Question bank — Law of cosines — proof and applications

The Law of Cosines only works for right triangles. ::: False. It works for every triangle; the right-triangle case is just where the correction term $-2ab\cos C$ happens to vanish because $\cos 90°=0$. If you know two angles and one side, the Law of Cosines is the tool to start with. ::: False. Two angles + a side (AAS/ASA) has no "two sides around a known angle" to plug in — that's a Law of Sines situation. Law of Cosines needs SAS or SSS. Increasing the included angle $C$ while keeping $a,b$ fixed always increases the opposite side $c$. ::: True. As $C$ grows from $0°$ to $180°$, $\cos C$ decreases monotonically, so $-2ab\cos C$ grows, so $c^2$ grows — the "hinge opens wider, gap gets bigger" picture. $c^2=a^2+b^2-2ab\cos C$ and $c^2=a^2+b^2+2ab\cos C$ are both valid, just with different sign conventions. ::: False. Only the minus version is correct. The sign is fixed by the derivation, not a convention; the "plus" case is already covered because $\cos C$ itself goes negative for obtuse angles. For a fixed $a$ and $b$, the largest possible value of $c$ occurs as $C\to 180°$. ::: True. At $C=180°$ the two sides lie flat in a straight line and $\cos 180°=-1$, giving $c^2=a^2+2ab+b^2=(a+b{)}^2$, so $c=a+b$ — the sides fully stretched out. The Law of Cosines can be used to check whether three given lengths form a valid triangle. ::: True. Compute $\cos C=\frac{a^2+b^2-c^2}{2ab}$; if this lands outside $[-1,1]$, no real angle exists, so the lengths cannot close into a triangle. If $\cos C$ comes out negative from the SSS formula, you made an arithmetic mistake. ::: False. A negative $\cos C$ is perfectly legal — it just means $C$ is obtuse ($>90°$), which happens whenever $c^2>a^2+b^2$. The Law of Cosines and the Dot Product formula $\vec{u}\cdot \vec{v}=|u||v|\cos \theta$ are secretly the same statement. ::: True. Writing $\vec{c}=\vec{b}-\vec{a}$ and expanding $|\vec{c}{|}^2=(\vec{b}-\vec{a})\cdot (\vec{b}-\vec{a})$ gives exactly $c^2=a^2+b^2-2ab\cos \theta$. The dot product is the correction term.

A student writes "$c^2=a^2+b^2-2ab\cos B$" for the side opposite $C$. Where's the slip? ::: The cosine must carry the angle opposite the lone side. Side $c$ pairs with angle $C$, not $B$. Correct: $-2ab\cos C$. Given $a=5$, $b=7$ and the angle $A=40°$ (opposite $a$), a student writes $a^2=b^2+c^2-2bc\cos 40°$ and solves for $c$ in one clean step. What's wrong? ::: They only know one side around angle $A$ (namely $b$); $c$ is unknown too, so this is one equation with the unknown appearing quadratically — it's the ambiguous SSA case, not clean SAS. May have 0, 1, or 2 solutions; Law of Sines or a quadratic is needed. "$\cos 120°=0.5$, so the correction just shrinks $c$ a bit." Spot the error. ::: $\cos 120°=-0.5$, not $+0.5$. Cosine is negative in the second quadrant. The correction $-2ab(-0.5)=+ab$ actually grows $c$. A student concludes that because Pythagoras needs a right angle, the Law of Cosines is "less general." Fix the logic. ::: It's backwards. Law of Cosines is more general — it contains Pythagoras as the special case $C=90°$. The one with fewer assumptions is the broader tool. "The included angle in SAS is any angle I happen to know." Spot the misconception. ::: The included angle must sit between the two known sides. An angle not wedged between them gives you no valid SAS input for the formula. Solving SSS, a student gets $\cos C=1.4$ and writes $C=\arccos (1.4)$. What went wrong upstream? ::: $\arccos$ is undefined for inputs outside $[-1,1]$. A value like $1.4$ signals the three side lengths violate the triangle inequality — no such triangle exists, so there's an error in the given data, not the method. "In $a^2=b^2+c^2-2bc\cos A$, the letters $b$ and $c$ can be swapped freely." Any harm? ::: No harm here — $b$ and $c$ are the two neighbours of angle $A$ and enter symmetrically as $2bc$ and in the sum $b^2+c^2$. Swapping them changes nothing. (But swapping the lone side with a neighbour breaks the formula.)

