}{\sqrt2} = \sqrt2. \end{aligned}$$Thus, the length of side $BC = \sqrt2.$
Next, by using the law of sines in triangle $BCD,$ we obtain
$$\begin{aligned} \dfrac{BD}{\sin \angle BCD} & = \dfrac{BC}{\sin \angle BDC} \\ \sin \angle BDC & = \dfrac{\sin \angle BCD}{BD} \cdot BC \\ \sin \angle BDC & = \dfrac{\sin 60^\circ}{4} \cdot \sqrt2 \\ \sin \angle BDC & = \dfrac{\frac12\sqrt3}{4} \cdot \sqrt2 = \dfrac18\sqrt6. \end{aligned}$$Therefore, the value of $\sin D$ is $\boxed{\dfrac18\sqrt6}.$
(Answer D)
In $\triangle ABC$, it is known that $(b+c) : (c + a) : (a + b)$ $= 4 : 5 : 6.$ The value of $\sin A : \sin B : \sin C = \cdots \cdot$
A. $7 : 5 : 3$ D. $4 : 5 : 6$
B. $3 : 5 : 7$ E. $6 : 5 : 4$
C. $7 : 3 : 5$
Suppose that for some positive integer $k,$ the following holds.
$$\begin{aligned} b+c & = 4k && (\cdots 1) \\ a+c & = 5k && (\cdots 2) \\ a+b & = 6k. && (\cdots 3) \end{aligned}$$Eliminate $c$ from Equations $(1)$ and $(2)$ to obtain $a-b = k.$ Call this Equation $(4).$
From Equations $(3)$ and $(4),$ we obtain $a = \dfrac72k$ and $b = \dfrac52k,$ so $c = \dfrac32k.$ Therefore, we get the ratio
$$\begin{aligned} a : b : c & = \dfrac72k : \dfrac52k : \dfrac32k \\ & = 7 : 5 : 3. \end{aligned}$$According to the law of sines, the ratio of the sines of the angles is equal to the ratio of their opposite sides, so $\boxed{\begin{aligned} \sin A : \sin B : \sin C & = a : b : c \\ & = 7 : 5 : 3. \end{aligned}}$
(Answer A)
In $\triangle ABC$, it is known that $\angle B = 70^{\circ}$, $\angle C = 80^{\circ}$, and $BC = 2$ cm. If $R$ is the radius of the circumcircle of triangle $ABC$, then the value of $R = \cdots~\text{cm}.$
A. $1$ C. $4$ E. $10$
B. $2$ D. $8$
According to the law of sines, we have
$$\boxed{\color{blue}{\dfrac{BC}{\sin \angle A}} = \dfrac{AB}{\sin \angle C} = \dfrac{AC}{\sin \angle B} = \color{blue}{2R}}$$where $R$ is the radius of the circumcircle of $\triangle ABC.$
Since $\angle B = 70^{\circ}$ and $\angle C = 80^{\circ},$ then $\angle A = (180-70-80)^{\circ} = 30^{\circ}$ so that
$$\begin{aligned} \dfrac{BC}{\sin \angle A} & = 2R \\ \dfrac{2}{\sin 30^{\circ}} & = 2R \\ \dfrac{2}{\frac12} & = 2R \\ R & = 2. \end{aligned}$$Therefore, the value of $\boxed{R = 2~\text{cm}}.$
(Answer B)
If the side lengths of triangle $ABC$ are respectively $AB=4~\text{cm}$, $BC=6~\text{cm},$ and $AC=5~\text{cm},$ while $\angle BAC = \alpha,$ $\angle ABC = \beta,$ and $\angle BCA = \gamma,$ then $\sin \alpha : \sin \beta : \sin \gamma = \cdots \cdot$
A. $4 : 5 : 6$ D. $4 : 6 : 5$
B. $5 : 6 : 4$ E. $6 : 4 : 5$
C. $6 : 5 : 4$
Consider the following sketch.
