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sir please solve these two question by using graph
Dear student,
Using CosA = [ (b2 + c2 – a2) / 2bc ] etc. , in the relation cosA + cosB + cosC = 3/2 we obtain
[ (b2 + c2 – a2) / 2bc ] + [ (c2 + a2 – b2) / 2ac ] + [ (a2 + b2 – c2) / 2ab ] = 3/2
or a.(b2 + c2 – a2) + b.(c2 + a2 – b2) + c.(a2 + b2 – c2) = 3abc
or a.(b2 + c2) + b.(c2 + a2 ) + c.(a2 + b2 ) = a3 + b3 + c3 + 3 abc
or ab2 + ac2 + bc2 + ba2 + ca2 + cb2 = a3 + b3 + c3 + 3 abc
Now subtract 6 abc from both the sides and rearraginf the terms we obtain
ab2 + ac2 - 2abc + bc2 + ba2 - 2abc + ca2 + cb2 - 2abc = a3 + b3 + c3 - 3 abc
or a ( b - c )2 + b ( c – a )2 + c ( a – b )2 = a3 + b3 + c3 - 3 abc
or a ( b - c )2 + b ( c – a )2 + c ( a – b )2 = (a + b + c) (a2 + b2 + c2 – ab – bc – ca) ... ( 1 )
Multiply both the sides by 2 we get
Left hand side = 2a ( b - c )2 + 2b ( c – a )2 + 2c ( a – b )2
& Right hand side = (a + b + c) (2a2 + 2b2 +2 c2 – 2ab – 2bc – 2ca)
or RHS = (a + b + c) (a2 + b2 + c2 – 2ab – 2bc – 2ca + a2 + b2 + c2 )
= ( a + b + c ) [ (a2 + b2 – 2ab) + ( b2 + c2 – 2 bc ) + (a2 + c2 - 2 ca) ]
= ( a + b + c ) [ ( a – b )2 + (b – c )2 + ( c – a )2 ]
Now eqn (1) becomes
2a ( b - c )2 + 2b ( c – a )2 + 2c ( a – b )2 = ( a + b + c ) [ ( a – b )2 + (b – c )2 + ( c – a )2 ]
or (a - b - c) ( b - c )2 + (b -a - c) ( c – a )2 + ( c- a - b) ( a – b )2 = 0 ... (2)
Now we know that sum of two sides of a triangle is greater than the third side. that is (a - b - c) is not equal to 0 similarly (b - a - c) is not equal to 0 and also (c - a - b) is not equal to 0
Hence in eqn (2) squared terms must be equal to zero.
ie. ( b - c )2 = 0 or b = c . Similarly c = a
Hence a = b = c
i.e. The triangle is Equilateral.
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Sagar Singh
B.Tech, IIT Delhi
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