Properties of triangle and circles connected with them
Theorem :
In any triangle ABC \(\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}}2R\) (sine rule) where a,b,c are the lengths of sides which are opposite of angles A, B and C respectively.
Prooof : Without loss of generality we consider A as acute or obtuse or right angle. Let ‘s’ be the centre of the circum circle and R be the circum radius of triangle ABC, let BD be circum diameter through B join CD. Since BCD is an angle in the sesmicircle \(\angle BCD = {90^0}\). Also BD = 2R
i) Let A be an acute angle
From figure ,\(\angle BAC=\angle BDC=\angle A\) (angles in the same segment are equal)
from the triangle BDC, \(\angle BDC=90^0\)
sin\(\angle BDC = \frac{{BC}}{{BD}} = \frac{a}{{2R}}\)
\(\angle BAC = \frac{a}{{2R}}\) \(\Rightarrow \sin A = \frac{a}{{2R}}(\because \angle BAC = \angle BDC)\)
a = 2R sinA
Similarly b = 2R sin B
c = 2R sinC
\(\therefore \,\,\,\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)
ii) If ‘A is an obtuse angle
ABDC is a cyclic quadrilateral
\( \angle {BDC} +\angle {BAC} = {180^0}\) { in a cycilc quadrilateral sum of the opposite angles is 1800}
\( \angle {BDC} = {180^0}-\angle {BAC} \)
\(\angle {BDC} \) = 1800-A
form triangle BDC
\(\sin \angle BDC = \frac{{BC}}{{BD}} = \frac{a}{{2R}}\)
\(\sin ({180^0} - \angle BAC) = \frac{a}{{2R}}\)
\(\sin (\angle BAC) = \frac{a}{{2R}}\)
\( \Rightarrow \sin A = \frac{a}{{2R}} \)
b = 2R sinB, c = 2R sinC
\(\therefore \,\,\,\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)
iii) if ‘A’ is right angle as
a = BC =2R = 2R.1 =2R sinA(A=900)
\( \Rightarrow \frac{a}{{\sin A}} = 2R,\,\,\,similarily\,\,\,\,\,\frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)
\(\therefore \,\,\,\frac{a}{{\sin A}} = \frac{b}{{\sin B}} = \frac{c}{{\sin C}} = 2R\)
Note : The above formula is also called “LAW OF SINES’’
Some examples on this formula
1) Problem: In triangle ABC if a =3, b =4, and sinA = 3/4 find anlge B
Solution :
from the sine rule,
\(\frac{a}{{\sin A}} = \frac{b}{{\sin B}}\)
\(\therefore \sin B = \frac{{b\sin A}}{a} = \frac{{4.3/4}}{3} = 1\)
2)If the length of the sides of a triangle are 3,4,5 find the circum radius of the triangle
Solution : Given that the sides of a triangle are 3,4,5
Now, 32 +42 = 52 hence the triangle is right angled and its hyprotenuse = 5
circum radius \(1 \over 2\) (hypotenuse)
=\(1 \over 2\) (5) =5/2
by using sine rule
= 2R
C = 2R sinC
C = 2R . sin900
C = 2R
Theorem : In triangle ABC
b2 = c2 +a2 - 2ac cosB
c2 = a2 +b2 - 2ab cosC
a2 = b2 + c2 - 2bc cosA
we know a =2 R sin A
Proof : a2 = (2R sinA)2
= 4R2(sin(180-(B+C))2
=4R2[sin(B+C)]2 =4R2[sinBcosC +sinCcosB]2
=4R2 [sin2B. cos2C + sin2C cos2B+2sinBsinC cosB cosC]
=4R2 \(\left\{ {{{\sin }^2}B(1 - {{\sin }^2}c) + {{\sin }^2}C(1 - {{\sin }^2}B) + 2\sin B.\sin C.\cos B\cos C} \right\}\)
=4R2 {sin2B+sin2c +2sinB sinc(cosB cosc - sinB sinc) }
= 4R2 {sin2B + sin2C + 2sinB sinC. cos(B+C)}
a2 = b2 +c2 - 2b.c.cosA
The proofs of the other two results are similar
Alternate method
Take the vertex B of triangle ABC as origin and its side BC along x - axis then
B=(0,0) and C = (a,0) angle made by the side
AB with x - axis = B
and also AB = c and A(ccosB,csinB)
b2 = CA2 = (ccosB -a)2 + (csinB-0)2
=c2cos2B+a2-2ac cosB +c2sin2B.
