Important Solutions of Triangle Formulas

In this article, we'll learn the key formulas that are essential for understanding triangles. A triangle is more special as compared to other polygons as it is the polygon having the least number of sides. A triangle has six main elements, three sides, and three angles. Triangles are basic yet fascinating shapes in geometry, offering a lot to learn about angles, side lengths, perimeter, and area.

In this article, we will cover the concept of the Important Solutions of Triangle Formulas. This category falls under the broader category of Trigonometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE `.

In a triangle, there are six variables, three sides (say a, b, c), and three angles (say A, B, C). If any three of these six variables (except all the angles A, B, C) is given, then the triangle can be known completely: the other three variables can be found using the formulae we learned in this chapter.

There are different cases that arise when a few components of the triangle are given.

Case 1

When three sides $(a, b,$ and $c)$ of a triangle are given

The remaining variables can be found by using the following formulae

Problems in which two sides and included angles are given

Case 2

When two sides (say $a, b)$ and included angle (angle $C$) of a triangle are given.

Problems in which one side and two angles are given

Case 3

When one side (say $a$) and two angles (say $A$ and $B$) are given

Case 4

When two side $a, b$ and an angle opposite to one of these sides is given (say angle $A$ is given)

Using the sine rule, we get
$
\sin B=\frac{b}{a} \sin A
$

Now following possibilities can occur
1. When $\frac{\mathrm{b}}{\mathrm{a}} \sin \mathrm{A}>1$ or $\mathrm{a}<\mathrm{b} \sin \mathrm{A}$

In the relation $\sin B=\frac{b}{a} \sin A$, which means $\sin B>1$, which is impossible
So, no such triangle exists
2. When $\frac{b}{a} \sin A=1$, then $\sin B=1 \Rightarrow \angle B=90^{\circ}$

So a unique triangle is possible which is right angle triangle with angle $B=90^{\circ}$
3. When $\frac{b}{a} \sin \mathrm{A}<1$ or $\mathrm{a}>\mathrm{b} \sin \mathrm{A}$

Here in $B<1$, which is possible, and hence a triangle will exist
Angle $C$ can be found out using $C=180-A-B$

Important Solutions of Triangle Formulas

Sine Rule

The ratio of the sine of one of the angles to the length of its opposite side will be equal to the other two ratios of the sine of the angle measured to the opposite side.

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

Cosine Rule
For a triangle with angles A, B, and C, and opposite corresponding sides a, b, and c, respectively, the Law of Cosines is given as three equations.

$
\cos A=\frac{b^2+c^2-a^2}{2 b c} \quad \cos B=\frac{a^2+c^2-b^2}{2 a c} \quad \cos C=\frac{a^2+b^2-c^2}{2 a b}
$

Tangent Rule or Napier's Analogy
For any $\triangle A B C$,

$
\begin{aligned}
& \tan \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}} \cot \frac{\mathrm{C}}{2} \\
& \tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}} \cot \frac{\mathrm{A}}{2} \\
& \tan \left(\frac{\mathrm{C}-\mathrm{A}}{2}\right)=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{c}+\mathrm{a}} \cot \frac{\mathrm{B}}{2}
\end{aligned}
$
Projection Formula

In the $\triangle A B C$,
1. $a=c \cos B+b \cos C$
2. $b=c \cos A+a \cos C$
3. $c=b \cos A+a \cos B$

Half-Angle Formula for Sine

$\begin{aligned} & \sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{b c}} \\ & \sin \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{a c}} \\ & \sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{a b}}\end{aligned}$

Half-Angle Formula for Cosine

$
\begin{aligned}
\cos \frac{\mathrm{A}}{2} & =\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{a})}{\mathrm{bc}}} \\
\cos \frac{\mathrm{B}}{2} & =\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{b})}{\mathrm{ac}}} \\
\cos \frac{\mathrm{C}}{2} & =\sqrt{\frac{(\mathrm{s})(\mathrm{s}-\mathrm{c})}{\mathrm{ab}}}
\end{aligned}
$

Half Angle Formula for tan

$
\begin{aligned}
& \tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}} \\
& \tan \frac{B}{2}=\sqrt{\frac{(s-a)(s-c)}{s(s-b)}} \\
& \tan \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{s(s-c)}}
\end{aligned}
$

Area of Triangle Formula
Area of $\Delta \mathrm{ABC}=\Delta=\frac{1}{2} b \cdot \mathrm{c} \sin \mathrm{A}$

$
\begin{aligned}
& =\frac{1}{2} a \cdot \mathrm{b} \sin \mathrm{C} \\
& =\frac{1}{2} c \cdot \mathrm{a} \sin \mathrm{B}
\end{aligned}
$

Area of a triangle in terms of sides (Heron’s Formula)

