C.4 Geometry and trigonometry

Syllabus
2021
Topic
Level
A2

Learning objectives

Use angle structure before calculating forces

Angles in a physical diagram come from the geometry of the structure and from the direction in which each vector acts. Establish these angle relationships before choosing a calculation.

Geometric fact Useful consequence
angles on a straight line total 180180^\circ adjacent direction angles can be found
angles around a point total 360360^\circ all vector directions at a joint can be checked
triangle angles total 180180^\circ a third angle follows from two known angles
perpendicular directions differ by 9090^\circ a plane's normal is perpendicular to the plane
parallel lines preserve corresponding/alternate angles an incline angle can transfer to a force triangle

On an inclined plane, weight remains vertically downward, the normal contact force is perpendicular to the plane, and friction is parallel to the plane. The plane angle therefore fixes the complementary angles used when resolving weight.

In a regular three-dimensional structure, identify which edges or axes are genuinely perpendicular or parallel. A perspective drawing may make a right angle look oblique, so use stated geometry rather than apparent page angle.

The angle marked in a diagram may be measured from the horizontal, vertical, plane or normal. Name its reference direction explicitly; using the correct number with the wrong reference swaps the relevant components.

Represent the object before representing its forces

A two-dimensional representation selects the plane and directions needed to solve a three-dimensional physical situation. A force diagram then isolates one object and shows only the external forces acting on it.

Step Representation decision
isolate replace the chosen object by a point or simple outline
choose axes align them with useful geometry, often horizontal/vertical or parallel/perpendicular to a plane
add forces draw an arrow from the object in each force's actual direction
label name each force and include a symbol or value when known
check include every external interaction once; keep geometry consistent

For a block on a rough incline, show weight vertically downward, normal contact force perpendicular to the surface and friction parallel to the surface opposing actual or impending relative motion.

A 2D projection can omit a third coordinate only when no required force or displacement component lies outside the chosen plane. Otherwise use separate perpendicular components or another view.

Velocity and acceleration arrows are not forces. Do not include the force the object exerts on another body in the same free-body diagram, and do not assume arrow lengths are to scale unless stated.

Geometry converts measured lengths into physical area and volume

Choose a geometry formula that matches the actual shape, convert every length to one unit system, and power the conversion factor with the dimension of the result.

Shape Length/area Surface area Volume
triangle A=12bhA=\tfrac12 bh - -
circle C=2πrC=2\pi r, A=πr2A=\pi r^2 - -
rectangular block - 2(lw+lh+wh)2(lw+lh+wh) lwhlwh
cylinder cross-section πr2\pi r^2 2πrh+2πr22\pi rh+2\pi r^2 πr2h\pi r^2h
sphere - 4πr24\pi r^2 43πr3\tfrac43\pi r^3

R=ρLA,Awire=πd24R=\frac{\rho L}{A},\qquad A_{\mathrm{wire}}=\frac{\pi d^2}{4}

For a wire of diameter 0.400mm0.400\,\mathrm{mm}, A=π(0.400×103)2/4=1.26×107m2A=\pi(0.400\times10^{-3})^2/4=1.26\times10^{-7}\,\mathrm{m^2}. With L=2.00mL=2.00\,\mathrm{m} and ρ=1.70×108Ωm\rho=1.70\times10^{-8}\,\Omega\,\mathrm{m}, R=0.270ΩR=0.270\,\Omega.

Diameter is twice radius, and 1mm2=106m21\,\mathrm{mm^2}=10^{-6}\,\mathrm{m^2} rather than 103m210^{-3}\,\mathrm{m^2}. For a composite object, divide it into non-overlapping standard shapes before adding areas or volumes.

Use right-triangle structure to combine perpendicular vectors

Pythagoras' theorem relates only the sides of a right-angled triangle. Perpendicular vector components form such a triangle, so their resultant magnitude is the hypotenuse.

