Question

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if tan(a+b)=√3 and tan(a-b)=1/√3 find a and b if 0°<A+B<90°..




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Answer 1

Answer:

⚡Question ⤵

if tan(a+b)=√3 and tan(a-b)=1/√3 find a and b if 0°<A+B<90°..

⚡Answer⤵

$given \: tan(a + b) = \sqrt{3}⟹ a + b = 60° .....(1) \\ tan(a - b) = \frac{1}{ \sqrt{3} } ⟹a - b = 30°..........(2) \\ (1) + (2) ⟹2a = 90 \\ a = \frac{90°}{2} \\ a = 45$

$put \: A \: value \: in(1) \\ b = 60° - 45° \\ b = 15°$

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Answer 2

tan (A + B) = √3 ⇒ A + B = 60° ………….. (i)

tan (A – B) = $\frac{1}{\sqrt{3} }$ ⇒ A – B = 30° ………… (ii)

From Eqn. (i) + Eqn.(ii), we have

2A = 90

∴ A = 45°

From eqn. (i), A + B = 60°

45° + B = 60°  

∴ B = 15

Therefore values of a and b are 45° and 15° respectively.