The Pythagoras theorem works only for right triangles: c 2 = a 2 + b 2 c^2 = a^2 + b^2 c 2 = a 2 + b 2 not right-angled. The Law of Cosines is Pythagoras
plus a correction term that accounts for the angle not being 90 ° 90° 90°
If the angle is exactly 90 ° 90° 90°
If the angle is acute (< 90 ° <90° < 90° shrinks c c c
If the angle is obtuse (> 90 ° >90° > 90° grows c c c
WHY it matters: with it you can solve any triangle when you know
(a) two sides + the included angle (SAS), or (b) all three sides (SSS) — cases
the Law of Sines alone cannot start.
Definition Law of CosinesIn any triangle with sides a , b , c a, b, c a , b , c A , B , C A, B, C A , B , C c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C a 2 = b 2 + c 2 − 2 b c cos A b 2 = a 2 + c 2 − 2 a c cos B a^2 = b^2 + c^2 - 2bc\cos A \qquad b^2 = a^2 + c^2 - 2ac\cos B a 2 = b 2 + c 2 − 2 b c cos A b 2 = a 2 + c 2 − 2 a c cos B correction term is = = − 2 a b cos C = = ==-2ab\cos C== == − 2 ab cos C ==
WHAT the letters mean: C C C between the two sides a a a b b b c c c opposite that angle. The angle and its opposite side always pair up.
Intuition Why the correction sign is "minus"cos C > 0 \cos C > 0 cos C > 0 subtract → c c c C → 90 ° C\to 90° C → 90° cos C → 0 \cos C\to 0 cos C → 0 C C C cos C < 0 \cos C<0 cos C < 0 − 2 a b cos C > 0 -2ab\cos C>0 − 2 ab cos C > 0 adds length.
Solve the definition for cos C \cos C cos C cos C = a 2 + b 2 − c 2 2 a b \cos C = \frac{a^2 + b^2 - c^2}{2ab} cos C = 2 ab a 2 + b 2 − c 2 Why useful? Given all three sides, this directly returns the angle via arccos \arccos arccos
Worked example SAS — find the third sideTriangle with a = 8 a = 8 a = 8 b = 5 b = 5 b = 5 C = 60 ° C = 60° C = 60° c c c c 2 = 8 2 + 5 2 − 2 ( 8 ) ( 5 ) cos 60 ° c^2 = 8^2 + 5^2 - 2(8)(5)\cos 60° c 2 = 8 2 + 5 2 − 2 ( 8 ) ( 5 ) cos 60° Why this step? SAS gives the two sides around C C C between them — exactly the inputs the formula needs.
c 2 = 64 + 25 − 80 ( 0.5 ) = 89 − 40 = 49 ⇒ c = 7 c^2 = 64 + 25 - 80(0.5) = 89 - 40 = 49 \Rightarrow c = 7 c 2 = 64 + 25 − 80 ( 0.5 ) = 89 − 40 = 49 ⇒ c = 7 Why cos 60 ° = 0.5 \cos 60°=0.5 cos 60° = 0.5 Standard value; makes the correction − 40 -40 − 40
Worked example SSS — find an angleSides a = 7 a=7 a = 7 b = 8 b=8 b = 8 c = 9 c=9 c = 9 C C C 9 9 9 cos C = 7 2 + 8 2 − 9 2 2 ( 7 ) ( 8 ) = 49 + 64 − 81 112 = 32 112 = 0.2857 \cos C = \frac{7^2 + 8^2 - 9^2}{2(7)(8)} = \frac{49+64-81}{112} = \frac{32}{112} = 0.2857 cos C = 2 ( 7 ) ( 8 ) 7 2 + 8 2 − 9 2 = 112 49 + 64 − 81 = 112 32 = 0.2857 Why this arrangement? C C C c = 9 c=9 c = 9 c c c a , b a,b a , b C = arccos ( 0.2857 ) ≈ 73.4 ° C = \arccos(0.2857) \approx 73.4° C = arccos ( 0.2857 ) ≈ 73.4°
Worked example Obtuse checka = 3 a = 3 a = 3 b = 4 b = 4 b = 4 C = 120 ° C = 120° C = 120° cos 120 ° = − 0.5 \cos120° = -0.5 cos 120° = − 0.5 c 2 = 9 + 16 − 2 ( 3 ) ( 4 ) ( − 0.5 ) = 25 + 12 = 37 ⇒ c ≈ 6.08 c^2 = 9 + 16 - 2(3)(4)(-0.5) = 25 + 12 = 37 \Rightarrow c \approx 6.08 c 2 = 9 + 16 − 2 ( 3 ) ( 4 ) ( − 0.5 ) = 25 + 12 = 37 ⇒ c ≈ 6.08 Why bigger than the 5 5 5 90 ° 90° 90° The obtuse angle adds + 12 +12 + 12
Recall Forecast before computingTriangle a = 6 , b = 6 , C = 90 ° a=6,\ b=6,\ C=90° a = 6 , b = 6 , C = 90° Predict c c c before reading on.
Verify: c 2 = 36 + 36 − 2 ( 36 ) ( 0 ) = 72 c^2 = 36+36-2(36)(0) = 72 c 2 = 36 + 36 − 2 ( 36 ) ( 0 ) = 72 c = 6 2 ≈ 8.49 c=6\sqrt2\approx8.49 c = 6 2 ≈ 8.49 90 ° 90° 90°
Common mistake Pairing the wrong angle with the wrong sideWrong feels right: "I'll use the angle I know, A A A b b b c c c − 2 b c cos B -2bc\cos B − 2 b c cos B Why tempting: all the letters are floating around and it's easy to mismatch.
