E-Storesolution of triangles
27 Jul 2007 20:14:30 IST
solution of triangles
Properties and solution of triangles :
r = 4R SinA/2.SinB/2.SinC/2
Sine rule :
a/Sina = b/Sinb = c/Sinc = 2R
Cosine rule :
CosA = b2+c2-a2/2bc
Projection rule :
a = bCosC + cCosB
Napier's rule :
tan(A-B/2) = [(a-b)/(a+b)]Cot C/2
Half angle Formulae :
Sin A/2 = [(s-b)(s-c)/bc]1/2
Cos A/2 = [s(s-a)/bc]1/2
tan A/2 = [(s-b)(s-c)/s(s-a)]1/2
Where 2s = a+b+c
Circumradius :
R = abc/4
Inradius :
r = /s = (s-a)tan A/2
/s = (s-a)tan A/2Exradius :
r0 = /(s-a) = s tan A/2
/(s-a) = s tan A/2Solution of triangles :
1)given a,b,c apply Cosine rule or half angle rule.
2)given b,c,A apply Napier's rule.
3)given a,B,C apply sine rule.
4)given b,c,B apply sine rule to obtain sinC (cases):
For acute B:
if SinC > 1 
No triangle
No triangleif SinC = 1 one right angled triangle.
one right angled triangle.if SinC < 1 & cSinB < b < c Two triangles.
Two triangles.if SinC < 1 & b 
c one triangle.
c one triangle.For Obtuse :
if b > c one triangle.
one triangle.if b 
c no triangle.
c no triangle.Area of triangle :
= 1/2(bcSinA) = [s(s-a)(s-b)(s-c)]1/2 Area of cyclic quadrilateral ABCD = [(s-a)(s-b)(s-c)(s-d)]1/2 Where 2s = a+b+c+d. Ptolmey's theorem : AC.BD = AB.CD + BC.AD Area of n sided regular polygon = nR2/2[Sin(2
/n)] Where R = radius of circumsircle. = nR2[tan(/n)] Hope you all find it useful. Cheers !!!!!!!!!!!!!!!!!
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