=c2(sin2B+cos2B)-2ac cosB+a2
=c2(1)+a2-2ac cosB
b2 = c2 +a2 - 2ac cosB
similarily we can prove the remaining 2 results.
Note :
1)This theorem is known as the “law of cosine”: and the results in it are called “cosine rules”.
2) from the cosine rules we can write
cosA =\(\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}},\cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ac}}\)
cos C=\(\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\)
Theorem : In triangle ABC a = b cosC + c cosB
\(\cos B = \frac{{{c^2} + {a^2} - {b^2}}}{{2ca}},\,\cos C = \frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}\)
bcosC+c cosB = \(b\left( {\frac{{{a^2} + {b^2} - {c^2}}}{{2ab}}} \right) + c\left( {\frac{{{c^2} + {a^2} - {b^2}}}{{2ca}}} \right)\)
= \(\frac{{{a^2} + {b^2} - {c^2} + {c^2} + {a^2} - {b^2}}}{{2a}}\) = \(\frac{{2{a^2}}}{{2a}} = a\) a = b cosC +- c cosB
Similarly we can prove that
b = c cosA + a cosC
c = a cosB + b cosA
Theorem : In triangle
Proof : from the sine rule
b = 2R sinB,c=2RsinC
Hence \(\frac{{b - c}}{{b + c}} = \frac{{2R(\sin B - \sin C)}}{{2R(\sin B + \sin C)}}\)
= \(\frac{{b - c}}{{b + c}} = \frac{{2R(\sin B - \sin C)}}{{2R(\sin B + \sin C)}}\)
=\(\cot \frac{{B + C}}{2}.\tan \frac{{B - C}}{2}\)
= \(\cot \frac{{B + C}}{2}.\tan \frac{{B - C}}{2}\)
= \(= \tan \left( {\frac{{B - C}}{2}} \right) = \left( {\frac{{b - c}}{{b + c}}} \right)\cot \frac{A}{2}\)
Similarly, we can prove that
\(= \tan \left( {\frac{{B - C}}{2}} \right) = \left( {\frac{{b - c}}{{b + c}}} \right)\cot \frac{A}{2}\) and \(\tan \left( {\frac{{C - A}}{2}} \right) = \frac{{c - a}}{{c + a}}\cot \frac{B}{2}\)
Note : These three results are called "Napier analogy " (or) trangent rules
Theroem :
In any triangle ABC, the values of \(sin \frac{A}{2},\cos \frac{A}{2},\tan \frac{A}{2}\) in terms of a,b,c are
i)\(sin \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \) ii)\(\cos \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{bc}}} \) iii)\(\tan \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c}}{{s(s - a)}}} \)
Proof : we know that
\(\sin \frac{A}{2} = \sqrt {\frac{{1 - \cos A}}{2}} \,\,\,\,\,\,\,\,\,\,\left( {\because \frac{A}{2}\,\,is\,\,a\,cute} \right)\)
1-cosA=\(1 - \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}\)
\(2{\sin ^2}\frac{A}{2} = \frac{{2bc - {b^2} - {c^2} + {a^2}}}{{2bc}}\)
\(= \frac{{{a^2} - {{\left( {b - c} \right)}^2}}}{{2bc}}\)
=\(\frac{{\left( {a + b - c} \right)\left( {a - b + c} \right)}}{{2bc}}\)
