Question

) If tan (A+B) = √3, tan (A-B) = 1, and A, B (B<A) are acute angles, find A and B.​

Answer 1

$\huge \sf{ \red{ \underline{ \underline{Question:-}}}}$

$\boxed{ \mathfrak{ \pink{If \: tan \: (A+B) = √3, \: tan (A-B) = 1, and \: A, B (B<A) \: are \: acute \: angles, \: find \: A \: and \: B.}}}$

Given:-

• tan(A+B)= √3

To find:-

• Value of A and B.

$\huge \rm{ \color{grey}{Solution:- }}$

⇒ tan(A+B)=tan60°ㅤㅤㅤㅤ [tan60°=√3]

⇒A+B=60° ㅤㅤㅤㅤㅤ (i)

Also, tan(A-B)=1

⇒A- B= 45°ㅤㅤㅤㅤㅤ (ii)

• From equation (i) and (ii)

$A + B = 60 \degree \\ A - B = 45 \degree \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ 2A = 105 \degree$

$⇒A= \frac{105 \degree}{2} = 52 \frac{1 \degree}{2}$

• Substituting the value of A equation 1 ,we get:-

$\frac{105 \degree}{2} + B = 60 \degree$

$B = 60 \degree - \frac{105 \degree}{2}$

$= \frac{120 \degree - 105 \degree}{2} = \frac{15 \degree}{2} = 7 \frac{1 \degree}{2 }$

$\boxed{ \underline{ \sf{ \pink{Hence,A = 52 \frac{1 \degree}{2} \: and \: B = 7 \frac{1 \degree}{2} }}}}$

Answer 2

Answer:

⇒ tan(A+B)=tan60°ㅤㅤㅤㅤ [tan60°=√3]

⇒A+B=60° ㅤㅤㅤㅤㅤ (i)

Also, tan(A-B)=1

⇒A- B= 45°ㅤㅤㅤㅤㅤ (ii)

From equation (i) and (ii)

A+B=60°A−B=45°‾2A=105°\begin{lgathered}A + B = 60 \degree \\ A - B = 45 \degree \\ \underline{ \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: \: } \\ 2A = 105 \degree\end{lgathered}

A+B=60°

A−B=45°

2A=105°

⇒A=105°2=521°2⇒A= \frac{105 \degree}{2} = 52 \frac{1 \degree}{2}⇒A=

2

105°

=52

2

1°

Substituting the value of A equation 1 ,we get:-

105°2+B=60°\frac{105 \degree}{2} + B = 60 \degree

2

105°

+B=60°

B=60°−105°2B = 60 \degree - \frac{105 \degree}{2}B=60°−

2

105°

2

120°−105°

=

2

15°

=7

2

1°

Hence,A=52

2

1°

andB=7

2

1°