$
\Delta=\frac{1}{2} b \cdot \mathrm{c} \sin \mathrm{A}=\frac{1}{2} b c \cdot 2 \sin \frac{\mathrm{A}}{2} \cos \frac{\mathrm{A}}{2}
$

use half angle formula

$
\begin{aligned}
& =\frac{1}{2} \cdot b c \cdot \sqrt{\frac{(s-b)(s-c)}{b c}} \sqrt{\frac{s(s-a)}{b c}} \\
& =\sqrt{s(s-a)(s-b)(s-c)}
\end{aligned}
$

Radius of Circumcircle
The radius of the circumcircle of a $\triangle A B C, R$ is given by the law of sines:

$
\mathrm{R}=\frac{a}{2 \sin \mathrm{A}}=\frac{b}{2 \sin \mathrm{B}}=\frac{c}{2 \sin \mathrm{C}}
$

Radius of Incircle
1. In - radius, $r$ is given by $=\frac{\Delta}{s}$
2. $r=(s-a) \tan \frac{\mathrm{A}}{2}=(s-b) \tan \frac{\mathrm{B}}{2}=(s-c) \tan \frac{\mathrm{C}}{2}$
3. $r=4 \mathrm{R} \sin \frac{\mathrm{A}}{2} \sin \frac{\mathrm{B}}{2} \sin \frac{\mathrm{C}}{2}$

Formulae for $r_1, r_2$ and $r_3$
1. $r_I=\frac{\Delta}{s-a}, r_2=\frac{\Delta}{s-b}, r_3=\frac{\Delta}{s-c}$
2. $r_1=s \tan \frac{A}{2}, r_2=s \tan \frac{B}{2}, r_3=s \tan \frac{C}{2}$
3. $r_t=4 R \sin \frac{A}{2} \cos \frac{B}{2} \cos \frac{C}{2}$
$r_2=4 R \cos \frac{A}{2} \sin \frac{B}{2} \cos \frac{C}{2}$
$r_3=4 R \cos \frac{A}{2} \cos \frac{B}{2} \sin \frac{C}{2}$

The distances of the special points from vertices and sides of the triangle

(i) circumcentre (O): $O A=R$ and $O_a=R \cos A$
(ii) Incentre (I): $\mathrm{IA}=\mathrm{r} \operatorname{cosec}(\mathrm{A} / 2)$ and $\mathrm{I}_{\mathrm{a}}=\mathrm{r}$
(iii) Excentre $\left(I_1\right): I_1 A=r_1 \operatorname{cosec}(A / 2)$
(iv) Orthocentre: $H A=2 R \cos A$ and $H_a=2 R \cos B \cos C$

Orthocentre

The triangle formed by joining the feet of the altitudes is called the Pedal Triangle.
(i) Its angles are $\pi-2 A, \pi-2 B$ and $\pi-2 C$.
(ii) The sides are a $\cos A=R \sin 2 A$
$a \cos B=R \sin 2 B$
a $\cos C=R \sin 2 C$
(iii) Circum radii of the triangle $\mathrm{PBC}, \mathrm{PCA}, \mathrm{PAB}$, and ABC are equal.

Excentral Triangle
The triangle formed by joining the three excentres $I_1, I_2$, and $I_3$ of $\triangle A B C$ is called the excentral triangle.
(i) $\triangle A B C$ is the pedal triangle of the $\Delta I_1 I_2 I_3$.
(ii) Angles are $\pi / 2-\mathrm{A} / 2, \pi / 2-\mathrm{B} / 2, \pi / 2-\mathrm{C} / 2$
(iii) Sides are $4 R \cos (A / 2), 4 R \cos (B / 2)$ and $4 R \cos (C / 2)$
(iv) $I I_1=4 R \sin (A / 2), I I_2=4 R \sin (B / 2), I I_3=4 R \sin (C / 2)$
(v) Incentre I of $\triangle A B C$ is the orthocentre of the excentral $\Delta I_1 I_2 I_3$.

Distance between special points:

(i) Distance between the circumcentre and orthocentre $\mathrm{OH}^2=\mathrm{R}^2(1-8 \cos \mathrm{A}$ $\cos B \cos C)$
(ii) Distance between the circumcentre and incentre $\mathrm{OI}^2=\mathrm{R}^2(1-8 \sin (\mathrm{A} / 2)$ $\sin (B / 2) \sin (C / 2)=R^2-2 R r$

Solved Examples Based on Important Solutions of Triangle Formulas

Example 1: In triangle ABC , if $\mathrm{a}=2, \mathrm{~b}=1$ and $\angle C=60^{\circ}$ then find the other two angles.

