Question

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

Pls answer it correctly..​

Answer 1

`\displaystyle A = {45}^( \circ) \\ B = {15}^( \circ)`

Explanation:

we are given two equation and condition

`\displaystyle \begin{cases}\tan(A+B)=√(3) \\ \tan(A - B) = (1)/( √(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( √(3) )`

`\displaystyle \: A+B = {60}^( \circ) \cdots \: \text{I}`

let's work with second equation:

`\displaystyle \: \tan(A - B) = (1)/( √(3) )`

`\displaystyle \: A - B = \arctan((1)/( √(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 \: (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)`

Answer 2

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°.