Solution of Triangles

The solution of a triangle involves finding its various elements such as angles, side lengths, and area, typically using trigonometric principles such as the law of sines and the law of cosines. This process is crucial in geometry and real-world applications, aiding in navigation, surveying, and engineering.

1.0Sine Rule

The Sine Rule, also known as the Law of Sines, is a fundamental principle in trigonometry used to relate the ratios of the sides of a triangle to the sines of its angles. Here's a detailed explanation:

Sine Rule Formula

In a triangle defined by the lengths of its sides a, b and c, along with the corresponding angles A, B & C, the Sine Rule asserts:

Key Points of Sine Rule

1. Equality of Ratios: The ratios of the lengths of the sides of a triangle to the sine of their corresponding angles are equal.
2. Ambiguity Case: The Sine Rule can be applied to solve triangles when:

• Given: ASA (Angle-Side-Angle) or SSA (Side-Side-Angle), where the angle given is not the angle between the given sides (ambiguous case).
• In the ambiguous case, there may be two possible triangles, one acute and one obtuse.

3. Finding Unknown Angles or Sides:

• Given two sides and a non-included angle (SSA), the Sine Rule can be used to find the remaining side or angle.
• Given one side and two angles (ASA), the Sine Rule can also be used to find the remaining sides and angles.

Example of Sine Formula

Angles of a triangle are in 4: 1: 1 ratio. The ratio between its greatest side and perimeter is -

(A) $\frac{3}{2+\sqrt{3}}$ (B) $\frac{\sqrt{3}}{2+\sqrt{3}}$

(C) $\frac{\sqrt{3}}{2-\sqrt{3}}$ (D) $\frac{1}{2+\sqrt{3}}$

Ans. (B)

Solution: Angles are in ration 4 : 1 : 1.

⇒ Angles are 120°, 30°, 30°.

If sides opposite to these angles are a, b, c respectively, then a will be the greatest side. Now from sine formula 

$\frac{a}{\sin 120^{\circ }}=\frac{b}{\sin 30^{\circ }}=\frac{c}{\sin 30^{\circ }}$

$\Rightarrow \frac{\mathrm{a}}{\sqrt{3}/2}=\frac{\mathrm{b}}{1/2}=\frac{\mathrm{c}}{1/2}$

$\Rightarrow \frac{\mathrm{a}}{\sqrt{3}}=\frac{\mathrm{b}}{1}=\frac{\mathrm{c}}{1}=\mathrm{k}\text{ (say) }$

Then a=$\sqrt{3}k$, perimeter =$(2+\sqrt{3})\mathrm{k}$

∴ required ratio = $\frac{\sqrt{3}k}{(2+\sqrt{3})k}=\frac{\sqrt{3}}{2+\sqrt{3}}$

The Sine Rule provides a powerful tool for solving triangles and understanding their properties, contributing significantly to various fields of mathematics and practical applications.

2.0Cosine Rule

The Cosine Rule, also known as the Law of Cosines, is a fundamental trigonometric principle used to relate the lengths of the sides of a triangle to the cosine of one of its angles.

Cosine Rule Formula

In a triangle characterized by the lengths of its sides, labeled as a, b, and c, and the angle opposite side c denoted as C, the Cosine Rule states:

For any triangle with sides a, b, and c, and the angle opposite side c denoted as C, the Cosine Rule states:

(a) $\cos \mathrm{C}=\frac{{\mathrm{a}}^2+{\mathrm{b}}^2-{\mathrm{c}}^2}{2\mathrm{ab}}$ or c² = a² + b² − 2ab cosC

Similarly, this rule can be expressed for the other sides of the triangle:

(b) $\cos B=\frac{c^2+a^2-b^2}{2ca}$ or b² = a² + c² − 2ac cosB

(c) $\cos A=\frac{b^2+c^2-a^2}{2bc}$ or a² = b² + c² − 2bc cosA

Key Points of Cosine Rule

1. Generalization of Pythagoras' Theorem: The Cosine Rule extends the concept of Pythagoras' Theorem to non-right-angled triangles.
2. Finding Unknown Side Lengths: Given two sides and the included angle (SAS), or all three sides (SSS), the Cosine Rule can be applied to find the length of the remaining side.
3. Finding Unknown Angles: Given all three sides, the Cosine Rule can be rearranged to find any of the angles using inverse trigonometric functions.
4. Ambiguity Case: Unlike the Sine Rule, there is no ambiguity when using the Cosine Rule to solve triangles.

Example of Cosine Rule

In a triangle ABC, If B = 30° and $\mathbf{C}=\sqrt{3}\mathbf{b}$, then A can be equal to

(A) 45° (B) 60° (C) 90° (D) 120°

Solution:

We have

$cos\mathrm{B}=\frac{c^2+a^2-b^2}{2ca}\Rightarrow \frac{3}{2}=\frac{3b^2+a^2-b^2}{2\times \sqrt{3}b\times a}$

⇒ a² – 3ab + 2b² = 0 

⇒ (a – 2b) (a – b) = 0

⇒ Either a = b 

⇒ A = 30° or a = 2b 

⇒ a² = 4b² = b² + c² 

⇒ A = 90°

Ans. (C)

The Cosine Rule provides a powerful tool for solving triangles and understanding their properties, contributing significantly to various fields of mathematics and practical applications.

