5.12 Trigonometry in context: vectors, kinematics and forces — Learn

Prerequisites: the sine and cosine rules; right-angled trigonometry; Pythagoras' theorem; the idea of a vector as a magnitude and a direction.

By the end of this section you can: resolve a force or velocity into two perpendicular components; find the resultant of two vectors at an angle; model a velocity in a current; resolve a weight on a slope.

1. Resolving a vector

A vector has a magnitude and a direction. Splitting a vector into two perpendicular parts is called resolving. For a vector of magnitude $F$ at an angle $\theta$ to the horizontal:

The component in a direction is the magnitude times the cosine of the angle between the vector and that direction. The perpendicular component uses the sine.

Example 1. A force of $50$ N acts at $30^\circ$ above the horizontal. Resolve it horizontally and vertically.

2. The resultant of two vectors

When two vectors act at a point, the single vector with the same effect is the resultant. Draw the two vectors nose to tail to form a triangle, then use the cosine rule and the sine rule.

If two vectors of magnitudes $p$ and $q$ have an angle $\theta$ between them, the resultant $R$ satisfies

The plus sign appears because the triangle's internal angle is $180^\circ - \theta$, and $\cos(180^\circ - \theta) = -\cos\theta$.

Example 2. Two forces of $6$ N and $8$ N act at $60^\circ$ to each other. Find the magnitude of the resultant.

To find the direction, use the sine rule in the same triangle. Let $\alpha$ be the angle between the resultant and the $8$ N force. Then

3. Velocity in a current

The same triangle method works for velocities. The velocity through the water and the velocity of the current add as vectors.

Example 3. A boat travels at $4$ m s$^{-1}$ due east. A current flows at $3$ m s$^{-1}$ due north. Find the size and the direction of the boat's resultant velocity.

The two velocities are perpendicular, so Pythagoras' theorem gives the size:

The angle north of east satisfies

The boat moves at $5$ m s$^{-1}$ at $36.9^\circ$ north of east.

4. Forces on a slope

A weight of magnitude $W$ on a slope of angle $\theta$ to the horizontal resolves into two perpendicular parts:

• down the slope: $W\sin\theta$;
• into the slope: $W\cos\theta$.

Example 4. A block of weight $200$ N rests on a slope of $25^\circ$. Find the components of the weight down the slope and into the slope.

Watch out. Three mistakes are common when resolving.

Measuring the angle from the wrong direction. The component in a direction uses the angle between the vector and that direction. A force at $30^\circ$ to the horizontal has horizontal component $F\cos 30^\circ$. If the angle given is to the vertical, the two components swap. Draw the triangle and mark the angle.

Using the wrong sign in the resultant. For two vectors at an angle $\theta$ to each other, the resultant uses $+2pq\cos\theta$. For two sides of a triangle with included angle $\theta$, the cosine rule uses $-2pq\cos\theta$. Decide which picture you are drawing before you choose.

Forgetting that components are signed. When a vector points left or down, its component in the positive direction is negative. Add the signed components, not the sizes. Re-check: the resultant must be at least as large as the larger component and no larger than the sum of the magnitudes.

You should now be able to

• resolve a force or velocity into two perpendicular components;
• find the resultant of two vectors at an angle with the cosine rule;
• find the direction of a resultant with the sine rule or with a tangent;
• model a velocity in a current;
• resolve a weight on a slope;
• check a resultant against the sizes of the two vectors.