Determining Vector Resultant Direction with Sine Rule

Resultant Direction Requires Another Angle

The cosine rule gives the resultant magnitude, but it does not fully give the direction. In vector problems, an answer such as $15.6\text{ N}$ is incomplete if the force direction is not stated.

For two vectors, the resultant direction can be found using the sine rule. This rule works on the resultant triangle formed by the two vectors and their resultant.

\frac{\sin\phi}{B}=\frac{\sin\theta}{R}

Here, $\theta$ is the included angle between $\vec{A}$ and $\vec{B}$ , $R$ is the resultant magnitude, and $\phi$ is the angle between $\vec{A}$ and resultant $\vec{R}$ .

The sine rule is usually used after the resultant magnitude is known.

Reading the Angle Being Asked For

Before calculating, decide the reference direction. If the question asks for the resultant direction relative to $\vec{A}$ , the required angle is from $\vec{A}$ toward $\vec{R}$ .

If the question asks for direction relative to $\vec{B}$ , the angle is different. Do not swap angles without checking the vector triangle.

A common relationship is:

\sin\phi=\frac{B\sin\theta}{R}

This formula finds the resultant angle relative to vector $\vec{A}$ . Because the side opposite angle $\phi$ has magnitude $B$ , $B$ appears in the numerator.

Resultant Force Direction

Two forces act at one point. $A=10\text{ N}$ , $B=8\text{ N}$ , and the included angle is $60^\circ$ . From the cosine rule, the resultant magnitude is $R\approx15.6\text{ N}$ .

Now find angle $\phi$ between $\vec{A}$ and $\vec{R}$ .

\begin{aligned} \sin\phi&=\frac{B\sin\theta}{R} \\ &=\frac{8\sin60^\circ}{15.6} \\ &=\frac{8(0.866)}{15.6} \\ &\approx0.444 \end{aligned}

The calculation gives:

\phi=\sin^{-1}(0.444)\approx26.4^\circ

So the resultant force is about $15.6\text{ N}$ at $26.4^\circ$ from $\vec{A}$ toward $\vec{B}$ .

Checking with Components

The same direction can be checked using components. If $\vec{A}$ is placed on the positive $x$ axis, then:

R_x=A+B\cos\theta

For the same example:

\begin{aligned} \tan\phi &=\frac{8\sin60^\circ}{10+8\cos60^\circ} \\ &=\frac{6.928}{14} \\ &\approx0.495 \end{aligned}

This gives $\phi\approx26.4^\circ$ , matching the sine rule. This check is useful when you are unsure which angle is being found.

Because $\sin\phi=\sin(180^\circ-\phi)$ , inverse sine alone can return an acute angle even when the triangle requires its obtuse supplement. Use the vector geometry or the component form $\phi=\operatorname{atan2}(R_y,R_x)$ to choose the correct quadrant.

Component Methods for More than Two Vectors

When there are more than two vectors, the component method puts every vector on the same $x$ and $y$ axes and then adds the components. This avoids constructing several triangles and deciding which triangle angle belongs in each sine-rule equation. The same component method also fits directions already measured from the coordinate axes.

Use the sine rule when the problem is clear: two vectors, a known included angle, a known resultant magnitude, and a requested resultant direction relative to one of the vectors. Always resolve the inverse-sine ambiguity before stating the direction.