Triangle law of vector addition

It states that: If two vector quantities acting simultaneously on a body are represented in both magnitude and direction by two sides of triangle taken in order, the resultant of the vector in both magnitude and direction is given by the third side taken in the opposite order.

Triangle law of vector addition

To find the magnitude and direction of the resultant of P and Q, a line OA is produced. From the point B, perpendicular is drawn on OA and produced to C. Let θ
be the angle between R⃗  and  P⃗  . Let ϕ be the angle made by the resultant R⃗  with  P⃗ .

Using the basic laws of vector addition,
In triangle OBC,
OB² = OC² + CB²
or, R² = (OA + AC)² + CB²
or, R² = OA² + 2OA ·AC + AC² + CB²……… (i)
In triangle BAC,

cosθ=ACBA=ACQ

∴AC=Qcosθ

Also,

sinθ=CBBA=CBQ

∴CB=Qcosθ

Substituting the value of OA, AC and CB in equation (i), we get:
R² = P² + 2PQCos  + (QCosθ)² + (QSinθ)²
or, R² = P² + 2PQCos  + Q² (Cos²θ + Sin²θ)
Since, Cos² θ + Sin²θ = 1,

R=P2+2PQcosθ+Q2−−−−−−−−−−−−−−−−√

The above equation gives the resultant of P and Q.
To find the direction of resultant R, suppose β
be the angle made by the resultant with direction of P.
In triangle OBC,

tanϕ=BCOC

or,tanϕ=BCOA+AC

tanϕ=QsinθP+Qcosθ

tan∴ϕ=tan−1(QsinθP+Qcosθ)……(ii)

The above equation gives the direction of resultant vector.