=\(\frac{{(2s - 2c)\left( {2s - 2b} \right)}}{{2bc}}\)
\(2{\sin ^2}\frac{A}{2} = \frac{{2\left( {s - b} \right).2(s - c)}}{{2bc}}\)
=\({\sin ^2}\frac{A}{2} = \frac{{\left( {s - b} \right)(s - c)}}{{bc}}\)
=\(\sin \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \)
Similarly,\(\sin \frac{B}{2} = \sqrt {\frac{{(s - a)(s - c)}}{{ac}}} \)
\(\sin \frac{C}{2} = \sqrt {\frac{{(s - a)(s - b)}}{{ab}}} \)
ii)We know that \(\cos \frac{A}{2} = + \sqrt {\frac{{1 + \cos A}}{2}} \)
\(1 + \cos A = 1 + \frac{{{b^2} + {c^2} - {a^2}}}{{2bc}} = \frac{{{{\left( {b + c} \right)}^2} - {a^2}}}{{2bc}}\)
\(2{\cos ^2}\frac{A}{2} = \frac{{(b + c + a)(b + c - a)}}{{2bc}}\)
\(2{\cos ^2}\frac{A}{2} = \frac{{2s.(s - a)}}{{2bc}}\)
\({\cos ^2}\frac{A}{2} = \frac{{s(s - a)}}{{bc}}\)
\(\cos \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{bc}}} \)
Similarily: we can prove that
\(\cos \frac{{\rm B}}{2} = \sqrt {\frac{{s(s - b)}}{{ac}}} ,\,\,\,\cos \frac{C}{2} = \sqrt {\frac{{s(s - c)}}{{ab}}} \) , \(\begin{gathered} \tan \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} \times \sqrt {\frac{{(s - b)(s - c)}}{{(s - b)(s - c)}}} \hfill \\ = \frac{{(s - b)(s - c)}}{\Delta } \hfill \\ \end{gathered} \)
by using the results (i) and (ii)
= \(\tan \frac{A}{2} = \frac{{\sin A/2}}{{csoA/2}} = \frac{{\sqrt {\frac{{(s - b)(s - c)}}{{bc}}} }}{{\sqrt {\frac{{s(s - a)}}{{bc}}} }}\)=\(\sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} \)
Similarily: \(\tan \frac{B}{2} = \sqrt {\frac{{(s - a)(s - c)}}{{s(s - b)}}} \) , \(\tan \frac{C}{2} = \sqrt {\frac{{(s - a)(s - b)}}{{s(s - c)}}} \)
Note :\(\tan \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} \)
=\(\sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} \times \sqrt {\frac{{s(s - a)}}{{s(s - a)}}} \)
=\(\frac{{\sqrt {s(s - a)(s - b)(s - c)} }}{{s(s - a)}}\)
\(\tan \frac{A}{2} = \frac{\Delta }{{s(s - a)}}\)
\(\therefore \cot \frac{A}{2} = \sqrt {\frac{{s(s - a)}}{{(s - b)(s - c)}}} = \frac{{s(s - a)}}{\Delta }\)
\(\tan \frac{B}{2} = \frac{\Delta }{{s(s - b)}},\,\,\,\cot \frac{B}{2} = \frac{{s(s - b)}}{\Delta }\)
\(\tan \frac{C}{2} = \frac{\Delta }{{s(s - c)}},\,\,\,\,\,\,\,\cot \frac{C}{2} = \frac{{s(s - c)}}{\Delta }\)
Theorem : In any triangle ABC, sinA, sinB, sinC in terms of a,b,c
Proof : sinA = 2sinA/2 cosA/2
\(2.\sqrt {\frac{{(s - b)(s - c)}}{{bc}}} \,\,\,\sqrt {\frac{{s(s - c)}}{{bc}}} \)
sinA=\(\frac{2}{{bc}}\sqrt {s(s - a)(s - b)(s - c)} = \frac{2}{{bc}}\Delta \)
