SSC MTS Model Paper 3

Question : 30 of 150

Marks: +1, -0

In a right-angled triangle ABC, ∠B is the right angled and AC = 2 √⎯5 cm. If AB − BC = 2 cm, thenthe value of (cos2A − cos2C) is

Solution:

In right angled ∆ABC

∠ B = 90 , B C = x , A B = x + 2 , A C = 2 √ { 5 }

∴ A B ^ { 2 } + B C ^ { 2 } = A C ^ { 2 }

∴ ( x + 2 ) ^ { 2 } + x ^ { 2 } = ( 2 √ { 5 } ) ^ { 2 }

\therefore x ^ { 2 } + 4 + 4 x + x ^ { 2 } = 20

2 x ^ { 2 } + 4 x = 16

x ^ { 2 } + 2 x - 2 x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0

x ^ { 2 } + 4 x - 2 x - 8 = 0 ( x + 4 ) ( x - 2 ) = 0

x + 4 ≠ 0

but

x - 2 = 0 ⇒ x = 2

\cos ^ { 2 } A = ( { x + 2 }/ { 2 √ { 5 } } ) ^ { 2 } =( { 2 + 2 }/ { 2 √{ 5 } } ) ^ { 2 } = { 16 }/ { 20 } = { 4 }/ { 5 }

\cos ^ { 2 } C =( { x }/ { 2 √ { 5 } } \) ^ { 2 } = ( { 2 }/ { 2 √{ 5 } } ) ^ { 2 } = { 1 } /{ 5 }

∴ \cos ^ { 2 } A - \cos ^ { 2 } C = { 4 }/ { 5 } - { 1 }/ { 5 } = { 3 }/ { 5 }

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