Question 3

We will use the given tangent values to find the angles $A+B$ and $A-B$. Then we will solve the resulting system of linear equations.

Step 1 — Find the angles

Key principle: If $\tan\theta_1 = \tan\theta_2$ and both angles are acute (between 0° and 90°), then $\theta_1 = \theta_2$. So matching the expression to a known tan value directly gives the angle.

We are given the value of $\tan(A+B)$. We know that $\tan 60^\circ$ is equal to $\sqrt{3}$. So, we can write the first equation.

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

$\tan (A + B) = \tan 60^\circ$

$\boxed{A + B = 60^\circ}$

Next, we use the second given value. We know that $\tan 30^\circ$ is equal to $\frac{1}{\sqrt{3}}$. So, we can write the second equation.

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

$\tan (A - B) = \tan 30^\circ$

$\boxed{A - B = 30^\circ}$

Step 2 — Solve for A and B

System of Linear Equations: Two equations with the same two unknowns (A and B). We use the elimination method — adding both equations cancels B and gives A directly, then we substitute back to find B.

Now we have a system of two linear equations. Let's call $A + B = 60^\circ$ as Equation (1). Let's call $A - B = 30^\circ$ as Equation (2). We can add Equation (1) and Equation (2) together.

$(A + B) + (A - B) = 60^\circ + 30^\circ$

$2A = 90^\circ$

$A = \frac{90^\circ}{2}$

$\boxed{A = 45^\circ}$

Now, let's substitute the value of A into Equation (1). Equation (1) is $A + B = 60^\circ$.

$45^\circ + B = 60^\circ$

$B = 60^\circ - 45^\circ$

$\boxed{B = 15^\circ}$

Answer

(i) $A = 45^\circ$ (ii) $B = 15^\circ$