Solution:

$
\begin{aligned}
& \cos C=\frac{a^2+b^2-c^2}{2 a b} \\
& \frac{1}{2}=\frac{4+1-c^2}{4} \\
& c=\sqrt{3}
\end{aligned}
$

Also, $\frac{\sin A}{a}=\frac{\sin B}{b}=\frac{\sin C}{c}$

$
\begin{aligned}
& \frac{\sin A}{2}=\frac{\sin B}{1}=\frac{\frac{\sqrt{3}}{2}}{\sqrt{3}} \\
& \sin A=1 \Rightarrow A=\frac{\pi}{2} \\
& \sin B=\frac{1}{2} \Rightarrow B=\frac{\pi}{6}
\end{aligned}
$

Hence, the other two angles $\frac{\pi}{6}, \frac{\pi}{2}$

Example 2: Consider a triangle with sides $3^{\prime \prime}, 6^{\prime \prime}$ and $9^{\prime \prime}$. Find the $\angle A$.
Solution
The triangle with sides $3^{\prime \prime}, 6^{\prime \prime}$ and $9^{\prime \prime}$.
Using the law of cosine,

$
\begin{aligned}
& \cos A=\frac{\left(b^2+c^2-a^2\right)}{2 b c} \\
& \cos A=\frac{(36+81-9)}{108} \\
& \cos A=1
\end{aligned}
$

$
A=0^{\circ}
$
Hence, the answer is 0

Example 3: Consider a triangle with sides $5^{\prime \prime}, 8^{\prime \prime}$ and $10^{\prime \prime}$. Find the $\angle A, \angle B$, and $\angle C$.
Solution: Given that, The triangle with sides $5^{\prime \prime}, 8^{\prime \prime}$ and $10^{\prime \prime}$.
Using the law of cosine,

$
\begin{aligned}
& \cos A=\frac{\left(b^2+c^2-a^2\right)}{2 b c} \\
& \cos A=\frac{(64+100-25)}{160} \\
& \cos A=0.7125 \\
& A=44.5^{\circ}
\end{aligned}
$

Using the law of sine to find the $\angle B$,

$
\begin{aligned}
& \frac{\sin B}{b}=\frac{\sin A}{a} \\
& \sin B=\frac{b}{a} \sin A \\
& \sin B=\frac{8}{5} \sin 44.5^{\circ} \\
& \sin B=\frac{8}{5} \times 0.7^{\circ} \\
& \sin B=1.12
\end{aligned}
$

$
B \approx 90^{\circ}
$

The third angle is,

$
\begin{aligned}
& C=180^{\circ}-A-B \\
& C=45.4^{\circ}
\end{aligned}
$

Hence, the answer is 45.4
Example 4: If $2 \sin ^3 x+\sin 2 \mathrm{x} \cos \mathrm{x}+4 \sin \mathrm{x}-4=0$ has exactly 3 solutions in the interval $\left[0, n \frac{\pi}{2}\right], \mathrm{n} \in \mathrm{N}$. Then the roots of the equation $x^2+\mathrm{nx}+(\mathrm{n}-3)=0$ belong to:
Solution: We know $\sin 2 x=2 \sin x \cos x$

$
\begin{aligned}
& 2 \sin ^3 x+2 \sin x \cos ^2 x+4 \sin x-4=0 \\
& \cos ^2 x=1-\sin ^2 x \\
& 2 \sin ^3 x+2 \sin x\left(1-\sin ^2 x\right)+4 \sin x-4=0 \\
& 2 \sin ^3 x+2 \sin x-2 \sin ^3 x+4 \sin x-4=0 \\
& 6 \sin x-4=0 \\
& 6 \sin x=4
\end{aligned}
$

sin x =$2/3$

$
\begin{aligned}
& x^2+5 x+2=0 \\
& \mathrm{a}, \mathrm{b}=\frac{-5 \pm \sqrt{17}}{2} \text { both negative hence }(- \text { infinity, } 0)
\end{aligned}
$

Hence, the answer is ( - infinity, 0 )
Example 5: The area of the circle in which a chord of length $\sqrt{2}$ makes an angle $\pi / 2$ at the center is
Solution: Let AB be the chord of length, O be the center of the circle, and let OC be the perpendicular from O on AB .
Then $\mathrm{AC}=\mathrm{BC}=\sqrt{2} / 2=1 / \sqrt{2}$

$
\text { In } \triangle O B C, \mathrm{OB}=\mathrm{BC} \operatorname{cosec} 450=(1 / \sqrt{2}) \times \sqrt{2}=1
$

Area of the circle $=\square(\mathrm{OB}) 2=\pi$
Hence, the answer is $\pi$

Summary

Application of the formula of trigonometry is important. As it gives us an idea about the use of different formulas in different cases. We get to know about different situations where we can use the formulas and find the values.