Why does the correction term vanish exactly at $90°$ and not, say, taper off gradually near it? ::: Because the term is $-2ab\cos C$ and $\cos 90°=0$ exactly. It does taper gradually as you approach $90°$ (cosine slides through zero), but it hits precisely zero only at $90°$ — that single point is where Pythagoras is exact. Why must the angle you cosine be the one opposite the lone side, geometrically? ::: In the coordinate proof, that angle sits at the origin between the two sides you placed, and the lone side is the gap across from it. The gap's length is governed by how wide that specific corner opens — no other angle controls it. Why is the Law of Cosines needed for SSS when the Law of Sines already relates sides and angles? ::: The Law of Sines $\frac{a}{\sin A}=\frac{b}{\sin B}$ needs at least one angle to get started. With only three sides you have zero angles, so Sines can't launch. The rearranged cosine formula produces the first angle from pure lengths. Why does using $\cos$ (rather than $\sin$ or $\tan$) fall out of the derivation? ::: The proof places $B$ at coordinates $(a\cos C,a\sin C)$ and squares the distance. Squaring and applying ${\cos }^{2}C+{\sin }^{2}C=1$ collapses the $\sin$ term away, leaving only a lone $\cos C$ — cosine is what survives the algebra. Why does the obtuse case make the triangle's opposite side longer than the Pythagorean estimate? ::: For obtuse $C$, $\cos C<0$, so $-2ab\cos C$ becomes a positive addition. Physically the two sides splay past a right angle, flinging their far ends apart, so the gap exceeds $\sqrt{a^2+b^2}$. Why can we relabel the formula three ways ($a^2=\dots$, $b^2=\dots$, $c^2=\dots$) without a new proof each time? ::: The naming of vertices $A,B,C$ is arbitrary — the geometry is symmetric under relabelling. Proving one version and rotating the letters gives the other two for free.

What does the formula give when $C=0°$ (the two sides fold onto each other)? ::: $\cos 0°=1$, so $c^2=a^2+b^2-2ab=(a-b{)}^2$, giving $c=|a-b|$. The gap is just the difference in stick lengths — the ends nearly coincide. What happens at $C=180°$ (sides in a straight line)? ::: $\cos 180°=-1$, so $c^2=(a+b{)}^2$ and $c=a+b$. This is the degenerate triangle: all three points collinear, "triangle" flattened to a segment. If $a=b$, does the formula simplify meaningfully? ::: Yes: $c^2=2a^2(1-\cos C)$. The triangle is isosceles, so $c$ depends only on the equal side and the apex angle — handy for regular polygons and Solving Triangles. What if one side is $0$ — say $a=0$? ::: Then $c^2=b^2$, so $c=b$: the triangle collapses because vertex $B$ coincides with $C$. Degenerate, but the formula still returns something consistent. Can $\cos C=0$ happen for a non-right triangle? ::: No. $\cos C=0$ forces $C=90°$ exactly, which is a right triangle. So the correction term is zero if and only if the triangle is right-angled. As $C$ ranges over all valid triangle angles $(0°,180°)$, what is the full range of $c$? ::: $c$ runs strictly between $|a-b|$ (limit as $C\to 0°$) and $a+b$ (limit as $C\to 180°$) — exactly the triangle-inequality bounds, with the endpoints only reached in the degenerate flat cases.