If sin(A + B + C) = 1, tan( A - B) = 1/√3
and
sec(A + C) = 2, then find 4, B and C respectively,
when they are acute angles.
answer me please
Answers
Explanation:
First case :
sin ( A + B + C ) = 1
= > sin ( A + B + C ) = sin 90
= > A + B + C = 90 ...................( 1 )
Second case :
tan ( A - B ) = 1 / √3
= > tan ( A - B ) = tan 30
= > A - B = 30 .........................( 2 )
Third case :
cos ( A + C ) = 1 / 2
= > cos ( A + C ) = cos 60
= > A + C = 60 ........................( 3 )
We get 3 equations :
A + B + C = 90
A - B = 30
A + C = 60
Add ( 1 ) and ( 2 )
A + B + C = 90
A - B = 30
------------------------
2 A + C = 90 + 30
= > 2 A + C = 120 .......................( 4 )
Subtract ( 3 ) from ( 4 )
2 A + C = 120
A + C = 60
--------------------
A = 60
Hence A = 60
A + C = 60
= > 60 + C = 60
= > C = 60 - 60
= > C = 0
A - B = 30
= > 60 - B = 30
= > B = 60 - 30
= > B = 30
ANSWER :
A = 60
B = 30
C = 0
All are angles . Don't forget to give degree signs I have not given !
Answer:
See the attachment !
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