Using the law of sines, we obtain the equation
$\dfrac{AB}{\sin \gamma} = \dfrac{BC}{\sin \alpha} = \dfrac{AC}{\sin \beta}.$
Based on this rule, the sine values of the angles are proportional to the lengths of their opposite sides. The side opposite angle $\alpha$ is $BC$, the side opposite angle $\beta$ is $AC$, and the side opposite angle $\gamma$ is $AB.$
In this case, we can write
$\boxed{\begin{aligned} \sin \alpha : \sin \beta : \sin \gamma & = BC : AC : AB \\ & = 6 : 5 : 4. \end{aligned}}$
(Answer C)
Read Also: Theory, Problems, and Solutions – law of cosines
Sukardi and Lili are standing on a beach, separated by a distance of $6$ km. The shoreline passing through them is a straight line. Both of them can see the same ship from where they stand. Suppose the angle formed by Sukardi’s position and the ship with the shoreline is $45^{\circ}$. Meanwhile, the angle formed by Lili’s position and the ship with the shoreline is $15^{\circ}$. If the distance between the ship and Lili’s position is $a\sqrt{b}$ km, where $a\sqrt{b}$ is in simplest radical form, then the value of $b-a = \cdots \cdot$
A. $0$ C. $3$ E. $6$
B. $2$ D. $4$
Consider the following sketch.
Point $C$ is the location of the ship. The measure of angle $C$ is $(180-45-15)^{\circ} = 120^{\circ}$. To find the distance between the ship and Lili, namely the length $BC$, use the law of sines.
$$\begin{aligned} \dfrac{AB}{\sin C} & = \dfrac{BC}{\sin A} \\ \dfrac{6}{\sin 120^{\circ}} & = \dfrac{BC}{\sin 45^{\circ}} \\ \dfrac{6}{\frac12\sqrt3} & = \dfrac{BC}{\frac12\sqrt2} \\ BC & = \dfrac{6}{\sqrt3} \times \sqrt2 \\ BC & = 2\sqrt6 \end{aligned}$$Therefore, we obtain $a = 2$ and $b = 6$ so that $\boxed{b-a=6-2=4}.$
(Answer D)
Essay Section
In triangle $ABC,$ it is known that $\angle A = 30^\circ,$ $\angle B = 45^\circ,$ and the side length $a = 8$ cm. Determine the length of side $b.$
The following information is known.
- $\angle A = 30^\circ.$
- $\angle B = 45^\circ.$
- $a = 8$ cm.
Using the law of sines, we obtain
$$\begin{aligned} \dfrac{a}{\sin A} & = \dfrac{b}{\sin B} \\ b & = \dfrac{a}{\sin A} \cdot \sin B \\ b & = \dfrac{8}{\sin 30^\circ} \cdot \sin 45^\circ \\ b & = \dfrac{8}{\cancel{\frac12}} \cdot \cancel{\dfrac12}\sqrt2 \\ b & = 8\sqrt2~\text{cm}. \end{aligned}$$Therefore, the length of side $b$ is $\boxed{8\sqrt2~\text{cm}}.$
Given triangle $ABC$ with $\angle A = 30^\circ,$ $\angle C = 105^\circ,$ and $BC = 10$ cm. Determine the length of $AC.$
The following information is known.
- $\angle A = 30^\circ.$
- $\angle C = 105^\circ.$
- $BC = 10$ cm.
Since the sum of the interior angles of a triangle is $180^\circ,$ we obtain
$$\begin{aligned} \angle B & = 180^\circ-(\angle A + \angle C) \\ & = 180^\circ-(30^\circ + 105^\circ) \\ & = 45^\circ. \end{aligned}$$Next, using the law of sines, we obtain
$$\begin{aligned} \dfrac{BC}{\sin A} & = \dfrac{AC}{\sin B} \\ AC & = \dfrac{BC}{\sin A} \cdot \sin B \\ AC & = \dfrac{10}{\sin 30^\circ} \cdot \sin 45^\circ \\ AC & = \dfrac{10}{\cancel{\frac12}} \cdot \cancel{\dfrac12}\sqrt2 \\ AC & = 10\sqrt2~\text{cm}. \end{aligned}$$Therefore, the length of $AC$ is $\boxed{10\sqrt2~\text{cm}}.$
Given triangle $PQR$ with $PR=3$ cm and $QR=\dfrac{3\sqrt6}{2}$ cm, and $\angle P=60^\circ.$ Determine the measure of angle $R.$
The following information is known.
- $PR = 3$ cm.
- $QR=\dfrac{3\sqrt6}{2}$ cm.
