## Projection formulas

The solution of triangles is the most important trigonometric problem to find out the characteristics of the triangles like measurements of the sides, angles of triangles while having some known values. The projection formulas and solutions of triangles is an important topic from the IIT JEE Main and JEE Advanced exams’ point of view. In this article we will summarize the various formulae and rules for the solution of triangles.

We can find out the angles and all the sides of the triangle using the following rules:

The law of sines.

The law of cosines.

3. The sum of all the angles of a triangle is always equal to 180 degrees or π radians.

To solve a particular triangle, multiple cases can occur for e.g., a side is given and an angle is to be found out, or sometimes angles are given while the side is to be calculated. The triangle given below is taken as an example to explain some basic rules to find out the solution of triangles, in which angles are denoted in capital letters (A, B and C) and the sides of the triangles are denoted in small letters (a, b and c) while the area of the given triangle is denoted with Δ or S.

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### Relations Between the Sides and the Angles of the Triangles

Sine Rule

In the same given triangle ABC, suppose if the angle A and angle B are given along with the side b, and we have to calculate the other sides and angles of the triangle. Then, the third angle C can be calculated using the angle sum property of a triangle, in which the sum of three angles of the triangle is equal to180 degrees.

While, to find the remaining sides of the triangle, certain fundamental relations between angles and sides are required. In this case, the Sine law.

The law of sine states that the side of the triangle ‘a’ when divided by the sine of angle ‘A’ is equal to the side of the triangle ‘b’ when divided by the sine of angle ‘B’ is equal to the side of the triangle ‘c’ when divided by the sine of angle ‘C’. Mathematically it can be stated as,

Sin A/a = Sin B/b = Sin C/c

Or, a/Sin A = b/Sin B = c/Sin C

### Cosine Rule (Two sides with one included angle)

When the two sides of a triangle are given with the angle which is between the two given sides,we use cosine law. Firstly, the side opposite to the given angle is calculated with the help of the cosine law. Similarly, for the sides given we can calculate the angles opposite to them with the help of the Cosine law.

The cosine rule can be stated as when the side of any triangle is given, then the angle opposite to any given side can be calculated and vice-versa i.e.

Cos A = (b2 + c2 – a2) / 2bc

Cos B = (a2 + c2 – b2) / 2ac

Cos C = (a2 + b2 – c2) / 2ab

### Angle- Sum Property

When the two angles of a triangle are known, then we can easily calculate the third angle of the triangle by knowing the fact that the total sum of three angles in the triangles combines to form 180°. So, let the angle A and angle B be known in the above shown triangle and angle C is to be calculated, then it can be shown as:

We know that the sum of all the angles of a triangle is 180°. Hence, we get:

∠A + ∠B + ∠C = 180°

∠C = 180° – (∠A + ∠B)

Through the above shown equation we can easily calculate the value of the missing third angle of the triangle.

### Three Sides and No Angles

In triangle ABC if the three sides (a, b and c) of the triangle are given and no angle is known then we will first try and calculate the angles (A, B and C) with the help of sine law Mathematically,

Sine rule:- Sin A/ a = Sin B/ b = Sin C/ c

But the law of sine fails due to the insufficient data provided to find any of the angles. So, we will try using the cosine rule to find the angles of the given triangle ABC,

Cosine rule:- Cos B = (a2 + c2 – b2) / 2ac

Or it can be rearranged as,

b2 = a2 + c2 – 2ac Cos B …….. (1)

By solving the above equation formed with the help of cosine law we can calculate angle B for the above given triangle.

By rearranging the above equation and bringing the terms with squares on the same side to get,

2ac Cos B = a2 + c2 – b2

Dividing by 2ac on both sides,

Cos B = (a2 + c2 – b2) / 2ac

Taking the inverse of cosine in the above equation on both the sides, cosine and inverse of cosine get canceled out on the LHS and inverse of cosine is added on the RHS of the above shown equation, we get

B = Cos-1(a2 + c2 – b2) / 2ac ………. (2)

From equation 2 we can calculate the value of angle B as we know all the three sides of the triangle ABC. Similarly, we can calculate the remaining angles.

From the above-shown processes, we can easily calculate the sides and angles of the triangle while any one of them is known and the other is not known.

Further, are some of the important formulas to find the area and other important components of the triangle including in solutions of the triangle.

### Area of Triangle

The area of a triangle can be calculated as half of the sine of the angle opposite to the two known sides with the length of the two sides denoted by Δ or S, where S is the semi-perimeter. It can be shown as follows,

Area of the triangle (Δ or S) = ½ (bc Sin A)

Or, Δ or S = ½ (ac Sin B)

Or, Δ or S = ½ (ab Sin C)

Also, Area of the triangle (Δ or S) = \[\sqrt{s(s-a)(s-b)(s-c)}\]

### Projection Formulas

Some of the projection formulae to find out the sides and angles of the triangle ABC can be shown as follows:

a = b Cos C + c Cos B

b = c Cos A + a Cos C

c = a Cos B + b Cos A

The solution of triangles includes a triangle inscribed inside the circle. When a triangle is inscribed within the circle then the base of the triangle forms the diameter of the circle making it the circum-radius of the triangle which is denoted by R.

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From the above-shown figure of a circle inscribing a triangle with angles A, B, C in which AC acts as the diameter of the circle. Its circum-radius can be calculated with the help of the following formula;

R = a/2 sin A = b/2 sin B = c/2 sin C

### Napier’s Analogy (Tangent Rule)

in ΔABC,

\[tan \frac{B-C}{2} = \frac{b-c}{b+c} cot \frac{A}{2}\]

\[tan \frac{C-A}{2} = \frac{c-a}{c+a} cot \frac{B}{2}\]

\[tan \frac{A-B}{2} = \frac{a-b}{a+b} cot \frac{C}{2}\]

So, IIT JEE aspirants can easily learn the formulas of the solution of triangles and solve their questions by reading this article thoroughly and revising these formulas on a regular basis.

## FAQs on Solution of Triangles for IIT JEE

1. What are the trigonometric ratios of half angles for finding solutions of triangles?

For a triangle with semiperimeter S, the the trigonometric ratios are:

(i) \[\sin\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{bc}}\]

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

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

(ii) \[\cos\frac{A}{2} = \sqrt{\frac{s(s-a)}{bc}}\]

\[\cos\frac{B}{2} = \sqrt{\frac{s(s-b)}{ac}}\]

\[\cos\frac{C}{2} = \sqrt{\frac{s(s-c)}{ab}}\]

(iii) \[\tan\frac{A}{2} = \sqrt{\frac{(s-b)(s-c)}{s(s-a)}} = \frac{\Delta}{s(s-a)}\]

2. What is the m-n theorem in the topic ‘solution of triangles’?

In the given triangle ABC, D is a point on side BC such that it divides BC in the ratio m:n. Then, the following trigonometric functions applies:

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ΔABC

(m + n) cot θ = m cot α – n cot ß.

Or,

(m + n) cot θ = n cot B – m cot C.