Why it works: Heron’s Formula is derived by combining the standard area formula \(A = \frac{1}{2}bh\) with the Law of Cosines to eliminate the height entirely.
Named after Hero of Alexandria (~60 AD), though likely known to Archimedes centuries earlier.
Heron’s Formula — Worked Example
A triangular plot of land has sides of 40 m, 50 m, and 70 m. Find its area.
Two forest rangers at stations \(A\) and \(B\), which are 10 miles apart, both spot a fire at point \(F\). Ranger \(A\) measures angle \(FAB = 52°\) and Ranger \(B\) measures angle \(FBA = 61°\). How far is the fire from Station \(A\)?
A telephone pole leans 8° from vertical. From a point on the ground 40 ft from the base, the angle of elevation to the top of the pole is 32°. Find the length of the pole.
The angle at the base of the pole (from the ground) is \(90° - 8° = 82°\).
A surveyor needs to locate a boundary marker. From point \(A\), she knows angle \(A = 38°\), side \(b = 20\) m, and side \(a = 14\) m. How many positions are possible for the third vertex?
\(h = 20 \sin 38° \approx 12.31\)
Since \(h \approx 12.31 < a = 14 < b = 20\): Two positions are possible.
\(\sin B = \frac{20 \sin 38°}{14} \approx 0.8791\)
Two tugboats pull a barge. Their cables make an angle of \(50°\) with each other. The cable lengths are 120 ft and 150 ft. How far apart are the tugboats?
\(d^2 = 120^2 + 150^2 - 2(120)(150)\cos 50°\)
\(= 14400 + 22500 - 36000(0.6428)\)
\(= 36900 - 23140.8 = 13759.2\)
\(d = \sqrt{13759.2} \approx 117.3 \text{ ft}\)
Law of Cosines — Problem 4
A parallelogram has sides of 12 cm and 17 cm. The angle between them is \(65°\). Find the length of the longer diagonal.
The longer diagonal is opposite the obtuse interior angle: \(180° - 65° = 115°\).
\(d^2 = 12^2 + 17^2 - 2(12)(17)\cos(115°)\)
\(= 144 + 289 - 408(-0.4226)\)
\(= 433 + 172.42 = 605.42\)
\(d = \sqrt{605.42} \approx 24.61 \text{ cm}\)
Law of Cosines — Problem 5
A golfer hits a ball 200 yards due north. A strong wind pushes it so its final position is 180 yards from the tee at a bearing of \(N\,25°\,E\). How far did the ball drift east from the intended landing spot?
The angle between the 200-yd path and the 180-yd actual distance is \(25°\).
\(d^2 = 200^2 + 180^2 - 2(200)(180)\cos 25°\)
\(= 40000 + 32400 - 72000(0.9063)\)
\(= 72400 - 65253.6 = 7146.4\)
\(d = \sqrt{7146.4} \approx 84.5 \text{ yards}\)
Choosing the Right Law
Use this decision tree to choose your approach for any oblique triangle.
html`<div style="font-family: Georgia, serif; font-size: 24px; line-height: 1.8; background: #f8f9fa; padding: 20px; border-radius: 10px; border-left: 5px solid #3498db;"><p><strong style="color:#2c3e50">What information do you have?</strong></p><ul> <li><strong style="color:#2980b9">AAS or ASA</strong> → <span style="color:#27ae60">Law of Sines</span></li> <li><strong style="color:#e67e22">SSA</strong> → <span style="color:#e74c3c">Law of Sines — Ambiguous Case (check for 0, 1, or 2 triangles)</span></li> <li><strong style="color:#8e44ad">SAS</strong> → <span style="color:#27ae60">Law of Cosines</span> (then Law of Sines for remaining angles)</li> <li><strong style="color:#c0392b">SSS</strong> → <span style="color:#27ae60">Law of Cosines</span> (find largest angle first)</li> <li><strong style="color:#16a085">SSS (area only)</strong> → <span style="color:#27ae60">Heron's Formula</span></li></ul><p><em>Always verify: angles sum to 180°, triangle inequality holds.</em></p></div>`