Similarly sinB=\(\frac{2}{{ca}}\sqrt {s(s - a)(s - b)(s - c)} = \frac{2}{{ca}}\Delta \)
sin C=\(
\frac{2}{{ab}}\sqrt {s(s - a)(s - b)(s - c)} = \frac{2}{{ab}}\Delta \)
Note : 1) From the above formulas
\(\Delta = \frac{1}{2}ab\sin C = \frac{1}{2}bc\sin A = \frac{1}{2}ca\sin B\)
2) \(\Delta = \frac{1}{2}ab\sin C = \) 2R2 sinA sin B sinC
Similarly \(\Delta = \frac{1}{2}bc\sin A = \frac{1}{2}.2R\sin A.2RsinCsinA\)
=2R2 sinAsinBsinC
\(\Delta = \frac{1}{2}ca\sin B\)
\(\Delta = \frac{1}{2}2R\sin C.2RsinA.\sin B\)
= 2R2 sinAsinBsinC
\(\Delta = \frac{1}{2}bc\sin A\)
\(\Delta = \frac{1}{2}.bc.\frac{a}{{2R}} = \frac{{abc}}{{4R}} \Rightarrow R = \frac{{abc}}{{4\Delta }}\)
\(\therefore \Delta = \frac{{abc}}{{4R}}\,\,\,and\,\,\,R = \frac{{abc}}{{4\Delta }}\)
4)\(\tan \frac{A}{2} = \sqrt {\frac{{(s - b)(s - c)}}{{s(s - a)}}} \times \frac{{\sqrt {(s - b)(s - c)} }}{{\sqrt {(s - b)(s - c)} }}\)
\(\tan \frac{{A\,}}{2} = \frac{{(s - b)(s - c)}}{{\sqrt {s(s - a)(s - b)(s - c)} }}\)=\(\frac{{(s - b)(s - c)}}{\Delta }\)
\(\cot \frac{A}{2} = \frac{\Delta }{{(s - b)(s - c)}}\)=\(\frac{{s(s - a)}}{\Delta }\)
Similarly\(\tan \frac{B}{2} = \frac{{(s - a)(s - c)}}{\Delta } = \frac{\Delta }{{s(s - b)}}\)
\(\cot \frac{B}{2} = \frac{\Delta }{{(s - a)(s - c)}} = \frac{{s(s - b)}}{\Delta }\)
\(\cot \frac{c}{2} = \frac{\Delta }{{(s - a)(s - b)}} = \frac{{s(s - c)}}{\Delta }\)
5) tanA=\(\frac{{\sin A}}{{\cos A}}\)
=\(\frac{{a/2R}}{{\left( {\frac{{{b^2} + {c^2} - {a^2}}}{{2bc}}} \right)}}\)
=\(= \frac{a}{{2R}}.\frac{{2bc}}{{{b^2} + {c^2} - {a^2}}}\)
=\(\frac{{abc}}{R}.\frac{1}{{{b^2} + {c^2} - {a^2}}}\)
=\(4\Delta .\left( {\frac{1}{{{b^2} + {c^2} - {a^2}}}} \right)\)
tanA=\(\frac{{4\Delta }}{{{b^2} + {c^2} - {a^2}}}\),\(\cot A = \frac{{{b^2} + {c^2} - {a^2}}}{{4\Delta }}\)
Similarly \(\tan B = \frac{{4\Delta }}{{{a^2} + {c^2} - {b^2}}}\)
\(\cot B = \frac{{{a^2} + {c^2} - {b^2}}}{{4\Delta }}\)
\(\tan C = \frac{{4\Delta }}{{{a^2} + {b^2} - {c^2}}}\)
\(\cot C = \frac{{{a^2} + {b^2} - {c^2}}}{{4\Delta }}\)
6) i) cotA, cotB, cotC are in A.P \( \Leftrightarrow \)a2, b2,c2 are in A.P
ii) cot \(A \over 2\) .cot \(B \over 2\).cot \(C \over 2\) are in A.P \( \Leftrightarrow \) a,b,c are in A.P
iii) if cot \(\frac{A}{2}\):cot \(\frac{B}{2}\):cot \(\frac{C}{2}\)= x:y:z
then a:b:c = y +z : z+x : x+y
iv) if a:b:c = x:y:z then cotA/2 : cotB/2:cotC/2
= y + z : z + x : x + y
7) If a+c= n , then tan \(\frac{A}{2}\) : tan \(C\over 2\) = \(\frac{{n - 1}}{{n + 1}}\)