I’m revising triangle basics to strengthen my fundamentals, and the cosine rule is the one that still trips me up in small but annoying ways. I think I remember the formula, but I get confused about when exactly I’m allowed to use it, how the sign works if the angle is obtuse, and how to label everything so I don’t accidentally plug in the wrong side/angle.
Here’s how I’m labeling: triangle ABC with sides a, b, c opposite angles A, B, C respectively. So the version I’m using is c^2 = a^2 + b^2 − 2ab cos C. That seems fine when I’m given two sides and the included angle. For example, if a = 7, b = 10, and the included angle C = 120°, I write c^2 = 7^2 + 10^2 − 2·7·10·cos 120°. Since cos 120° is negative, this turns into something like “149 minus a negative number,” which effectively adds. That feels weirdly like “Pythagoras plus an extra chunk,” which I guess matches the idea that an obtuse angle makes the opposite side longer. But I’m not fully confident I’m interpreting the sign correctly. Is my setup here solid, and is thinking of it as “the minus times a negative becomes a plus” the right mental model, or am I oversimplifying?
Another place I wobble is solving for an angle when all three sides are known. Say sides are 5, 6, and 7, and I want the angle opposite 7. I rearranged to cos C = (a^2 + b^2 − c^2)/(2ab) with c = 7, a = 5, b = 6. Numerator is 25 + 36 − 49, denominator is 2·5·6. That gives me a cosine that looks reasonable (positive), so I expect an acute angle. But I’m nervous about two things: (1) does choosing a different angle first ever change things due to rounding, and is there a “safer” order to compute angles to avoid cosine values nudging outside [−1, 1]? and (2) is there a quick way to predict obtuse vs acute from the side lengths before I compute, just to sanity-check the sign of the cosine?
Where I get most stuck is when the known angle isn’t included between the two known sides. For example, if I know a = 8, b = 11, and angle A = 43° (so the angle is opposite a, not between a and b), I keep wanting to use a^2 = b^2 + c^2 − 2bc cos A, but then c is also unknown, so I have two unknowns in one equation. That makes me think cosine rule just isn’t the right starting tool for SSA data and I should go to the sine rule first. Is that the correct logic? Is there ever a way to start with cosine rule in an SSA situation without introducing an extra unknown?
Could someone sanity-check the two worked setups above and point out if I’ve mislabeled anything? I’d also love a step-by-step way to decide which side/angle to call a, b, c so I don’t accidentally treat a non-included angle as included. And, about the sign: is “negative cosine means the −2ab cos term effectively adds” the clean way to think about obtuse cases? I’m aiming to get the reasoning straight, not just memorize plug-and-chug.
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3 Responses
Your setup is spot-on: the cosine rule a^2 = b^2 + c^2 − 2bc cos A is symmetric, and for an obtuse C we get cos C < 0 so the −2ab cos C term adds (so c^2 > a^2 + b^2); for SSS, find the angle opposite the longest side first to tame rounding, and predict obtuse/acute by comparing c^2 with a^2 + b^2 (> obtuse, = right, < acute). For SSA, use the sine rule (cosine adds a second unknown) and watch the ambiguous case, and for labeling always match sides to opposite angles and, when using cosine, pick the included angle as the letter-does the “longest side ↔ largest angle” and “SAS/SSS → cosine, SSA → sine” checklist help lock it in?
Yep, your setups are solid-label the angle you care about as C (opposite c), remember “included” means between the two known sides, and in c^2 = a^2 + b^2 − 2ab cos C an obtuse C makes cos C negative so the −2ab cos term adds; when all three sides are known use cos C = (a^2 + b^2 − c^2)/(2ab), find the largest angle first (opposite the longest side) to reduce rounding, predict obtuse/right/acute by comparing the longest side (c^2 > a^2 + b^2 ⇒ obtuse, = ⇒ right, < ⇒ acute), and for SSA start with the sine rule because cosine would introduce another unknown (and the case can be ambiguous). Example: a=7, b=10, C=120° ⇒ cos120°=−1/2 so c^2=49+100−2·7·10·(−1/2)=219 ⇒ c≈14.8; with a=5, b=6, c=7 ⇒ cos C=(25+36−49)/60=0.2 ⇒ C≈78.5° (acute since 7^2<5^2+6^2); more detail: https://www.khanacademy.org/math/geometry/hs-geo-trig/hs-geo-law-of-cosines/a/law-of-cosines
Your setup and instincts are sound. The law of cosines c^2 = a^2 + b^2 − 2ab cos C is valid for any triangle as long as C is the angle opposite c (and similarly for the other letters). For your SAS example (a=7, b=10, C=120°), cos 120° = −1/2, so c^2 = 49 + 100 − 2·7·10·(−1/2) = 149 + 70 = 219, which is indeed “Pythagoras plus extra,” matching that an obtuse angle makes the opposite side longer-so the “minus times a negative becomes plus” reading is exactly right. For SSS to find an angle, your rearrangement cos C = (a^2 + b^2 − c^2)/(2ab) is correct; with (5,6,7) and c=7, cos C = 12/60 = 0.2, so C ≈ 78.5°, acute. Two practical tips here: (1) compute the largest angle first using the side of greatest length (opposite it) to avoid rounding issues and ambiguity, then use the sine rule for the others; and (2) quick obtuse/acute test: with c the largest side, C is obtuse iff c^2 > a^2 + b^2, right if equal, acute if less. For SSA (e.g., a=8, b=11, A=43°), cosine rule introduces an extra unknown, so start with the sine rule: sin B / b = sin A / a, which here gives sin B ≈ (11/8) sin 43° ≈ 0.938, so B could be about 69.7° or 110.3° (the classic ambiguous case); once you fix B, finish with the sine rule or a single cosine-rule step to get the remaining side. Labeling advice: always pair opposite letters; if you intend to use cosine rule, name the angle you’re solving for as C and set c opposite it; if it’s SSA, commit the known opposite pair to A–a and use the sine rule first; if it’s SSS, label the longest side as c and find C first. Would a short “which law when” checklist for SAS/SSA/SSS/ASA cases be useful to keep by your notes?