Question

If tan (A + B) = √3 and tan (A - B) =1, 0° <(A + B) < 90° and
A > B, then find A and B.
​

Answer 1

Given :-

● tan(A+B)=√3

● tan(A-B)=1

and 0°<(A+B)<90°

To find :-

The value of A and B

A.T.Q :-

$\sf\:\implies tan(A+B)=\sqrt{3}\\ \\ \sf\:\:\:We \:know\:that\: tan60{}^{\circ}=\sqrt{3}\\ \\ \sf\implies So,\: tan(A+B)=tan60{}^{\circ}\\ \\ \sf\bullet A+B=60{}^{\circ}--------(i)$

$\sf\implies tan(A-B)=1\\ \\ \sf\:\: \bullet tan45{}^{\circ}=1\\ \\ \sf\implies tan(A-B)=tan 45{}^{\circ}\\ \\ \sf\:\:\:\bullet A-B=45{}^{\circ} ------(ii)$

Now add equation (i) and (ii)

$\sf\implies A\cancel{+B}+A\cancel{-B}=60{}^{\circ}+45{}^{\circ}\\ \\ \sf\implies 2A=105{}^{\circ}\\ \\ \sf\implies A=\cancel{\dfrac{105}{2}}\\ \\ \sf\implies A=52.5{}^{\circ}$

• Now the value of B

$\sf\implies A+B=60{}^{\circ}\\ \\ \sf\implies 52.5{}^{\circ}+B=60{}^{\circ}\\ \\ \sf\implies B=60{}^{\circ}-52.5{}^{\circ}\\ \\ \sf\bullet B=7.5{}^{\circ}$

$\boxed{\sf{A=52.5{}^{\circ}}}$

$\boxed{\sf{B=7.5{}^{\circ}}}$

Answer 2

${\underline{\huge{\mathbf{\color{pink}{question}}}}}$

If tan (A + B) = √3 and tan (A - B) =1, 0° <(A + B) < 90° and

A > B, then find A and B.

___________________________❤

Given

• tan(A+B)=√3
• tan(A-B)=1
• 0°<(A+B)<90°

To find :-

• The value of A and B

step by step solution:-

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

• $tan 60 = \sqrt 3$
• $(A+ B) = 60$-------------(i)

$tan (A-B) = 1$

• $tan 45 = 1$
• $tan(A-B)=tan 45$-----------(ii)

Now add equation (i) and (ii)

$\sf A\cancel{+B}+A\cancel{-B}=60{}^{\circ}+45{}^{\circ}\\ \\ \sf2A=105{}^{\circ}\\ \\ \sfA=\cancel{\dfrac{105}{2}}\\ \\ \sf A=52.5{}^{\circ}$

put value of A in eq 1

$A+B=60$

${}^{\circ}\\ \\ \sf 52.5{}^{\circ}+B=60{}^{\circ}\\ \\ \sf B=60{}^{\circ}-52.5{}^{\circ}\\ \\ \sf B=7.5{}^{\circ}$

answers :-

$\boxed{\sf{A=52.5{}^{\circ}}}$

$\boxed{\sf{B=7.5{}^{\circ}}}$

________________❤