## Relation between the sides and the angles of a triangle, laws of sines and cosines

Except Pythagoras theorem there is little help in the area of establishing a relationship between the sides and the angles of a triangle. The laws of sines and the laws of cosines meet this important gap.

In this session we will go through the mechanisms of these two important group of concepts.

### Laws of sines

This law states,

The three ratios of a side and the sine function of its opposite angle in a triangle will be equal to each other.

Let us explain this group of laws with the aid of the following figure.

In the above figure by convention we call the sides in short names as,

$AB=c$,

$BC=a$, and

$CA=b$.

The internal angles also are referred to as,

$\angle BAC=\angle A$,

$\angle CBA=\angle B$, and

$\angle ACB=\angle C$.

$AD$ is the perpendicular from A to the opposite side BC.

According to the laws of sines the ratios of sides to sines of corresponding opposite angles are equal. In expression form this means,

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

We will first see how the ratio equality $\displaystyle\frac{b}{sin B}=\frac{c}{sin C}$ happens.

#### The mechanism or proof of laws of sines

In right $\triangle ABD$,

$sin B=\displaystyle\frac{AD}{AB}=\frac{AD}{c}$.

Using the common element AD in two right triangles $\triangle ABD$ and $\triangle ACD$ as link element (which links the two triangles), in right $\triangle ACD$,

$sin C=\displaystyle\frac{AD}{AC}=\frac{AD}{b}$.

Taking the ratio of the two and eliminating AD in the process (which was the purpose of using AD as link element in the first place),

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

Or, $\displaystyle\frac{b}{sin B}=\frac{c}{sin C}$.

Similarly using perpendicular BE as the link element we will get,

$\displaystyle\frac{a}{sin A}=\frac{c}{sin C}$, and from these two equations finally,

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

In solving many problems involving sides and sines of angles, laws of sines prove to be an extremely important problem solving resource.

### Laws of cosines

Using the terminologies of the above figure, we can state these laws in the form of,

$c^2=a^2+b^2-2ab \text{cosC}$,

$a^2=b^2+c^2-2bc \text{cosA}$, and

$b^2=c^2+a^2-2ca \text{cosB}$.

These are laws in squared sides and so intuitively can be assumed to have the Pythagoras relation embedded in them.

Let us see the mechanisms behind this group of laws.

#### Proof of laws of cosines based on compound angle concept

We will use a figure very similar to the above for explaining the mechanisms behind the laws.

In $\triangle ABD$

$BD=x=c \text{cosB}$,

And in $\triangle ACD$,

$CD=y=b \text{cosC}$, where $x+y=a$.

Adding the squaring,

$(BD+CD)^2=BC^2=a^2=b^2cos^2C+2bc\text{cosB}\text{cosC}+c^2 cos^2B$,

Or, $a^2=b^2(1-sin^2 C)+2bc\text{cosB}\text{cosC}+c^2(1-sin^2 B)$

$=b^2+c^2-(b^2sin^2C-2bc\text{cosB}\text{cosC}+c^2sin^2B)$.

From laws of sines,

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

Or, $b\text{sinC}=c\text{sinB}$.

So,

$b^2sin^2C=b sin C\times{b sin C}=b sin C\times{c sin B}=bc\text{sinB}\text{sinC}$.

Similarly,

$c^2sin^2B=bc\text{sinB}\text{sinC}$.

using these results we get,

$a^2=b^2+c^2-(2bc\text{sinB}\text{sinC}-2bc\text{cosB}\text{cosC})$

$=b^2+c^2+2bc(cos B cos C-sin B sin C)$

$=b^2+c^2+2bc\text{cos(B+C)}$

$=b^2+c^2+2bc cos(\pi-A)$

Or, $a^2=b^2+c^2-2bc\text{cosA}$, as $cos(\pi-A)=-cos A$.

In this proof we have used both parts $x$ and $y$ of CD as well as rich concepts of compound angle function expressions and function sign change. We have not used the Pythagoras theorem, but again, this core theorem helps to create Trigonometry itself.

#### Proof of laws of cosines based on Pythagoras theorem and link element use technique

Using the same figure we will now use a different method to arrive at the other two laws of cosines,

$b^2=c^2+a^2-2ca\text{cosB}$, and

$c^2=a^2+b^2-2ab\text{cosC}$.

Here we will use one single variable $x$ or $y$ and Pythagoras theorem along with link element use technique and substitution technique.

Let us first prove,

$b^2=c^2+a^2-2ca\text{cosB}$.

To establish this relation we will use $x$ as the single unknown variable and $y$ as, $y=a-x$. We will call the height AD as $h$.

From $\triangle ABD$,

$h=csin B$, and

$x=ccos B$.

From the two,

$h^2=c^2-x^2$.

In $\triangle ACD$,

$b^2=h^2+(a-x)^2$

$=h^2+a^2-2ax+x^2$

$=c^2-x^2+a^2-2ax+x^2$

$=c^2+a^2-2ax$

$=c^2+a^2-2ac\text{cosB}$,

Or, $b^2=c^2+a^2-2ca\text{cosB}$.

Instead of $x$ as the single unknown variable, if we take $y$ as the single unknown variable and start working from $\triangle ACD$ instead of $\triangle ABD$, we will get the third relation following the same method,

$c^2=a^2+b^2-2ab\text{cosC}$.

This second method should generally be felt as the easier one. Nevertheless, to us the first proof is direct and more matured in terms of showing the embedded rich concepts in the laws of cosines.

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