Fix: The angle in the correction term must be the one opposite the side on the left. In c 2 = ⋯ − 2 a b cos C c^2=\dots-2ab\cos C c 2 = ⋯ − 2 ab cos C c c c C C C a , b a,b a , b neighbours .
Common mistake Using the wrong angle in SASWrong: given a , b a,b a , b non-included angle, plugging it straight in.
Fix: Law of Cosines' SAS needs the angle between the two given sides. If the angle is not between them, it's the ambiguous SSA case — use the Law of Sines instead.
Common mistake Dropping the sign for obtuse anglesWrong: treating cos 120 ° \cos120° cos 120° + 0.5 +0.5 + 0.5 Fix: cos \cos cos negative in the second quadrant. Keeping the sign is exactly what makes obtuse triangles come out longer.
Recall Explain to a 12-year-oldImagine two sticks joined at a hinge. The far ends have a gap between them. If you open the hinge
a little (small angle), the ends are close, so the gap is small. Open it wide (big angle) and the
ends swing far apart, so the gap is large. The Law of Cosines is just a formula that tells you the
gap's length from the two stick lengths and how wide you opened the hinge. When the hinge is a
perfect corner (90 ° 90° 90°
"SOS — Same, Opposite, Subtract cos."
Left side = S quare of the lone side; right = sum of the other two S quares; then
S ubtract 2 × 2\times 2 × × cos \times\cos × cos O pposite angle).
Also: "The angle you cosine is the one facing the loner side."
State the Law of Cosines for side c c c c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C Which angle appears in c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2+b^2-2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C The angle
C C C opposite side
c c c (between sides
a a a and
b b b ).
What does the Law of Cosines reduce to when C = 90 ° C=90° C = 90° Pythagoras:
c 2 = a 2 + b 2 c^2=a^2+b^2 c 2 = a 2 + b 2 (since
cos 90 ° = 0 \cos90°=0 cos 90° = 0 ).
Rearrange to find angle C C C cos C = a 2 + b 2 − c 2 2 a b \cos C = \dfrac{a^2+b^2-c^2}{2ab} cos C = 2 ab a 2 + b 2 − c 2 Which two triangle cases require Law of Cosines to start? SAS (two sides + included angle) and SSS (three sides).
Why is the correction term negative for acute angles? cos C > 0 \cos C>0 cos C > 0 so
− 2 a b cos C < 0 -2ab\cos C<0 − 2 ab cos C < 0 , shrinking
c c c below the Pythagorean value.
For an obtuse angle, why is c c c cos C < 0 \cos C<0 cos C < 0 , so
− 2 a b cos C > 0 -2ab\cos C>0 − 2 ab cos C > 0 adds length.
In the coordinate proof, what are B B B B = ( a cos C , a sin C ) B=(a\cos C, a\sin C) B = ( a cos C , a sin C ) .
Given a = 8 , b = 5 , C = 60 ° a=8,b=5,C=60° a = 8 , b = 5 , C = 60° c c c c 2 = 64 + 25 − 40 = 49 c^2=64+25-40=49 c 2 = 64 + 25 − 40 = 49 , so
c = 7 c=7 c = 7 .
Pythagoras Theorem — the C = 90 ° C=90° C = 90° Law of Sines — complements this; used for AAS/ASA and the ambiguous SSA.Dot Product — u ⃗ ⋅ v ⃗ = ∣ u ∣ ∣ v ∣ cos θ \vec{u}\cdot\vec{v}=|u||v|\cos\theta u ⋅ v = ∣ u ∣∣ v ∣ cos θ Solving Triangles — decision tree: which law for which given data.Unit Circle and Cosine Values — why cos \cos cos 90 ° 90° 90°
extended by correction term
two sides plus included angle
Intuition Hinglish mein samjhoDekho, Law of Cosines ka basic idea simple hai: Pythagoras theorem sirf right-angle triangle
ke liye chalta hai (c 2 = a 2 + b 2 c^2 = a^2+b^2 c 2 = a 2 + b 2 c 2 = a 2 + b 2 − 2 a b cos C c^2 = a^2 + b^2 - 2ab\cos C c 2 = a 2 + b 2 − 2 ab cos C C C C 90 ° 90° 90° cos 90 ° = 0 \cos90°=0 cos 90° = 0
Yeh formula tab kaam aata hai jab tumhe do side aur beech ka angle pata ho (SAS), ya teeno
side pata ho (SSS). In cases mein Law of Sines akela shuru nahi kar sakta. Ek zaroori baat yaad
rakho: correction term mein jo angle aata hai wo hamesha wahi hota hai jo opposite side ke saamne
hai. Yaani c 2 c^2 c 2 C C C c c c a a a b b b
Proof bilkul basic hai — triangle ko coordinate axis par rakho, C C C B B B ( a cos C , a sin C ) (a\cos C, a\sin C) ( a cos C , a sin C ) cos 2 + sin 2 = 1 \cos^2+\sin^2=1 cos 2 + sin 2 = 1 120 ° 120° 120° cos \cos cos negative hota hai, toh sign flip ho jaata hai aur c c c