R=Rx2+Ry2R=\sqrt{R_x^2+R_y^2}

For components Rx=6.0NR_x=6.0\,\mathrm{N} east and Ry=8.0NR_y=8.0\,\mathrm{N} north, R=6.02+8.02=10.0NR=\sqrt{6.0^2+8.0^2}=10.0\,\mathrm{N}. The direction is found separately from the component triangle.

A triangle's interior angles total 180180^\circ. To check whether measured sides aa, bb and longest side cc make a right angle, test whether a2+b2=c2a^2+b^2=c^2 within measurement uncertainty.

Do not use Pythagoras for non-perpendicular vectors; resolve them onto perpendicular axes first or use a more general triangle rule. Squared components lose their signs, but signs remain essential when the components are first combined along each axis.

Resolve a vector from the angle's reference axis

Sine, cosine and tangent connect a vector to a right-triangle representation. First identify the angle's reference axis; the adjacent component uses cosine and the opposite component uses sine.

Relationship Use
sinθ=opposite/hypotenuse\sin\theta=\text{opposite}/\text{hypotenuse} component opposite the stated angle
cosθ=adjacent/hypotenuse\cos\theta=\text{adjacent}/\text{hypotenuse} component beside the stated angle
tanθ=opposite/adjacent\tan\theta=\text{opposite}/\text{adjacent} angle or ratio of perpendicular components

A 50.0N50.0\,\mathrm{N} force at 30.030.0^\circ above the horizontal has Fx=50.0cos30.0=43.3NF_x=50.0\cos30.0^\circ=43.3\,\mathrm{N} and Fy=50.0sin30.0=25.0NF_y=50.0\sin30.0^\circ=25.0\,\mathrm{N}.

For a resultant with components RxR_x and RyR_y, tanθ=Ry/Rx\tan\theta=R_y/R_x. Use the signs of both components to choose the correct quadrant and state the direction relative to an axis.

If the supplied angle is measured from the vertical, the horizontal and vertical sine/cosine assignments swap. Keep the calculator in the angle mode used by the question and never drop component signs.

Small angles turn trigonometry into simple ratios

For a sufficiently small angle measured in radians, sinθθ\sin\theta\approx\theta, tanθθ\tan\theta\approx\theta and cosθ1\cos\theta\approx1. These approximations replace a curved trigonometric relationship by a simple linear one.

Check Requirement
angle unit θ\theta must be in radians
geometry transverse displacement is much smaller than distance to the screen
use retain the approximation sign and check the resulting scale is plausible

wλDsw\approx\frac{\lambda D}{s}

For wavelength λ=600nm\lambda=600\,\mathrm{nm}, screen distance D=2.00mD=2.00\,\mathrm{m} and slit separation s=0.500mms=0.500\,\mathrm{mm}, the fringe spacing is w=(600×109)(2.00)/(0.500×103)=2.40mmw=(600\times10^{-9})(2.00)/(0.500\times10^{-3})=2.40\,\mathrm{mm}.

The approximations are not identities and fail as the angle grows. Applying sin3030\sin30^\circ\approx30 is meaningless: convert degrees to radians before comparing the angle with its sine or tangent.

Radians measure angle by arc length

One radian is the central angle that subtends an arc equal in length to the radius. This definition makes angular relationships such as arc length and phase naturally dimensionless.

θ=sr,2π rad=360\theta=\frac{s}{r},\qquad 2\pi\ \mathrm{rad}=360^\circ

Conversion Rule
degrees to radians multiply by π/180\pi/180
radians to degrees multiply by 180/π180/\pi
phase fraction of one cycle θ/(2π)\theta/(2\pi) in radians or θ/360\theta/360^\circ in degrees

A phase difference of 25.025.0^\circ is 25.0π/180=0.436rad25.0\pi/180=0.436\,\mathrm{rad}. Conversely, 1.20rad=1.20(180/π)=68.81.20\,\mathrm{rad}=1.20(180/\pi)=68.8^\circ.

Radians are dimensionless but write rad when it prevents ambiguity. Match calculator mode to the angle supplied, and do not multiply by 2π/3602\pi/360 twice: that expression is already the degree-to-radian factor.