Question

if tan ( A+B)=√3 and tan (A-B) =1/√3 , A>B, then value A is​

Answer 1

Answer:

A = 45° and B = 15°

Step-by-step explanation:

tan (A+B) = √3

√3 = tan 60

tan (A+B) = tan 60°

tan function gets cancelled....

A+B = 60°.......(i)

tan(A-B) = 1/√3

1/√3 = tan 30°

tan(A-B) = tan 30°

tan function gets cancelled...

A-B = 30°.....(ii)

equating (i) and (ii) :

A + B = 60°

B = 60° - A...(iii)

substituting the value of B in equation (ii)

A-B = 30°

A - (60° - A)[from (iii)] = 30°

A - 60° + A = 30°

2A = 30°+60°

2A = 90°

A = 90/2

A = 45°

for B ,

substituting the value of A in eqn (iii) to find B

B = 60° - A

B = 60° - 45°

B = 15°

hence, B is smaller than A (A>B)

you can equate the values in any equation to find A and B respectively ,..... you will get the same answer as this :)

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Answer 2

Answer:

tan(A+B)=√3

tan60°=√3

tan(A-B)=1/√3

tan30°=1/√3

Now, A>B

A+B=60°

A-B=30°

2A=30°

A=15°

Step-by-step explanation:

I hope this answer may help you