Angles and Their Measure

Study Sheet

Angles and Their Measure

Standard position, radians, arc length, sector area, and speed

Angles in Standard Position

Concept
Standard Position
55°

An angle is in standard position when its vertex sits at the origin and its initial side lies along the positive xx-axis. The terminal side is where the ray ends after rotating.

  • [leftmargin=5mm]
  • A positive angle rotates counterclockwise.
  • A negative angle rotates clockwise.

The quadrant of the angle is the quadrant containing its terminal side.

An angle measures a turn.

A positive angle θ\theta in standard position (terminal side in Quadrant II).

A negative angle −θ-\theta (clockwise rotation, terminal side in Quadrant IV).

Concept
Quadrantal Angles

A quadrantal angle has its terminal side lying on an axis. The common ones are

0∘,90∘,180∘,270∘,360∘0^\circ,\quad 90^\circ,\quad 180^\circ,\quad 270^\circ,\quad 360^\circ

(equivalently 0, π2, π, 3π2, 2π0,\ \dfrac{\pi}{2},\ \pi,\ \dfrac{3\pi}{2},\ 2\pi radians). These angles are in no quadrant.

Coterminal Angles

Concept
Coterminal Angles
55°

Two angles are coterminal if they share the same terminal side. Add or subtract full revolutions:

θ±360∘k(degrees)orθ±2πk(radians),k=1,2,3,…\theta \pm 360^\circ k \quad(\text{degrees})\qquad\text{or}\qquad \theta \pm 2\pi k \quad(\text{radians}),\quad k=1,2,3,\dots

An angle measures a turn.

60∘60^\circ and −300∘-300^\circ are coterminal: 60∘−360∘=−300∘60^\circ-360^\circ=-300^\circ.

Example
Example: Coterminal Angles

Find one positive and one negative angle coterminal with θ=210∘\theta=210^\circ. Positive: 210∘+360∘=570∘210^\circ+360^\circ=\mathbf{570^\circ}.  Negative: 210∘−360∘=−150∘210^\circ-360^\circ=\mathbf{-150^\circ}.

Tip

Tip: To find the coterminal angle between 0∘0^\circ and 360∘360^\circ, keep adding or subtracting 360∘360^\circ until you land in that range. (For radians, use 2π2\pi.)

Degree Measure and DMS

Concept
Degrees, Minutes, Seconds
55°

One full revolution is 360∘360^\circ. A degree splits further:

1∘=60′ (minutes),1′=60′′ (seconds),1∘=3600′′.1^\circ = 60' \ (\text{minutes}),\qquad 1' = 60'' \ (\text{seconds}),\qquad 1^\circ = 3600''.

An angle measures a turn.

Example
Example: Decimal →\to DMS

Convert 24.6∘24.6^\circ to DMS. Multiply the decimal part by 6060:

0.6×60=36′.⇒24.6∘=24∘36′0.6\times 60 = 36'.\qquad\Rightarrow\quad 24.6^\circ = \mathbf{24^\circ 36'}
Example
Example: DMS →\to Decimal

Convert 18∘45′36′′18^\circ 45' 36'' to decimal degrees.

18+4560+363600=18+0.75+0.01=18.76∘18 + \frac{45}{60} + \frac{36}{3600} = 18 + 0.75 + 0.01 = \mathbf{18.76^\circ}
Tip

Tip: Going to DMS, multiply the leftover decimal by 6060 at each step. Going from DMS, divide minutes by 6060 and seconds by 36003600, then add.

Radian Measure and Conversions

Concept
Radians
55°

One radian is the central angle that subtends an arc equal in length to the radius. A full circle is 2π2\pi radians, so

 180∘=π radians \boxed{\,180^\circ = \pi \text{ radians}\,}
degrees→radians: ×π180∘radians→degrees: ×180∘π\text{degrees}\to\text{radians: } \times\dfrac{\pi}{180^\circ}\qquad\qquad \text{radians}\to\text{degrees: } \times\dfrac{180^\circ}{\pi}

An angle measures a turn.

Example
Example: Degrees →\to Radians

Convert 150∘150^\circ to radians.

