Question

If tan(A+B) =√3 and tan (A-B) =1/√3 0degree <(A+B) i. e 90 degree A>B then find the value of A and B​

Answer 1

Step-by-step explanation:

tan 60=√3

hence a+b=60. (1)

now tan 30=1/√3

hence a-b=30 (2)

solving 1 and 2 we get

2a=30

hence a=15°

now substuiting a=15° in (1) we get b=45°

hence value of a and b are 15° and 45° respectivley

Answer 2

Answer:

Value of A is 45 and B is 15.

Step-by-step explanation:

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

$\sf{ \tan(A - B) = \dfrac{1}{ \sqrt{3} } }$

$\begin{array}{ |c |c|c|c|c|c|} \sf\angle A & \sf{0}^{ \circ} & \sf{30}^{ \circ} & \sf{45}^{ \circ} & \sf{60}^{ \circ} & \sf{90}^{ \circ} \\ \\ \sf \: tan \: A & 0 & \dfrac{1}{ \sqrt{3} }& 1 & \sqrt{3} & \sf Not \: de fined \\ \: \end{array}$

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

We know that : $\sf{\tan( {60}^{\circ}) = \sqrt{3}}$

$\longrightarrow\sf{ A+B= 60}$

$\longrightarrow\sf{ A = 60 - B} \: - - (i)$

$\longrightarrow\sf{ \tan(A - B) = \dfrac{1}{\sqrt{3}}}$

We know that : $\sf{\tan( {30}^{\circ}) = \dfrac{1}{ \sqrt{3}}}$

$\longrightarrow\sf{ A - B = {30}^{\circ}}$

Substitute eq. i

$\longrightarrow\sf{ (60 - B) - B = {30}^{\circ}}$

$\longrightarrow\sf{ - 2B = {30 - 60}}$

$\longrightarrow\sf{ - 2B = { - 15}}$

$\longrightarrow\sf{ B = { \dfrac{30}{2} }}$

$\longrightarrow\sf{ B = {15}}$

Value of A,

Put value of B in eq. i

$\longrightarrow\sf{ A = 60 - 15}$

$\longrightarrow\sf{ A = 45}$

Therefore, value of A is 45 and B is 15.