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Grade: 11

                        

A particle is thrown over a triangle from one end of a horizontal base and grazing the vertex falls on the other end of the base. If A and B be the base angles and C the angle of projection then prove that tan C= tan A+ tan B.

2 years ago

Answers : (1)

Arun
24738 Points
							
tan A = h / x₁ … tan B = h / x₂ … x₁ + x₂ = ( h / tan A ) + ( h / tan B ) … 
…………….…………….…………….……….. = h [ tan B + tan A ] / [ tan A tan B ] … 
… h [ tan A + tan B ] = ( x₁ + x₂ ) tan A tan B = [ ( u cos C ) t₁ ] tan A tan B … 
… y = ( u sin C ) t₁ - ½ g t₁ ² = 0 …-->… 2 u sin C = g t₁ … so that … 
… t₁ = 2 ( v₀ / g ) sin C … substituting in the last expression involving h … 
… h [ tan A + tan B ] = ( u cos C ) [ 2 ( u / g ) sin C ] tan A tan B … 
…………….………… = 2 ( u² / g ) sin C cos C tan A tan B … 
… h = 2 ( u² / g ) sin C cos C tan A tan B / [ tan A + tan B ] … 
which is the first expression for h … and using … h = ( u sin C ) t₂ - ½ g t₂ ² … 
… where … x₁ = ( u cos C ) t₂ = h / tan A …-->… t₂ = h / [ u cos C tan A ] … 
… we get another expression for h … 
… h = ( h u sin C ) / [ u cos C tan A ] - ½ g h ²/ [ u cos C tan A ] ² … 
Cancelling h, the last equation reduces to … 
… 1 = ( u sin C ) / [ u cos C tan A ] - ½ g h / [ u cos C tan A ] ² 
… 2 [ u cos C tan A ] ² = 2 ( u sin C ) [ u cos C tan A ] – g h 
… h = [ 2 ( u ² / g ) cos C tan A ] { sin C - cos C tan A } … which is the second 
expression for h … using the two expressions obtained for h … 
… 2 ( u² / g ) sin C cos C tan A tan B / [ tan A + tan B ] 
…………….……………… = [ 2 ( u ² / g ) cos C tan A ] { sin C - cos C tan A } … 
… which reduces to … sin C tan B / [ tan A + tan B ] = { sin C - cos C tan A } … 
after cancelling the common factors … rewriting the last equation … 
… sin C tan B = { sin C - cos C tan A } [ tan A + tan B ] 
…………….… = sin C [ tan A + tan B ] - cos C tan A [ tan A + tan B ] 
…………….… = sin C tan A + sin C tan B ] - cos C tan A [ tan A + tan B ] 
Cancelling sin C tan B on both sides of the equation … 
… 0 = sin C tan A - cos C tan A [ tan A + tan B ] … where … tan A is another 
common factor that cancels out … 0 = { sin C - cos C [ tan A + tan B ] } … or 
… sin C = cos C [ tan A + tan B ] …-->… tan C = tan A + tan B …
2 years ago
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