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23 Maths -- Trigonometry

The Sine LawStatement: In any triangle the sides are proportional to the Sine of the opposite angles.In other words, in any ΔABC, 

The Sine Law

Statement: In any triangle the sides are proportional to the Sine of the opposite angles.
In other words, in any ΔABC, 

1 Answers

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Proof:
Let ABC be a triangle with a = BC, b = CA and c = AB, then three cases are possible. The angle C is either acute or right or obtuse angle.

Draw AD perpendicular to BC ( produce BC if necessary )
In ΔABC, SinB =  for all figures
         
or, AD = cSinB → (i)

Also in ΔACD,
1 { SinC in fig (i), Sin90 = SinC in fig (ii) and Sin(π-C) = SinC in fig (iii) }

Therefore for all the figures,
SinC = 

AD = bSinC → (ii)

From (i) and (ii) we can write,

AD = bSinC = cSinB
↣ 
Similarly we can show,


Combining these we get,
Proved.

More questions on Trigonometry

If a4+b4+c4 = 2c2(a2+b2), prove that C = 45o or 135o.

a4 + b4 +c4= 2c2(a2+b2)

or, (a2+b2)2- 2a2b2 + c4=2(a2+b2)c2 [a2+ b2= (a+b)2- 2ab]

or, (a2+b2)2 -2(a2+b2)c2 + (c2)2=2a2b2

or, (a2+ b2-c2)2=2a2b2

or, a2+ b2-c2=2 ab

or,=2

or,=

or, cos C =

or, cos C = cos 45 or cos 135

or, C =45 or 135




The Sine Law in terms of Circum-Radius R.Statement: In any ΔABC, where R is the Circum-Radius of ΔABC.

Proof:
Consider the circum-circle of ΔABC with center O and radius R. Then we can have three possible figures.

Where the angle A is acute in fig i, or right angled in fig ii or obtuse in fig iii.
Let O be the center of the circum-circle of ΔABC and R be the circum-radius. Join BO and produce it to meet the circle at D.

Now in figure i,
BAC = BDC = A, BCD = 90º (angle at the semicircle).

= Sin(BDC)
= SinA
2R

In fig iii,
1 = Sin90º = SinA
SinA ( Since BC = BD )
SinA
2R

In fig iii,SinA
SinA
2R

Therefore in each...

The Cosine Law

Proof:
We consider a triangle ABC then we can have three possible figures where angle C is acute in figure i, right angle in figure ii and obtuse in figure iii.

From A draw AD perpendicular to BC ( Produce BC if necessary )

In ΔABC, AB2= AD2+ BD2
c2=AD2+ BD2 Ⅰ

In figure (i),
AD = bSinC

Also,
DC = ACCosC = bCosC
BD = BC - DC = a-bCosC

Now from equation Ⅰ,
c2= AD2+ BD2(becomes)
c2= (bSinC)2+ (a-bCosC)2
c2= b2Sin2C+ a2- 2abCosC + b2CosC2
c2= b2+ a2- 2abCosC
2abCosC = b2+ a2- c2


Similarly in figure ii,

AB2= BC2...

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If 8R² = a² + b² + c², prove that the triangle is right angled.

a2+b2+c2=8R2sin2A+sin2B+sin2C=23(cos2A+cos2B+cos2C)=4cos2A+cos2B+cos2C=114cosAcosBcosC=1cosAcosBcosC=0cosA=0orcosB=0orcosC=0A=90orB=90orC=90the triangle is right angled
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