3.0Law of Tangents

The Law of Tangents is a trigonometric law used to determine the length of a side that is not known or to ascertain the measure of an angle that is unknown When two sides of a triangle and the angle between them are provided, or when two angles and a non-included side are given. It is particularly useful when the Law of Sines or the Law of Cosines cannot be directly applied. Here is how it works:

Law of Tangents Formula

In a triangle characterized by the lengths of its sides, labeled as a, b, and c, and the angles opposite those sides denoted as A, B, and C, the Law of Tangents states:

Napier’s Analogy (Tangent Rule)

(a) $\tan \left(\frac{\mathrm{B}-\mathrm{C}}{2}\right)=\frac{\mathrm{b}-\mathrm{c}}{\mathrm{b}+\mathrm{c}}\cot \frac{\mathrm{A}}{2}$

(b) $\tan \left(\frac{\mathrm{C}-\mathrm{A}}{2}\right)=\frac{\mathrm{c}-\mathrm{a}}{\mathrm{c}+\mathrm{a}}\cot \frac{\mathrm{B}}{2}$

(c)  $\tan \left(\frac{\mathrm{A}-\mathrm{B}}{2}\right)=\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}\cot \frac{\mathrm{C}}{2}$

$\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\frac{\tan \left(\frac{1}{2}(\mathrm{A}-\mathrm{B})\right)}{\tan \left(\frac{1}{2}(\mathrm{A}+\mathrm{B})\right)}$

Key Points Law of Tangents

1. Usage: The Law of Tangents is typically used when you have two sides and the included angle or when you have two angles and a non-included side.
2. Ambiguity: Like the Law of Sines, the Law of Tangents has an ambiguity case where multiple triangles may be possible or no triangle at all.
3. Finding Unknown Sides or Angles: The Law of Tangents can be rearranged to solve for an unknown side or angle by algebraic manipulation.

Example of Tangents Formula

Given a triangle with sides a = 6, b = 8 and A = 45°, find the length of side c.

Using the Law of Tangents:

$\frac{6-8}{6+8}=\frac{\tan \left(\frac{1}{2}\left(45^{\circ }-\mathrm{B}\right)\right)}{\tan \left(\frac{1}{2}\left(45^{\circ }+B\right)\right)}$

Solving for B, we find ≈ 64.5°.

Once B is found, C can be calculated using C = 180° – A –B.

Finally, the length of side C can be found using the Law of Sines or the Law Cosines.

The Law of Tangents provides an alternative method for solving triangles and can be especially useful in certain scenarios where other trigonometric laws may not be directly applicable.

4.0Projection Rule/ Projection Formulae

1. b cos C + c cos B = a
2. c cos A + a cos C = b
3. a cos B + b cos A = c

5.0Area of Triangle

$\Delta =\sqrt{\mathrm{s}(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})(\mathrm{s}-\mathrm{c})}$

$=\frac{1}{2}\mathrm{bc}\sin \mathrm{A}=\frac{1}{2}\mathrm{ca}\sin \mathrm{B}=\frac{1}{2}\mathrm{ab}\sin \mathrm{C}=\frac{1}{2}{\mathrm{ap}}_1=\frac{1}{2}{\mathrm{bp}}_2=\frac{1}{2}{\mathrm{cp}}_3$,

Where p₁, p₂, p₃ are altitudes from vertices A, B, C respectively.

6.0Trigonometric Ratios of Half Angle Formulae

$\mathrm{s}=\frac{\mathrm{a}+\mathrm{b}+\mathrm{c}}{2}$ = semi-perimeter of triangle.

1. (i) $\sin \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{bc}}$

(ii) $\sin \frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{ca}}$

(iii) $\sin \frac{C}{2}=\sqrt{\frac{(s-a)(s-b)}{ab}}$

2. (i) $\cos \frac{\mathrm{A}}{2}=\sqrt{\frac{s(\mathrm{s}-\mathrm{a})}{\mathrm{bc}}}$

(ii) $\cos \frac{\mathrm{B}}{2}=\sqrt{\frac{\mathrm{s}(\mathrm{s}-\mathrm{b})}{\mathrm{ca}}}$

(iii) $\cos \frac{\mathrm{C}}{2}=\sqrt{\frac{s(s-c)}{ab}}$

3. (i) $\tan \frac{A}{2}=\sqrt{\frac{(s-b)(s-c)}{s(s-a)}}=\frac{\Delta }{s(s-a)}$