- $\angle P = 60^\circ.$
Using the law of sines, we will first determine $\angle Q.$
$$\begin{aligned} \dfrac{PR}{\sin Q} & = \dfrac{QR}{\sin P} \\ \sin Q & = \dfrac{\sin P}{QR} \cdot PR \\ \sin Q & = \dfrac{\sin 60^\circ}{\frac{\cancel{3}\sqrt6}{2}} \cdot \cancel{3} \\ \sin Q & = \dfrac{\frac{1}{\cancel{2}}\sqrt3}{\frac{\sqrt6}{\cancel{2}}} \\ \sin Q & = \dfrac{\sqrt3}{\sqrt6} = \dfrac12\sqrt2. \end{aligned}$$This means that the possible values of angle $Q$ are $45^\circ$ or $135^\circ.$ However, $\angle P = 60^\circ,$ so the sum of the interior angles of the triangle would exceed $180^\circ$ if $\angle Q = 135^\circ.$ Therefore, $\angle Q = 45^\circ.$
Next, to determine $\angle R,$ use the fact that the sum of the interior angles of a triangle is always $180^\circ.$
$$\begin{aligned} \angle R & = 180^\circ-(\angle P + \angle Q) \\ & = 180^\circ-(60^\circ + 45^\circ) \\ & = 180^\circ-105^\circ \\ & = 75^\circ. \end{aligned}$$Therefore, the measure of angle $R$ is $\boxed{75^\circ}.$
In triangle $ABC,$ it is known that $\angle A = 45^\circ,$ $AC = 2$ cm, and $BC = 2\sqrt2$ cm. Determine the value of $\cos B.$
The following information is given.
- $\angle A = 45^\circ.$
- $AC = 2$ cm.
- $BC = 2\sqrt2$ cm
Using the law of sines, we will first determine the value of $\sin B.$
$$\begin{aligned} \dfrac{BC}{\sin A} & = \dfrac{AC}{\sin B} \\ \sin B & = \dfrac{\sin A}{BC} \cdot AC \\ \sin B & = \dfrac{\sin 45^\circ}{\cancel{2}\sqrt2} \cdot \cancel{2} \\ \sin B & = \dfrac{\frac12\cancel{\sqrt2}}{\cancel{\sqrt2}} \\ \sin B & = \dfrac12. \end{aligned}$$This means that the possible values of angle $B$ are $30^\circ$ or $150^\circ.$ However, $\angle A = 45^\circ,$ so the sum of the angles in the triangle would exceed $180^\circ$ if $\angle B = 150^\circ.$ Therefore, $\angle B = 30^\circ.$
Hence, $\cos B = \cos 30^\circ = \dfrac12\sqrt3.$
Therefore, the value of $\cos B$ is $\boxed{\dfrac12\sqrt3}.$
Sigma Boy is initially at point $A.$ On a map with scale $1 : 100,$ he is shown walking $6$ cm toward point $B.$ Turning $30^\circ$ clockwise, he sees a children’s slide located at point $C.$ Help Sigma Boy determine the actual distance (not the distance on the map) from his starting position (point $A$) to the slide (point $C$) if $\angle ACB=120^\circ.$
The following information is given.
- $AB = c = 6$ cm.
- $\angle B = 30^\circ.$
- $\angle C = 120^\circ.$
Using the law of sines, we will determine the length of $AC = b$ on the map.
$$\begin{aligned} \dfrac{b}{\sin B} & = \dfrac{c}{\sin C} \\ b & = \dfrac{c}{\sin C} \cdot \sin B \\ b & = \dfrac{6}{\sin 120^\circ} \cdot \sin 30^\circ \\ b & = \dfrac{6}{\frac12\sqrt3} \cdot \dfrac12 \\ b & = \dfrac{6}{\sqrt3} = 2\sqrt3~\text{cm}. \end{aligned}$$On the map, the distance from Sigma Boy’s starting position to the slide is $2\sqrt3$ cm. Since the map uses a scale of $1 : 100,$ the actual distance is $2\sqrt3 \times 100 = 200\sqrt3$ cm, or $2\sqrt3$ m.
