if tan(A+B)=√3 and tanA=1,then find angle B

Tan (A+B) = √3
tan (A+B) = tan 60 degrees
A+B = 60 degrees ------------------- 1

given that,
tan A = 1
tan A = tan 45
A = 45 ------------------------ 2

Substituting 2 in 1,
45 + B = 60
B = 60-45
B = 15 degrees

Answer Link

Asterinn Asterinn · Answer 2 · 2020-08-30T13:46:55+05:30

GIVEN:

tan(A+B)=√3

tanA=1

TO FIND :

tan B

FORMULA USED :

[tex]tan(A+B)= \frac{tanA + tanB}{1 - tanAtanB} [/tex]

SOLUTION :

tan(A+B)=√3

tanA=1

Now by using the formula =>

[tex]⟶tan(A+B)= \frac{tanA + tanB}{1 - tanAtanB} [/tex]

put the given values:-

[tex]⟶ \sqrt{3} = \frac{1+ tanB}{1 -( 1 \times tanB)} [/tex]

now cross multiply :-

[tex]⟶ \sqrt{3} (1 - tanB )= 1+ tanB[/tex]

[tex]⟶ \sqrt{3} - \sqrt{3}tanB = 1+ tanB[/tex]

[tex]⟶ \sqrt{3} - 1 = tanB + \sqrt{3}tanB[/tex]

[tex]⟶ \sqrt{3} - 1 = (1 + \sqrt{3})tanB[/tex]

[tex]⟶ \frac{ \sqrt{3} - 1}{ (1 + \sqrt{3})} =tanB[/tex]

we know :-

[tex]⟶ \frac{ \sqrt{3} - 1}{ (1 + \sqrt{3})} =tan15°[/tex]

Therefore , TanB = tan15°

therefore angle B = 15°

ANSWER :

angle B = 15°

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Other Formulae :-

1. Sin(A+B) = Sin A CosB + Cos A SinB

2. Cos (A+B) = Cos A CosB - Sin ASinB

3. Cos (A-B) = Cos A CosB + Sin ASinB