(ii) $\tan \frac{B}{2}=\sqrt{\frac{(s-c)(s-a)}{s(s-b)}}=\frac{\Delta }{s(s-b)}$

(iii) $\tan \frac{\mathrm{C}}{2}=\sqrt{\frac{(\mathrm{s}-\mathrm{a})(\mathrm{s}-\mathrm{b})}{\mathrm{s}(\mathrm{s}-\mathrm{c})}}=\frac{\Delta }{\mathrm{s}(\mathrm{s}-\mathrm{c})}$

7.0Solutions of Triangles Formulas

Solution of Triangles Formulas are-

1. Sine Rule:  $\frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}$
2. Cosine Rule:

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

$\cos B=\frac{c^2+a^2-b^2}{2ca}$

$\cos C=\frac{a^2+b^2-c^2}{2ab}$

3. $\frac{\mathrm{a}-\mathrm{b}}{\mathrm{a}+\mathrm{b}}=\frac{\tan \left(\frac{1}{2}(\mathrm{A}-\mathrm{B})\right)}{\tan \left(\frac{1}{2}(A+B)\right)}$

8.0Solutions for Area of Triangle

Question 1:  In a Δ ABC if c.sin A (a. cos C + c. cos A) = 100, then area of  ΔABC is

Solution : IF a, b and c are the side opposite to ∠ A, ∠ B and ∠ C respectively.

Projection rule : a cos C + Cos A = b 

Area of triangle Δ = $\frac{1}{2}bcsinA$

Given that, C sin A (a cos C + C. cos A) = 100

∴ a cos c + c. cos A = b

⇒ bc sin A = 100

$\Rightarrow \frac{1}{2}\mathrm{bc}\sin \mathrm{A}=\frac{100}{2}=50$

Hence, area of triangle is 50 sq. unit

Question 2: If a ΔABC if c² = a² + b² , then Find 4 s(s–a) (s–b) (s–c)

Solution: If c² = a²+ b²  then ΔABC is a right angled at C (By Pythagoras theorem)

∴ Area of triangle =$\frac{1}{2}ab\sin c=\frac{1}{2}ab$

⇒ Area of triangle =$\sqrt{s(s-a)(s-b)(s-c)}$

$\Rightarrow \sqrt{s(s-a)(s-b)(s-c)}=\frac{1}{2}ab$

$\Rightarrow s(s-a)(s-b)(s-c)=\frac{a^2{b}^2}{4}$

⇒ 4s (s–a) (s–b) (s–c) = a²b²   Ans.

Question 3: The angles A, B and C of a triangle ABC are in A.P and a : b = 1: $\sqrt{3}$

If c = 4 cm, then the area (in sq cm) of the triangle is

Solution: Given that the angles A, B and C of a triangle ABC are in AP, then

2B = A + C ………………..(1)

By angle sum property of a triangle the sum of all angles in a triangle is equal to 180°.

⇒ ∠A + ∠ B + ∠ C = 180°

⇒ 2∠ B + ∠ B = 180°        (From (1))

⇒ 3 ∠ B = 180°

⇒∠B =$\frac{180^{\circ }}{3}$ = 60°

Therefore ∠ B = 60°

Now we have to find ∠A

Using sine formula

$\Rightarrow \frac{a}{\sin A}=\frac{b}{\sin B}$

$\Rightarrow \frac{a}{b}=\frac{\sin A}{\sin B}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sin \mathrm{A}}{\sin 60^{\circ }}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{\sin \mathrm{A}}{\frac{\sqrt{3}}{2}}$

$\Rightarrow \frac{1}{\sqrt{3}}=\frac{2\sin \mathrm{A}}{\sqrt{3}}$

⇒ 2 sin A = 1.

$\Rightarrow \sin A=\frac{1}{2}$

⇒ sin A = sin 30°

⇒∠Α = 30° 

Now we have to find ∠ C

By angle sum property

⇒∠ A + ∠ B + ∠ C = 180°

⇒ 30° + 60° + ∠C = 180°

⇒ ∠ C = 180° – 90° = 90° 

Therefore ∠C = 90°

Now we have to find area of a triangle

Using sine formula

$\Rightarrow \frac{a}{\sin A}=\frac{c}{\sin C}$

$\Rightarrow \frac{a}{c}=\frac{\sin A}{\sin C}$

$\Rightarrow \frac{a}{4}=\frac{\sin 30^{\circ }}{\sin 90^{\circ }}$

$\Rightarrow \frac{\mathrm{a}}{4}=\frac{\frac{1}{2}}{1}$

$\Rightarrow a=\frac{4}{2}=2\mathrm{cm}$

Also, $\frac{\mathrm{a}}{\mathrm{b}}=\frac{1}{\sqrt{3}}$

$\frac{2}{b}=\frac{1}{\sqrt{3}}\Rightarrow \mathrm{b}=2\sqrt{3}\mathrm{cm}$

Now Area of triangle =$\frac{1}{2}\sin \mathrm{c}$

=$\frac{1}{2}\times 2\times 2\sqrt{3}\times \sin 90^{\circ }$

=$2\sqrt{3}{\mathrm{cm}}^2$ Ans.