# Solution of Triangle

There are three sides and three angles in a triangle. They are called the six elements of a triangle. Given any three elements (at least one side) in any triangle, remaining three elements can be determined. This process of finding the unknown three elements is called solution of triangle.

If each of the three parts given is an angle, then in this case only the ratios of the sides can be determined. And hence, there exists an infinite number of triangles with the same set of angles.

In solving a triangle, the following different cases arise:

1. Three angles given
2. Three sides given
3. Two angles and one side given
4. Two sides and the included angle given
5. Two sides and an angle opposite to one of them is given

Let $ABC$ be a triangle. Let us denote the angle $BAC$, $CBA$, $ACB$ of the triangle $ABC$ by $A$, $B$, $C$ respectively; and the lengths of the sides $BC$, $CA$, $AB$ by $a$, $b$, $c$ respectively. Let $s$ be the semi perimeter of the triangle $ABC$. $\therefore s=\frac{a+b+c}{2}$

## Case I: Given Three Angles

In this case, no unique solution is possible. We can only determine the ratios of the sides. From sine law, we know, $a:b:c=\sin A:\sin B:\sin C$ Hence, only the shape of the triangle can be determined but not the size.

## Case II: Given Three Sides

Applying any one of the following formulae, we can determine the angle $A$, $\cos A=\frac{b^2+c^2-a^2}{2bc}$ $\cos\frac{A}{2}=\sqrt{\frac{s(s-a)}{bc}}$ $\sin A=\frac{2}{bc}\sqrt{s(s-a)(s-b)(s-c)}$ $\sin\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$ $\tan\frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}$

Once the value of $A$ is obtained, we can now determine the angle $B$ from the relation $\sin A:\sin B=a:b$ Then, angle $C$ may be found from the relation $C=π-(A+B)$

## Case III: Given Two Angles and a Side

Let $A$ and $B$ be two given angles and $a$ be the given side. Then, the angle $C$ can be obtained from the relation $C=π-(A+B)$

Now, by using the sine law, we can determine the other two sides $b$ and $c$ i.e. $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$

## Case IV: Given Two Sides and an Included Angle

Let $b$ and $c$ be the two given sides and the angle be $A$. Then, we may find the side $a$ by using the cosine law, $a^2=b^2+c^2-2bc\cos A$

Then, we can determine $B$ from $\frac{a}{\sin A}=\frac{b}{\sin B}$ Finally, we can find $C$ from $C=π-(A+B)$

## Case V: Given Two Sides and an Opposite Angle

Let $b$ and $c$ be the given two sides and $B$ be the given angle. Then, we can determine the side $c$ from the sine law, $\frac{b}{\sin B}=\frac{c}{\sin C}$ Then, we can find $A$ from $A=π-(B+C)$

Once the value of $A$ is obtained, we can find the side $a$ from $\frac{a}{\sin A}=\frac{b}{\sin B}$

Thus, the solution of the triangle depends upon the possibility the determination of $c$ from $\frac{b}{\sin B}=\frac{c}{\sin C}$ $\text{or,}\;\;\sin C=\frac{c\sin B}{b}$

In the determination of $c$, the following three cases arise:

### Case I:

If $c\sin B>b$, then $\sin C>1$ which is not possible and hence no triangle is possible.

### Case II:

If $c\sin B=b$, then $\sin C=1$ or $C=90°$. Then, a right angled triangle is the required solution. Then, $A=\frac{π}{2}-B$ and, we can determine $a$ from Pythagorean theorem, $a=\sqrt{c^2-b^2}$

### Case III:

If $c\sin B<b$, then two supplement values of $C$ are possible, one acute and one obtuse. In this case, the following three sub-cases arise:

#### Sub-case I:

If $c<b$, then $C<B$, hence $C$ must be acute and only one triangle is possible.

#### Sub-case II:

If $c=b$, then $C=B$, hence $C$ can not be obtuse otherwise there will be two obtuse angles in a triangle. So, only one triangle is possible.

#### Sub-case III:

If $c>b$, then the angle $C$ is not restricted and it can have either acute or obtuse value as long as $B$ is less than $90°$. Since $C$ can have either acute or obtuse value, two triangles are possible with given parts.

This is usually known as the Ambiguous Case in the solution of triangle.

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