Therefore, the actual distance (not the distance on the map) from Sigma Boy’s starting position (point $A$) to the slide (point $C$) is $\boxed{2\sqrt3~\text{m}}.$
In a triangle $ABC$, angle $\angle C$ is three times angle $\angle A$ and angle $\angle B$ is twice angle $\angle A.$ Determine the ratio of the lengths $AB$ and $BC.$
In a triangle, the sum of the angles is always $180^{\circ}$ so we can write $\angle A + \angle B + \angle C = 180^{\circ}.$
Since $\angle C = 3\angle A$ and $\angle B = 2\angle A,$ we obtain
$\begin{aligned} \angle A + 2 \angle A + 3 \angle A & = 180^{\circ} \\ 6 \angle A & = 180^{\circ} \\ \angle A & = 30^{\circ}. \end{aligned}$
This means that $\angle B = 60^{\circ}$ and $\angle C = 90^{\circ}.$ Therefore, triangle $ABC$ is a right triangle at $C.$
Using the law of sines, we obtain
$\begin{aligned} \dfrac{AB}{\sin C} & = \dfrac{BC}{\sin A} \\ \dfrac{AB}{BC} & = \dfrac{\sin C}{\sin A} \\ AB : BC & = \sin 90^{\circ} : \sin 30^{\circ} \\ & = 1 : \dfrac12 = 2 : 1. \end{aligned}$
Therefore, the ratio of the lengths $AB$ and $BC$ is $\boxed{2 : 1}.$
Prove that in triangle $ABC,$ the relation $\dfrac{a-b}{c} = \dfrac{\sin A-\sin B}{\sin C}$ holds.
In triangle $ABC$, the following relation holds: $\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}.$
In another form, this can be written as
$a = \dfrac{b \sin A}{\sin B}$ and $c = \dfrac{b \sin C}{\sin B}.$
Thus, we obtain
$\begin{aligned} \dfrac{a-b}{c} & = \dfrac{\dfrac{b \sin A}{\sin B}-b}{\dfrac{b \sin C}{\sin B}} \\ & = \dfrac{\dfrac{b \sin A}{\cancel{\sin B}}-\dfrac{b \sin B}{\cancel{\sin B}}}{\dfrac{b \sin C}{\cancel{\sin B}}} \\ & = \dfrac{b \sin A-b \sin B}{b \sin C} \\ & = \dfrac{\sin A-\sin B}{\sin C}. \end{aligned}$
Therefore, it is proven that in triangle $ABC,$ the relation $\dfrac{a-b}{c} = \dfrac{\sin A-\sin B}{\sin C}$ holds.
In $\triangle ABC$, prove that
$$c(\sin^2 A + \sin^2 B) = \sin C(a \cdot \sin A + b \cdot \sin B).$$
We will prove
$$\underbrace{c(\sin^2 A + \sin^2 B)}_{\text{left-hand side}} = \underbrace{\sin C(a \cdot \sin A + b \cdot \sin B)}_{\text{right-hand side}}.$$Use the law of sines.
$$\boxed{\dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}}$$From the equation above, we obtain
$$\begin{aligned} \sin A & = \dfrac{a \sin B}{b} \\ \sin B & = \dfrac{b \sin A}{a}. \end{aligned}$$Next, the proof will begin from the left-hand side.
$$\begin{aligned} c(\sin^2 A + \sin^2 B) & = c\left(\sin A \cdot \sin A + \sin B \cdot \sin B\right) \\ & = c\left(\dfrac{a \sin B}{b} \cdot \sin A + \dfrac{b \sin A}{a} \cdot \sin B\right) \\ & = c\left(\dfrac{\sin B}{b} \cdot (a \sin A) + \dfrac{\sin A}{a} \cdot (b \sin B)\right) \\ & = c\left(\dfrac{\sin C}{c} \cdot (a \sin A) + \dfrac{\sin C}{c} \cdot (b \sin B)\right) && (\text{Law of sines}) \\ & = \cancel{c} \cdot \dfrac{\sin C}{\cancel{c}}(a \sin A + b \sin B) && (\text{Factored}) \\ & = \underbrace{\sin C(a \cdot \sin A + b \cdot \sin B)}_{\text{right-hand side}}. \end{aligned}$$Thus, it is proven that in $\triangle ABC$, the following identity holds: $$c(\sin^2 A + \sin^2 B) = \sin C(a \cdot \sin A + b \cdot \sin B).$$