Finding supplementary angles from a tangent equation

Key takeaway: This problem combines an equation in tangent with the angle relationship formed by intersecting lines, so the main task is to solve for one angle and then convert it to the other possible angle.

Identify the relationship between the angles

When two lines meet, the two adjacent angles form a straight line, so they are supplementary. That means

$A+B=180^\circ.$

If one angle is found first, the other angle is determined by subtracting from $180^\circ$. Because tangent is periodic, the equation for $A$ may give more than one angle in the valid range, which is why there are two possible values of $B$.

Solve for angle A

Given

$2+3\tan A=-0.5,$

subtract 2 from both sides:

$3\tan A=-2.5,$

so

$\tan A=-\frac{5}{6}.$

This tangent value corresponds to an acute reference angle of

$\alpha=\tan^{-1}\left(\frac{5}{6}\right).$

Since the tangent is negative, $A$ can lie in either Quadrant II or Quadrant IV depending on the context and allowed angle range.

Find the two possible values of B

If $A$ is one angle formed by the intersecting lines, the supplementary angle is

$B=180^\circ-A.$

Using the two possible directed-angle interpretations gives two matching values for $B$. In practical school geometry, the intended pair is often the acute and obtuse supplementary pair.

So the two possible values of $B$ are determined by the two possible corresponding angles for $A$ that satisfy $\tan A=-\frac{5}{6}$ and the supplementary-angle rule.

Key trig fact

The equation $\tan\theta=t$ repeats every $180^\circ$, so when a problem asks for possible angle values, always check the interval implied by the diagram or by standard angle conventions. Then use $A+B=180^\circ$ to move from one angle to the other.

---

Pitfalls the pros know 👇 The biggest pitfall is solving only for a reference angle and forgetting that tangent is negative here. That can lead to a single answer when the problem expects two possible values. Another common issue is mixing up supplementary angles with vertically opposite angles. Adjacent angles on a straight line add to $180^\circ$, but vertically opposite angles are equal. Students also sometimes leave the tangent equation as a decimal and never connect it back to the angle pair, which makes it hard to justify the final result. Always state the geometric relationship before giving the numerical angle.

What if the problem changes? If the equation were changed to $2+3\tan A=0.5$, then $\tan A=-\frac{1.5}{3}=-\frac{1}{2}$, so the reference angle would change. The method would stay the same: solve for $\tan A$, identify the valid angle(s) for $A$, then use $B=180^\circ-A$. If the lines were not forming a straight line but instead crossed as vertically opposite angles, then the relationship would change completely and you would set $A=B$ rather than $A+B=180^\circ$.

Tags: supplementary angles, tangent ratio, reference angle