Grade 11 Mathematics Note
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.
So, ΔΔABC = 1212BC.AD = 1212BC.AB sinB [sinB = ADABADAB]
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}}} $