Question

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

Answer 1

Sol➜

We have, tan(A + B) = √3

⇒ tan(A + B) = tan60°

∴ A + B = 60° -----[i]

Again, tan(A - B) = 1/√3

⇒ tan(A - B) = tan30°

∴ A - B = 30° -----[ii]

Now, Adding [i] and [ii], we get

A + B + A - B = 60° + 30°

2A = 90° ⇒ A = 45°

Putting the value of A in [i], we have

45° + B = 60°

∴ B = 60° - 45° = 15°

Hence, A = 45° and B = 15°.

Answer 2

Answer:

$\displaystyle A = {45}^{ \circ} \\ B = {15}^{ \circ}$

Step-by-step explanation:

we are given two equation and condition

$\displaystyle \begin{cases}\tan(A+B)=\sqrt{3} \\ \tan(A - B) = \dfrac{1}{ \sqrt{3} } \end{cases}( {0}^{ \circ} < A+B < {90}^{ \circ} ) \: \text{and} \: A > B$

let's work with first equation

recall unit circle so A and B should be in Q:I

$\displaystyle \: A+B = \arctan( \sqrt{3} )$

$\displaystyle \: A+B = {60}^{ \circ} \cdots \: \text{I}$

let's work with second equation:

$\displaystyle \: \tan(A - B) = \frac{1}{ \sqrt{3} }$

$\displaystyle \: A - B = \arctan(\frac{1}{ \sqrt{3} } )$

$\displaystyle \: A - B = {30}^{ \circ} \cdots \: \text{II}$

now let's use elimination method to figure out A and B

to do so combine equation I and II

$\displaystyle \underline{\begin{array}{c c c}A+B = {60}^{ \circ} \\ A - B = {30}^{ \circ} \end{array}} \\ 2A = {90}^{ \circ}$

divide both sides by 2:

$\displaystyle \: \frac{2 A}{2} = \frac{ {90}^{ \circ} }{2} \\ A = {45}^{ \circ}$

substitute the value of A to the second equation:

$\displaystyle \: {45}^{ \circ} - B = {30}^{ \circ}$

cancel 60° from both sides:

$\displaystyle \: - B = { - 15}^{ \circ}$

divide both sides by -1

$\displaystyle \: B = {15 }^{ \circ}$

hence,

$\displaystyle \: A = {45}^{ \circ} \\ B = {15}^{ \circ}$