Trigonometry for Electricians

Electricians use trigonometry more than almost any other trade. From bending conduit to calculating impedance in AC circuits, trig shows up every day on the job. If you have ever measured a phase angle, looked up a power factor, or figured the offset multiplier for a conduit bend, you were doing trig — whether you called it that or not.

This page connects the right-triangle trig you learned in SOH CAH TOA and Finding Missing Angles to real electrical work.

Impedance Triangles

In an AC circuit, impedance ($Z$) is the total opposition to current flow. It combines two components:

• Resistance ($R$) — opposition from resistive loads (heaters, incandescent lamps)
• Reactance ($X$) — opposition from inductors ($X_L$) or capacitors ($X_C$)

These form a right triangle where:

• $R$ is the adjacent side (horizontal)
• $X$ is the opposite side (vertical)
• $Z$ is the hypotenuse

$Z = \sqrt{R^2 + X^2}$

The phase angle $\theta$ between voltage and current is:

$\theta = \arctan\!\left(\frac{X}{R}\right)$

Worked Example: Impedance and Phase Angle

A motor circuit has resistance $R = 30 \;\Omega$ and inductive reactance $X_L = 40 \;\Omega$. Find the impedance $Z$ and the phase angle $\theta$.

Step 1 — Find impedance:

$Z = \sqrt{R^2 + X_L^2} = \sqrt{30^2 + 40^2} = \sqrt{900 + 1600} = \sqrt{2500} = 50 \;\Omega$

Step 2 — Find the phase angle:

$\theta = \arctan\!\left(\frac{X_L}{R}\right) = \arctan\!\left(\frac{40}{30}\right) \approx \arctan(1.3333) \approx 53.13°$

Answer: The impedance is 50 ohms and the phase angle is 53.13 degrees. This is a 3-4-5 right triangle scaled by 10 — the same ratio you see in basic trig.

Power Triangles and Power Factor

The power triangle works exactly like the impedance triangle, but with power quantities instead of ohms:

• Real power $P$ (watts, W) — the useful work, analogous to $R$ (adjacent)
• Reactive power $Q$ (volt-amperes reactive, VAR) — energy stored and returned by inductors/capacitors, analogous to $X$ (opposite)
• Apparent power $S$ (volt-amperes, VA) — total power the source must deliver, analogous to $Z$ (hypotenuse)

$S = \sqrt{P^2 + Q^2}$

The power factor (PF) is the cosine of the phase angle:

$\text{PF} = \cos\theta = \frac{P}{S}$

A power factor of 1.0 means all power is doing useful work. A power factor of 0.8 means real power is 80% of the apparent power, with the remaining apparent power caused by reactive loads.

Worked Example: Power Triangle

A motor has an apparent power of $S = 5 \text{ kVA}$ and a power factor of 0.8. Find the real power, the reactive power, and the phase angle.

Step 1 — Find the phase angle from PF:

$\theta = \arccos(0.8) \approx 36.87°$

Step 2 — Find real power:

$P = S \times \cos\theta = 5{,}000 \times 0.8 = 4{,}000 \text{ W}$

Step 3 — Find reactive power:

$Q = S \times \sin\theta = 5{,}000 \times \sin(36.87°) = 5{,}000 \times 0.6 = 3{,}000 \text{ VAR}$

Verification: $S = \sqrt{P^2 + Q^2} = \sqrt{4{,}000^2 + 3{,}000^2} = \sqrt{16{,}000{,}000 + 9{,}000{,}000} = \sqrt{25{,}000{,}000} = 5{,}000 \text{ VA}$ . Correct.

Answer: Real power is 4,000 W, reactive power is 3,000 VAR, and the phase angle is 36.87 degrees. Notice this is another 3-4-5 triangle (scaled by 1,000).

AC Sine Waves

Alternating current follows a sine wave. In North America, standard wall outlet voltage is described by:

$v(t) = 170 \sin(120\pi t)$

Where $v(t)$ is the instantaneous voltage in volts and $t$ is time in seconds.

From this equation you can read off the key parameters using what you know about Graphs of Sine and Cosine:

• Peak voltage ($V_{\text{peak}}$) = 170 V (the amplitude)
• Angular frequency ($\omega$) = $120\pi$ rad/s
• Frequency = $\frac{\omega}{2\pi} = \frac{120\pi}{2\pi} = 60 \text{ Hz}$
• Period = $\frac{1}{60} \approx 0.01667 \text{ s}$

Peak vs. RMS Voltage

The voltage you measure with a multimeter is not the peak — it is the RMS (root mean square) value:

$V_{\text{RMS}} = \frac{V_{\text{peak}}}{\sqrt{2}} = \frac{170}{1.414} \approx 120 \text{ V}$

That is why we call it “120-volt” power even though the sine wave actually peaks at 170 volts. For 240 V circuits, the peak is approximately $240 \times \sqrt{2} \approx 339$ V.

Conduit Bending

Every offset bend uses a trig-based multiplier. When you bend conduit at an angle $\theta$, the relationship between the offset depth and the travel distance comes from the sine function:

$\text{Multiplier} = \frac{1}{\sin(\theta)}$

The 30-degree bend is the most popular on the job because the multiplier is exactly 2 — easy mental math.

For full details on offset bends, saddle bends, shrink constants, and stub-ups, see Conduit Bending Math.

Common Mistakes

1. Confusing impedance with resistance. $Z$ and $R$ are only equal when reactance is zero (a purely resistive circuit). In any circuit with motors, transformers, or capacitors, $Z$ is always larger than $R$.

2. Forgetting to convert power factor to an angle. Power factor is $\cos\theta$, not $\theta$ itself. If PF = 0.8, the phase angle is $\arccos(0.8) \approx 36.87°$, not 0.8 degrees and not 80 degrees.

3. Mixing degrees and radians with angular frequency. The $\omega$ in $v(t) = V_{\text{peak}} \sin(\omega t)$ is in radians per second. If your calculator is set to degrees, the sine calculation will be wrong. When working with AC waveforms, set your calculator to radians.

4. Treating VA the same as watts. Apparent power (VA) and real power (W) are only equal when the power factor is 1.0. A 5 kVA generator with a 0.8 PF load only delivers 4 kW of real power. Sizing equipment based on watts alone can lead to overloaded circuits.

Practice Problems

Test your understanding with these problems. Click to reveal each answer.

Key Takeaways

• The impedance triangle relates resistance, reactance, and impedance using the same right-triangle trig as SOH CAH TOA
• The power triangle uses the same geometry: real power (W), reactive power (VAR), and apparent power (VA)
• Power factor = $\cos\theta$ — it tells you what fraction of apparent power does useful work
• AC voltage follows a sine wave; the RMS value you measure is peak divided by $\sqrt{2}$
• Conduit bending multipliers come directly from $1 / \sin(\theta)$
• Every one of these calculations is a right triangle problem — if you know SOH CAH TOA, you can do electrical trig

Return to Trigonometry for more topics, or see Math for Electricians for more electrical math.