Q.Explain in detail the triangle law of addition.

Step 1. Statement of the law. To add two vectors $\vec A$ and $\vec B$ inclined at angle $\theta$: represent $\vec A$ and $\vec B$ as two sides of a triangle taken in the same order — draw $\vec A$ from O to P, then draw $\vec B$ starting from P (the head of $\vec A$) to a point Q. The resultant $\vec R=\vec A+\vec B$ is then the third side of the triangle, drawn directly from O to Q.

Step 2. Magnitude. Extend OP to a point N so that PN is along OP extended, and drop a perpendicular from Q onto this extended line, meeting it at N. In right triangle PQN (with $\angle$QPN $=180°-\theta$'s supplement, giving the angle at P equal to $\theta$ measured appropriately), $PN=B\cos\theta$ and $QN=B\sin\theta$. Then $ON=OP+PN=A+B\cos\theta$, and applying Pythagoras to right triangle OQN:

Expanding, $R^2=A^2+2AB\cos\theta+B^2\cos^2\theta+B^2\sin^2\theta=A^2+B^2+2AB\cos\theta$ (using $\cos^2\theta+\sin^2\theta=1$), so

Step 3. Direction. If $\vec R$ makes angle $\alpha$ with $\vec A$, then in right triangle OQN, $\tan\alpha=\dfrac{QN}{ON}=\dfrac{B\sin\theta}{A+B\cos\theta}$.

Step 4. Special cases (checks). For $\theta=0$ (parallel vectors), $R=A+B$, the ordinary sum. For $\theta=180°$ (anti-parallel), $R=|A-B|$. For $\theta=90°$, $R=\sqrt{A^2+B^2}$, ordinary Pythagoras — all consistent with simple intuition.

Construct the triangle, drop a perpendicular to form a right triangle, apply Pythagoras for magnitude and tan for direction.

• Drawing both vectors from the same tail point (that is the parallelogram law's construction, not the triangle law's)
• Sign errors when expanding (A+Bcosθ)² + (Bsinθ)²