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Law of Sines

TheLaw of Sinesis the relationship between the sides and angles of non-right (oblique)triangles. Simply, it states that the ratio of the length of a side of a triangle to the sine of the angle opposite that side is the same for all sides and angles in a given triangle.

在$\Delta ABC$is an oblique triangle with sides$a,b$and$c$, then$\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$.

To use the Law of Sines you need to know either two angles and one side of the triangle (AAS or ASA) or two sides and an angle opposite one of them (SSA). Notice that for the first two cases we use the same parts that we used to prove congruence of triangles in geometry but in the last case we could not provecongruent trianglesgiven these parts. This is because the remaining pieces could have been different sizes. This is called the ambiguous case and we will discuss it a little later.

The third angle of the triangle is

$m\angle C=180°-m\angle A-m\angle B=180°-30°-20°=130°$

由Law of Sines,

$\frac{45}{\sin 30°}=\frac{b}{\sin 20°}=\frac{c}{\sin 130°}$

由Properties ofProportions

$b=\frac{45\sin 20°}{\sin 30°}\approx 30.78\text{m}$and$c=\frac{45\sin 130°}{\sin 30°}\approx 68.94\text{m}$

The third angle of the triangle is:

$m\angle C=180°-m\angle A-m\angle B=180°-42°-75°=63°$

由Law of Sines,

$\frac{a}{\sin 42°}=\frac{b}{\sin 75°}=\frac{22}{\sin 63°}$

由Properties of Proportions

$a=\frac{22\sin 42°}{\sin 63°}\approx 16.52\text{cm}$and$b=\frac{22\sin 75°}{\sin 63°}\approx 23.85\text{cm}$

The Ambiguous Case

If two sides and an angle opposite one of them are given, three possibilities can occur.

(1) No such triangle exists.

(2) Two different triangles exist.

(3) Exactly one triangle exists.

Consider a triangle in which you are given$a,b$and$A$. (The altitude$h$from vertex$B$to side$\overline{AC}$, by the definition of sines is equal to$b\sin A$.)

(1) No such triangle exists if$A$is acute and$a<h$or$A$is obtuse and$a\le b$.

(2) Two different triangles exist if$A$is acute and$h<a<b$.

(3) In every other case, exactly one triangle exists.

$h=b\sin A=25\sin 80°\approx 24.6$

Notice that$a<h$. So it appears that there is no solution. Verify this using the Law of Sines.

$\begin{array}{l}\frac{a}{\sin A}=\frac{b}{\sin B}\\ \frac{15}{\sin 80°}=\frac{25}{\sin B}\\ \sin B=\frac{25\sin 80°}{15}\approx 1.641>1\end{array}$

This contrasts the fact that the$-1\le \sin B\le 1$. Therefore, no triangle exists.

$h=b\sin A=7\sin 30°=3.5$

$h<a<b$therefore, there are two triangles possible.

由Law of Sines,$\frac{a}{\sin A}=\frac{b}{\sin B}$

$\sin B=\frac{b\sin A}{a}=\frac{7\sin 30°}{6}\approx 0.5833$

There are two angles between$0°$and$180°$whose sine is approximately 0.5833, are$35.69°$and$144.31°$.

$\begin{array}{cccc}\begin{array}{l}\text{If}B\approx 35.69°\\ C\approx 180°-30°-35.69°=114.31°\\ c=\frac{a\sin C}{\sin A}\approx \frac{6\sin 114.31°}{\sin 30°}\approx 10.94\end{array}& & & \begin{array}{l}\text{If}B\approx 144.31°\\ C\approx 180°-30°-144.31°=5.69°\\ c\approx \frac{6\sin 5.69°}{\sin 30°}\approx 1.19\end{array}\end{array}$

$a>b$

由Law of Sines,$\frac{a}{\sin A}=\frac{b}{\sin B}$

$\begin{array}{l}\sin B=\frac{b\sin A}{a}=\frac{12\sin 40°}{22}\approx 0.3506\\ B\approx 20.52°\end{array}$

$B$is acute.

$m\angle C=180°-m\angle A-m\angle B=180°-40°-20.52°=29.79°$

由Law of Sines,

$\begin{array}{l}\frac{c}{\sin 119.48°}=\frac{22}{\sin 40°}\\ c=\frac{22\sin 119.48°}{\sin 40°}\approx 29.79\end{array}$

If we are given two sides and an included angle of a triangle or if we are given$3$sides of a triangle, we cannot use the Law of Sines because we cannot set up any proportions where enough information is known. In these two cases we must use theLaw of Cosines.