150∘×π180∘=150π180=5π6 rad150^\circ \times \frac{\pi}{180^\circ} = \frac{150\pi}{180} = \mathbf{\dfrac{5\pi}{6}\ \text{rad}}
Example
Example: Radians →\to Degrees

Convert 3π4\dfrac{3\pi}{4} radians to degrees.

3π4×180∘π=3×180∘4=135∘\frac{3\pi}{4}\times\frac{180^\circ}{\pi} = \frac{3\times 180^\circ}{4} = \mathbf{135^\circ}
Tip

Remember: When no degree symbol appears, the angle is in radians. Keep answers exact with π\pi unless a decimal is requested.

Arc Length and Sector Area

Concept
Arc Length and Sector Area

For a circle of radius rr and a central angle θ\theta measured in radians:

Arc length: s=rθSector area: A=12r2θ\text{Arc length: } s = r\theta\qquad\qquad \text{Sector area: } A = \tfrac{1}{2}r^2\theta

If θ\theta is given in degrees, convert to radians first.

A sector with radius rr, central angle θ\theta, arc length ss, and area AA.

Example
Example: Arc Length and Area

A circle has r=6r=6 and central angle θ=π3\theta=\dfrac{\pi}{3}.

s=rθ=6⋅π3=2πA=12r2θ=12(36)π3=6πs = r\theta = 6\cdot\frac{\pi}{3} = \mathbf{2\pi}\qquad A=\tfrac12 r^2\theta = \tfrac12(36)\frac{\pi}{3} = \mathbf{6\pi}
Tip

Trap: s=rθs=r\theta and A=12r2θA=\tfrac12 r^2\theta only work when θ\theta is in radians. A degree measure must be converted first.

Linear Speed and Angular Speed

Concept
Two Kinds of Speed

Suppose a point moves along a circle of radius rr, sweeping angle θ\theta (radians) in time tt.

Angular speed: ω=θtLinear speed: v=st=rω\text{Angular speed: } \omega = \frac{\theta}{t}\qquad\qquad \text{Linear speed: } v = \frac{s}{t} = r\omega

Revolutions: 11 revolution =2π=2\pi radians, so NN RPM =2πN= 2\pi N rad per minute.

Example
Example: Wheel Speed

A wheel of radius 0.30.3 m spins at 6060 RPM. Find its angular and linear speed.

ω=60×2π=120π rad/min\omega = 60\times 2\pi = 120\pi \ \text{rad/min}
v=rω=0.3×120π=36π≈113.1 m/minv = r\omega = 0.3\times 120\pi = \mathbf{36\pi \approx 113.1 \ \text{m/min}}
Tip

Tip: Belts and gears in contact share the same linear speed vv, so r1ω1=r2ω2r_1\omega_1 = r_2\omega_2. A point at the center has ω\omega but zero linear speed.

Complementary and Supplementary Angles

Concept
Complements and Supplements
55°

Two positive angles are

  • [leftmargin=5mm]
  • complementary if they add to 90∘90^\circ (or π2\dfrac{\pi}{2} rad),
  • supplementary if they add to 180∘180^\circ (or π\pi rad).

So the complement of θ\theta is 90∘−θ90^\circ-\theta, and its supplement is 180∘−θ180^\circ-\theta.

An angle measures a turn.

Example
Example

Find the complement and supplement of 40∘40^\circ.

Complement=90∘−40∘=50∘Supplement=180∘−40∘=140∘\text{Complement} = 90^\circ-40^\circ = \mathbf{50^\circ}\qquad \text{Supplement} = 180^\circ-40^\circ = \mathbf{140^\circ}
Tip

Remember: Only angles smaller than 90∘90^\circ have a complement. “Complement” pairs with 9090 (a Corner); “Supplement” pairs with 180180 (a Straight line).

Going Deeper: Advanced Angle Ideas

Concept
Gear, Pulley, and Belt Chains
55°

When two wheels are connected, one of two rules applies:

  • [leftmargin=5mm]
  • In contact (meshed gears, or a belt/pulley) they share the same linear speed at the rim: v1=v2v_1=v_2, so r1ω1=r2ω2r_1\omega_1 = r_2\omega_2.
  • On a common axle (rigidly fixed together) they share the same angular speed: ω1=ω2\omega_1=\omega_2.

