Question

Plzz answer fast..

If tan(A+B ) =√3 and
tan(A-B) = 1/√3 find A and B.
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Answer 1

SOLUTION ☺️

QUES.=) If tan (A+B)= √3 and tan (A-B) = 1/√3, find A and B.

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✳️tan(a+b) = √3

✳️tan(a-b)= 1/√3

We know tan 60 = √3 and tan 30 = 1/√3

So, a+b = 60, a-b = 30

Adding the two gives

=a+a+b-b=60+30

=2a=90

=a = 45

=a+b=60

=45+b=60

=b=15 ans.

hope it helps

Answer 2

$\huge \boxed{ \underline{ \underline{ \bf{Answer}}}}$

$\implies \: \: tan \: (A + B) = \sqrt{3}$

We know, tan 60° = √3, so we can write

$\implies \: \: tan \: (A + B) = tan \: 60 \degree$

$\implies \: A + B \: = \: 60 \degree \: \: \: \rightarrow \: (1)$

$\implies \: \: tan \: (A - B) = \frac{1}{ \sqrt{3} }$

We know, tan 30° = 1/√3, so we can write

$\implies \: \: tan \: (A - B) = tan \: 30 \degree$

$\implies \: A - B \: = \: 30 \degree \: \: \: \rightarrow \: (2)$

Now, Adding (1) and (2), we get

=> A + B + A - B = 60° + 30°

=> 2A = 90°

=> A = 45°

and,

=> A + B = 60°

=> 45° + B = 60°

=> B = 60° - 45°

=> B = 15°

$\boxed{ \bf{Hence, \: A = 45 \degree \: and \: B = 15 \degree}}$