Question

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Answer 1

Answer:

A = 120° , B = 45° , C = 15°

Step-by-step explanation:

Given----> In Δ ABC , ∠A > 90° and,

Sec ( B + C ) = Cosec ( B - C ) = 2

To find-----> Value of A , B and C .

Solution-----> If ,

Sec ( B + C ) = Cosec ( B - C ) = 2

Now , Sec ( B + C ) = 2

=> Sec ( B + C ) = Sec 60°

=> B + C = 60° ..............( 1 )

Now, Cosec ( B - C ) = 2

=> Cosec ( B - C ) = Cosec 30°

=> B - C = 30° ...................( 2 )

Now solving equation ( 1 ) and ( 2 ) , for this we add both equation ,

B + C + B - C = 60° + 30°

=> 2 B = 90°

=> B = 90° / 2

=> B = 45°

Putting B = 45° , in equation ( 1 ) , we get,

B + C = 60°

=> 45° + C = 60°

=> C = 60° - 45°

=> C = 15°

Now , by angle sum property of triangle ,

A + B + C = 180°

=> A +45° + 15° = 180°

=> A + 60° = 180°

=> A = 180° - 60°

=> A = 120°