To trace a chain, alternate these rules link by link. For meshed gears the tooth counts obey N1ω1=N2ω2N_1\omega_1 = N_2\omega_2, since teeth are equally spaced around the rim.

An angle measures a turn.

Two meshed wheels touch at the rim, so their rim (linear) speeds match: r1ω1=r2ω2r_1\omega_1=r_2\omega_2.

Example
Example: A Two-Stage Gear Train

Gear A (rA=8r_A=8 cm) meshes with gear B (rB=3r_B=3 cm). Fixed rigidly to gear B on the same axle is gear C (rC=6r_C=6 cm), which meshes with gear D (rD=2r_D=2 cm). Gear A turns at 3030 RPM. Find the angular speed of gear D. First convert: ωA=30×2π=60π\omega_A = 30\times 2\pi = 60\pi rad/min. A--B in contact: rAωA=rBωB⇒ωB=83(60π)=160πr_A\omega_A = r_B\omega_B \Rightarrow \omega_B = \dfrac{8}{3}(60\pi)=160\pi rad/min. B--C on one axle: ωC=ωB=160π\omega_C=\omega_B = 160\pi rad/min. C--D in contact: rCωC=rDωD⇒ωD=62(160π)=480π rad/minr_C\omega_C = r_D\omega_D \Rightarrow \omega_D = \dfrac{6}{2}(160\pi)=\mathbf{480\pi\ \text{rad/min}}. As RPM: ωD=480π2π=240 RPM\omega_D = \dfrac{480\pi}{2\pi}=\mathbf{240\ \text{RPM}}.

Concept
Clock-Hand Angles

A clock face is 360∘360^\circ, split into 1212 hours of 30∘30^\circ each. The hands sweep at constant rates:

minute hand: 6∘ per min,hour hand: 0.5∘ per min.\text{minute hand: } 6^\circ \text{ per min},\qquad \text{hour hand: } 0.5^\circ \text{ per min}.

At H:MH{:}M the hands make angles (measured clockwise from 1212)

θmin⁡=6M,θhr=30H+0.5M.\theta_{\min}=6M,\qquad \theta_{\text{hr}}=30H+0.5M.

The angle between them is ∣ 30H−5.5M ∣|\,30H-5.5M\,|; if this exceeds 180∘180^\circ, subtract from 360∘360^\circ.

The angle ϕ\phi between the hour and minute hands.

Example
Example: Angle at 3:40, and When Hands Overlap

(a) Angle at 3:40. Use ∣ 30H−5.5M ∣|\,30H-5.5M\,| with H=3, M=40H=3,\ M=40:

∣ 30(3)−5.5(40) ∣=∣ 90−220 ∣=130∘.|\,30(3)-5.5(40)\,| = |\,90-220\,| = 130^\circ.

Since 130∘<180∘130^\circ<180^\circ, the angle between the hands is 130∘\mathbf{130^\circ}. (b) Overlaps. The hands coincide when 6M=30H+0.5M6M = 30H+0.5M, i.e. 5.5M=30H5.5M=30H, so M=60H11M=\dfrac{60H}{11}. Starting from 12:0012{:}00, successive overlaps are spaced

Δt=360∘6∘−0.5∘ per min=3605.5=72011≈65.45 min\Delta t = \frac{360^\circ}{6^\circ-0.5^\circ \text{ per min}}=\frac{360}{5.5}=\mathbf{\dfrac{720}{11}\approx 65.45\ \text{min}}

apart, giving 1111 overlaps every 1212 hours (not 1212). Equivalently the minute hand gains a full 360∘360^\circ on the hour hand every 36011\dfrac{360}{11} of an hour.

