5.12 Trigonometry in context: vectors, kinematics and forces — Practise with help

Draw the vector triangle before you calculate. Mark the angle you have been given, and check whether it is between the two vectors or between a vector and a direction.

Warm-up: remember

Work out the hypotenuse of a right-angled triangle with shorter sides $3$ and $4$ .
For an angle $\theta$ in a right-angled triangle, write down $\tan\theta$ in terms of the opposite and adjacent sides.
Work out $10\cos 60^\circ$ and $10\sin 60^\circ$ .

Answers

$\sqrt{9 + 16} = 5$ .
$\tan\theta = \dfrac{\text{opposite}}{\text{adjacent}}$ .
$10\cos 60^\circ = 10 \times \dfrac{1}{2} = 5$ and $10\sin 60^\circ = 10 \times \dfrac{\sqrt{3}}{2} = 5\sqrt{3} \approx 8.66$ .

If question 2 was hard, revise right-angled trigonometry before you continue.

Practise with help

1. Resolve a force of $20$ N at $40^\circ$ above the horizontal into its horizontal and vertical components, to three significant figures.

Hint

The horizontal component uses cosine. The vertical component uses sine.

Next step

Work out $20\cos 40^\circ$ and $20\sin 40^\circ$ .

Full working

\text{horizontal} = 20\cos 40^\circ \approx 15.3 \text{ N}.

\text{vertical} = 20\sin 40^\circ \approx 12.9 \text{ N}.

2. Two forces of $5$ N and $9$ N act at $90^\circ$ to each other. Find the magnitude of the resultant.

Hint

At $90^\circ$ the cosine is zero, so the resultant uses Pythagoras' theorem.

Next step

$R = \sqrt{5^2 + 9^2} = \sqrt{25 + 81}$ .

Full working

R = \sqrt{25 + 81} = \sqrt{106} \approx 10.3 \text{ N}.

3. Two forces of $4$ N and $7$ N act at $50^\circ$ to each other. Find the magnitude of the resultant, to three significant figures.

Hint

Use $R^2 = p^2 + q^2 + 2pq\cos\theta$ with $\theta = 50^\circ$ .

Next step

$R^2 = 16 + 49 + 2 \times 4 \times 7 \times \cos 50^\circ = 65 + 56 \times 0.6428$ .

Full working

R^2 = 16 + 49 + 56 \times 0.6428 \approx 65 + 36.0 = 101.0.

R \approx \sqrt{101.0} \approx 10.0 \text{ N}.

4. A swimmer heads due east at $1.2$ m s $^{-1}$ in a current of $0.5$ m s $^{-1}$ due south. Find the size of the resultant velocity.

Hint

The two velocities are perpendicular. Draw them nose to tail.

Next step

$v = \sqrt{1.2^2 + 0.5^2}$ .

Full working

v = \sqrt{1.44 + 0.25} = \sqrt{1.69} = 1.3 \text{ m s}^{-1}.

5. A block of weight $120$ N is on a slope of $20^\circ$ . Find the component of the weight down the slope.

Hint

The component down the slope uses the sine of the slope angle.

Next step

Work out $120\sin 20^\circ$ .

Full working

120\sin 20^\circ \approx 120 \times 0.3420 = 41.0 \text{ N}.

6. Two forces of $6$ N and $8$ N act at $60^\circ$ to each other. The resultant has magnitude $12.2$ N. Find the angle between the resultant and the $8$ N force, to one decimal place.

Hint

Use the sine rule in the vector triangle: $\dfrac{\sin\alpha}{6} = \dfrac{\sin 60^\circ}{12.2}$ .

Next step

$\sin\alpha = \dfrac{6\sin 60^\circ}{12.2} = \dfrac{5.196}{12.2} \approx 0.4259$ .

Full working

\alpha = \sin^{-1}(0.4259) \approx 25.2^\circ.