# Properties of Triangle.

A triangle consists of three sides and three angles. They are collectively known as the six elements of the triangle.

In a triangle ABC, the side opposite to angle A is denoted by ‘a’, that opposite to angle B is denoted by ‘b’ and similarly the one opposite to angle C is denoted by ‘c’.

Sine law:

In any ΔΔABC,the sine law states that,

a. asinA = bsinB = csinC and

b. asinA = bsinB = csinC = 2R.

Where R is the radius of the circum–circle of ΔΔABC.

Now,

a. In ΔΔABC, AD is perpendicular to BC, BC = a.

CA = b and AB = c.

Or, ΔABC = 12 a. c sinB …(i)

Also, ΔABC = 12 BC.AD = 12 BC.AC sinC [sinC =DAC]

So, ΔABC = 12ab sinC …(ii)

Similarly, ΔABC = 12 b.c.sinA …(iii)

Now, from (i),(ii) and (iii) we have,

Or, 12 b.c. sinA = 12 a.c. sinB = 12 a.b. sinC

Or, sinAa = sinBb=sinCc [dividing by abc]

So, asinA = bsinB = csinC.

b.

Let O be the centre and R be the circum–radius of the circum – radius ABC circumscribed on the ΔABC.

So, OA = OB = OC = R.

Let OD is perpendicular to BC when BC = a.

So, ΔOBD −~ΔOCD. [OD is perpendicular to BC]

So, ∠BOD = ∠COD and

BD = DC = BC2 = a2.

And ∠BAC = 12∠BOC = 12 (∠BOD + ∠COD)

= 12.2 ∠BOD (∠BOD = ∠COD)

= ∠BOD.

So, ∠BAC = ∠BOD [BOD = A]

Now, in right–angled triangle BOD, we have,

Sin(BOD) = BDOBBDOB = a2.Ra2.R [BD = a2a2, OB = R]

Or, sinA = a2R

So, asinA = 2R …(i)

Since, we have,

Or, asinA = bsinA = csinC …(ii)

So, from (i) and (ii) we have,

Or, asinA = bsinB = csinC = 2R.

The projection law:

Projection Formula gives the relation between angles and sides of a triangle. We can find the length of a side of the triangle if other two sides and corresponding angles are given using projection formula. If a, b and c be the length of sides of a triangle and A, B and C are angles opposite to the sides respectively, then projection formula is given below:

a = b cos C + c cos B

b = c cos A + a cos C

c = a cos B + b cos A

The Area of Triangle:

The area of a triangle is always half the product of the height and base.

Area = $\frac{1}{2}{\rm{base*height}}$

In order to express the area of a triangle in terms of the sides of a triangle, we shall appeal to the formula

Δ = $\frac{1}{2}{\rm{bc}}\sin {\rm{A}} = \frac{1}{2}{\rm{ca}}\sin {\rm{B}} = \frac{1}{2}{\rm{ab\dag sinC}}$

And the half-angle formulae

$\sin \frac{1}{2}{\rm{A}} = \sqrt {\frac{{\left( {{\rm{s}} - {\rm{b}}} \right)\left( {{\rm{s}} - {\rm{c}}} \right)}}{{{\rm{bc}}}}}$

$\cos \frac{1}{2}{\rm{A}} = \sqrt {\frac{{{\rm{s}}\left( {{\rm{s}} - {\rm{a}}} \right)}}{{{\rm{bc}}}},{\rm{\dag etc}}}$

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