Q.Prove that in any triangle ABC, $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$, where $a$, $b$, $c$ are the magnitudes of the sides opposite to the vertices A, B, C, respectively.

Concept: Law of Cosines (derived via coordinate geometry or vector dot product).

Place vertex $A$ at the origin and side $AB$ along the positive $x$-axis. Then:

• $B = (c, 0)$
• $C = (b \cos A, \, b \sin A)$

The side $a = BC$ is the distance between $B$ and $C$:

Expanding:

Since $\cos^2 A + \sin^2 A = 1$, this simplifies to: …

Why This Formula Works — The Core Intuition

The cosine rule is essentially Pythagoras’ theorem with a correction term for non-right triangles. In a right triangle where $A = 90^\circ$, we have $a^2 = b^2 + c^2$. When $A$ is not $90^\circ$, the side $a$ is either shorter (if $A < 90^\circ$) or longer (if $A > 90^\circ$) than the Pythagorean sum. The term $-2bc\cos A$ accounts for this adjustment.

The formula $\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ is just a rearrangement — it expresses the cosine of the angle in terms of the three side lengths. This is powerful because it lets us find an angle without constructing any perpendiculars.

Step-by-Step Proof

1. Set up a coordinate system

Place vertex $A$ at the origin $(0,0)$ and vertex $C$ on the positive $x$-axis at $(b, 0)$. This is always possible — we can rotate and translate the triangle without changing its side lengths or angles.

Now vertex $B$ must lie somewhere such that $AB = c$ and $AC = b$. Since $AC$ is along the $x$-axis, the coordinates of $B$ are:

Why? Because the distance from $A$ to $B$ is $c$, and the angle between $AB$ and $AC$ is exactly $A$. The $x$-coordinate is the projection of $c$ onto $AC$, and the $y$-coordinate is the perpendicular component.

2. Write the distance $BC$ in terms of coordinates

Side $a$ is the distance between $B$ and $C$. Point $C$ is at $(b, 0)$, so:

3. Expand and simplify

Group the $c^2$ terms:

The identity $\cos^2 A + \sin^2 A = 1$ collapses the $c^2$ terms into a single $c^2$.

4. Rearrange to isolate $\cos A$

Bring the $b^2 + c^2$ term to the left:

Multiply both sides by $-1$:

Finally, divide by $2bc$ (which is non-zero since sides are positive lengths):

$\cos A = \frac{b^2 + c^2 - a^2}{2bc}$ …

Method: Proving the cosine rule by squaring a vector triangle relation

Use this pattern to prove side/angle identities like $\cos A=\dfrac{b^2+c^2-a^2}{2bc}$ using vectors.

Steps

Step 1: Represent the sides as vectors from the angle's vertex.

Put $\vec{AB}$ and $\vec{AC}$ at vertex $A$, with $|\vec{AB}|=c$, $|\vec{AC}|=b$, and angle $A$ between them. Then the opposite side is $\vec{BC}=\vec{AC}-\vec{AB}$, with $|\vec{BC}|=a$.

Step 2: Square by dotting the vector with itself.

$a^2=|\vec{AC}-\vec{AB}|^2=|\vec{AC}|^2+|\vec{AB}|^2-2\,\vec{AC}\cdot\vec{AB}.$ …

Common Mistakes

Mistake 1: Using the wrong pair of adjacent sides for angle $A$.

Why it's wrong: angle $A$ sits between sides $b$ and $c$, so the numerator is $b^2+c^2-a^2$ and the denominator $2bc$; writing $\tfrac{a^2+b^2-c^2}{2ab}$ is actually $\cos C$. Correct approach: match the angle to the two sides meeting at its vertex.

Mistake 2: Sign error in $\vec{BC}=\vec{AC}-\vec{AB}$.

Why it's wrong: reversing it to $\vec{AB}-\vec{AC}$ still squares to the same length, but the cross term's role in $-2\,\vec{AC}\cdot\vec{AB}$ must stay negative to reach the cosine rule. Correct approach: keep $\vec{BC}=\vec{AC}-\vec{AB}$ and expand the square carefully. …