I’m solid with Pythagoras and basic trig, but the cosine rule keeps tripping me up in dumb ways. I know c^2 = a^2 + b^2 − 2ab cos C, but I keep mixing up which side/angle go together and when the angle should be obtuse.
Example: sides 5, 7, and 10. The longest side is 10, so the angle opposite 10 should be the biggest. Quick check: 10^2 = 100 is bigger than 5^2 + 7^2 = 74, so that big angle should be > 90°. But when I plug into the formula, my calculator gives me an angle that looks way too small. So I’m obviously pairing things wrong (or rounding something I shouldn’t).
I tried rearranging to cos C = (a^2 + b^2 − c^2)/(2ab) and setting c = 10 so C is opposite it. I also tried relabeling the triangle so the given angle is C if it’s the included angle, but then I get tangled when the given angle isn’t between the known sides. Not sure which approach is actually the ‘right’ habit.
Can someone give me the simple, foolproof way to label and plug numbers so I don’t mismatch sides/angles? Also, what’s a quick sanity check to decide if the angle should come out obtuse before I hit arccos? And any trick to stop the cos value drifting slightly over 1 or under −1 from rounding so the calculator doesn’t complain?
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3 Responses
I get tangled by the labels too, so my “foolproof” habit is: pick the angle you want, call it C, and then c is the side directly opposite it; the sides that touch C are a and b, and that’s the version c^2 = a^2 + b^2 − 2ab cos C you should use. Think of it like place cards at a dinner table: angle C sits across from seat c, and the two neighbors of C are a and b. For your 5, 7, 10 triangle, put the longest side opposite the angle you’re finding: let c = 10, a = 5, b = 7, so cos C = (5^2 + 7^2 − 10^2)/(2·5·7) = (25 + 49 − 100)/70 = −26/70 ≈ −0.3714, hence C ≈ 111.8°, which matches the quick check 10^2 > 5^2 + 7^2 so the big angle is obtuse. General sanity checks: biggest side ↔ biggest angle; if c^2 > a^2 + b^2 then C is obtuse, = gives right, < gives acute; and before pressing arccos, a negative cosine means the angle will be > 90°. To avoid rounding grief, carry extra digits and then clamp: if your computed cosine is, say, 1.0000003, replace it by 1; if it’s −1.0000003, replace by −1. One last habit that keeps me from mixing things up: when given two sides and an included angle, name that angle C so a and b are the sides you know; when given three sides, always solve first for the angle opposite the longest side using cos C = (a^2 + b^2 − c^2)/(2ab), which also minimizes the chance you’ll mispair things.
Here’s the foolproof habit: pick the angle you care about first, call it C, and label the side directly opposite it as c. The other two sides are a and b (their order doesn’t matter because the formula is symmetric in a and b). If you’re given two sides and the included angle, that angle is C and the unknown opposite side is c, so use c^2 = a^2 + b^2 − 2ab cos C. If you’re given three sides and want an angle, choose the angle opposite the longest side (that’ll be the biggest angle), label that side c, and use cos C = (a^2 + b^2 − c^2)/(2ab). Swapping a and b won’t change the answer; the only way to go wrong is to mismatch C with a side that isn’t opposite it.
For your 5, 7, 10 example, set c = 10 and a = 5, b = 7. Then cos C = (25 + 49 − 100)/(2·5·7) = −26/70 = −13/35 ≈ −0.371, so C ≈ arccos(−0.371) ≈ 111.8°, which matches the “longest side means obtuse angle” gut check. Quick sanity checks I like: compare c^2 with a^2 + b^2-if c^2 > a^2 + b^2, C must be obtuse; if equal, C is 90°; if less, C is acute. Also, before dividing, the sign of (a^2 + b^2 − c^2) already tells you the sign of cos C. For rounding gremlins, carry extra digits and only round at the end; if your computed cosine is, say, 1.0000002 or −1.0000003 from rounding, just clip it to 1 or −1 before hitting arccos. Hope this helps!
Here’s the foolproof habit: always rename the triangle so the angle you want is C and the side directly opposite it is c; then the other two sides are a and b opposite A and B. The cosine rule doesn’t care which letters you pick, but the opposite-pairing must match (C ↔ c, A ↔ a, B ↔ b). Quick sanity checks: longest side ↔ largest angle, and the Pythagoras-style test c^2 ? a^2 + b^2 tells you the vibe-if c^2 > a^2 + b^2 then angle C is obtuse; equal means right; less means acute. Also, cos C should be negative for an obtuse angle (anything past a right angle-roughly 100° or more-makes cos go negative), so if you get a small positive cosine with the longest side opposite C, something’s mismatched. Worked example: with sides 5, 7, 10, set c = 10, a = 5, b = 7. Then cos C = (a^2 + b^2 − c^2)/(2ab) = (25 + 49 − 100)/(2·5·7) = −26/70 ≈ −0.3714, so C ≈ arccos(−0.3714) ≈ 112°, nicely obtuse as expected. A practical rounding guard: after computing the cosine, clamp it to the interval [−1, 1] before arccos (e.g., if you get 1.0000002, treat it as 1). One more mental cue: when you know C is obtuse, you can think “minus times negative equals plus,” so the −2ab cos C term effectively adds something; some people even flip the minus to a plus in their head to remember they should get a bigger c than Pythagoras would predict. If the given angle isn’t the included one, I like to relabel so it is, since cosine rule ‘likes’ the included angle best-then it’s just plug-and-chug with matching opposite pairs and a quick sign check on cos to confirm the angle’s nature.