Question

if sin(A+B+C) =1 ,tan(A-B)=1/√3,cos(A+C)=1/2, then find the value of A,B,C using the values of trigonometric ratio at angle 0°,30°,45°,60°,90°

Answer 1

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 2

A=60 ,B=30,C=0

hope this is helpful