Question

From ch 8 INTRODUCTION TO TRIGONOMETRY
1) If tan(A+B)=1/√3and tan( A - B) = √3,0° <(A+B) < 90' and A > B, then find A and B.​

Answer 1

Correct question :-

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

Solution :-

tan (A + B) = √3

We know,

tan 60° = √3

So we get :-

A + B = 60         . . .(i)

_____________________

tan (A - B) = 1/√3

We know,

tan 30° = 1/√3

So we get :-

A - B = 30          . . .(ii)

◘ Solving equation (i) and (ii) :-

From (ii),

A = 30 + B

Putting the value in (i) :-

⟶ 30 + B + B = 60

⟶ 30 + 2B = 60

⟶ 2(15 + B) = 60

⟶ 15 + B = 30

⟶ B = 30 - 15

⟶ B = 15°

_________________

A = 30 + B

⟶ A = 30 + 15

⟶ A = 45°

Answer 2

★ Question :-

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

★Solution:-

➪ A - B = 1 / √3

• ${\sf{\underline{\boxed{\bold{ tan 30 = \dfrac{1}{\sqrt 3}}}}}}$

➪ A - B = 30 ----------( i )

➪ A + B = √3

• ${\sf{\underline{\boxed{\bold{ tan 60 = {}{\sqrt 3}}}}}}$

➪ A + B = 60 ----------( ii )

Solving simultaneously :-

• adding eq i and ii

➪ A - B + A + B = 60 + 30

➪ 2A = 90

➪ A = 45

• putting value of " A " in eq i

➪ A - B = 30

➪45 - B = 30

➪B = 45 - 30

➪B = 15

________________________

${\sf{\underline{\boxed{\bold{ A = 45 \: and \: B = 15 }}}}}$