Last lesson's triangle worked because a pure resistance and a pure reactance sit exactly 90° apart. Real loads are rarely that tidy. A motor winding is part resistance and part reactance, so its impedance points somewhere between 0° and 90°, and a line reactor in series with it points somewhere else. Each impedance is a phasor: an arrow with a length in ohms and an angle. Putting two in series means adding the arrows, not the lengths.

$Z = \sqrt{Z_1^{2} + Z_2^{2} + 2 Z_1 Z_2 \cos\Delta\theta}$, read aloud Z equals the root of Z-one squared plus Z-two squared plus two Z-one Z-two cos delta-theta. $Z_1$ and $Z_2$ are the magnitudes of the two series impedances, in ohms; subscript 1 is one element and 2 is the other, and the order does not matter. $\Delta\theta$ (say delta-theta) is the angle BETWEEN the two phasors, in degrees: one impedance's own angle minus the other's. $Z$ is the magnitude of the pair, in ohms, and it is what you are solving for. Only the difference of the angles matters, and its sign does not, because the cosine of −40° is the cosine of 40°.

Three limits are worth carrying, because they tie the whole chapter together. At 0° the cosine is 1, the quantity under the root is a perfect square, and $Z = Z_1 + Z_2$: that is why series resistors simply add. At 90° the cosine is 0, the cross term vanishes, and you are back to $\sqrt{R^{2} + X^{2}}$. At 180° the cosine is −1 and $Z = |Z_1 - Z_2|$: an inductor against a capacitor, and the doorway to resonance. Every other angle lands in between, and the resultant is always shorter than the plain sum. Adding the magnitudes head-on is the marquee trap of this lesson, and it always comes out too big.

One warning for anyone who remembers the law of cosines from geometry class, where it carries a MINUS: that version uses the interior angle of the triangle. Between two phasors drawn tail to tail, the angle is the supplement of that interior angle, and the sign flips to a plus. Same triangle, measured from the other side. Read backwards, three bridge readings give up the angle: $\Delta\theta = \arccos\!\left(\dfrac{Z^{2} - Z_1^{2} - Z_2^{2}}{2 Z_1 Z_2}\right)$. And keep the calculator in degrees, or the cosine will quietly answer a different question.