Concept
Sector Systems with Two Solutions

A sector is fixed by any two of {r, θ, s, A}\{r,\ \theta,\ s,\ A\}. Because arc length is linear in rr but area is quadratic in rr, combining an arc (or perimeter) condition with an area condition often yields a quadratic, hence two valid sectors. The sector perimeter is

P=s+2r=rθ+2r,P = s + 2r = r\theta + 2r,

the arc plus the two radii. Pair this with A=12r2θA=\tfrac12 r^2\theta to solve.

Example
Example: Two Sectors, Same Perimeter and Area Data

A sector has perimeter P=16P=16 and area A=15A=15. Find the possible radii and central angles. From A=12r2θ=15A=\tfrac12 r^2\theta=15 we get rθ=30rr\theta = \dfrac{30}{r}, so the arc s=rθ=30rs=r\theta=\dfrac{30}{r}. Substitute into P=s+2r=16P=s+2r=16:

30r+2r=16  ⟹  2r2−16r+30=0  ⟹  r2−8r+15=0  ⟹  (r−3)(r−5)=0.\frac{30}{r}+2r = 16 \;\Longrightarrow\; 2r^2-16r+30=0 \;\Longrightarrow\; r^2-8r+15=0 \;\Longrightarrow\; (r-3)(r-5)=0.

So r=3r=3 or r=5r=5. If r=3r=3:  θ=30r2=309=103 rad\ \theta = \dfrac{30}{r^2}=\dfrac{30}{9}=\mathbf{\dfrac{10}{3}\ \text{rad}} (about 191∘191^\circ). If r=5r=5:  θ=3025=65 rad\ \theta = \dfrac{30}{25}=\mathbf{\dfrac{6}{5}\ \text{rad}} (about 69∘69^\circ). Both are genuine sectors, so the data admits two solutions.

Concept
Revolutions per Minute and Unit Conversions

Rotating machinery is usually rated in RPM (revolutions per minute). Convert deliberately, one factor at a time:

RPM  →×2π  rad/min  →×r  (rim) distance/min  →÷60  per second.\text{RPM}\;\xrightarrow{\times 2\pi}\;\text{rad/min}\;\xrightarrow{\times r}\;\text{(rim) distance/min}\;\xrightarrow{\div 60}\;\text{per second}.

For ground speed of a rolling wheel, distance per revolution is the circumference 2πr2\pi r. Watch that every quantity uses the same length and time units before combining.

Example
Example: Satellite Ground Track and Wheel Speed

(a) Satellite. A satellite completes one orbit of radius r=7000r=7000 km every 100100 minutes. Its angular speed is

ω=2π100=π50 rad/min,v=rω=7000⋅π50=140π≈439.8 km/min.\omega = \frac{2\pi}{100}=\frac{\pi}{50}\ \text{rad/min},\qquad v = r\omega = 7000\cdot\frac{\pi}{50}=\mathbf{140\pi\approx 439.8\ \text{km/min}}.

(b) Wheel. A car tire of radius 0.350.35 m turns at 500500 RPM. Distance per minute is

v=(2πr)×RPM=2π(0.35)(500)=350π m/min.v = (2\pi r)\times \text{RPM} = 2\pi(0.35)(500)=350\pi\ \text{m/min}.

Converting to km/h:   350π mmin×60 min1 h×1 km1000 m=21π≈66.0 km/h\;350\pi\ \dfrac{\text{m}}{\text{min}}\times\dfrac{60\ \text{min}}{1\ \text{h}}\times\dfrac{1\ \text{km}}{1000\ \text{m}}=21\pi\approx\mathbf{66.0\ \text{km/h}}.

Tip

Coterminal with a condition: to find the angle coterminal with θ\theta that lands in a required window, add or subtract full turns until it fits. For example, the value of 29π6\dfrac{29\pi}{6} in [0,2π)[0,2\pi) is 29π6−2⋅2π=29π−24π6=5π6\dfrac{29\pi}{6}-2\cdot 2\pi=\dfrac{29\pi-24\pi}{6}=\dfrac{5\pi}{6}. To force a specific quadrant or a [−π,π)[-\pi,\pi) range, keep the same ±2πk\pm 2\pi k step and stop in the target interval.

Tip

Master trap: s=rθs=r\theta, A=12r2θA=\tfrac12 r^2\theta, and v=rωv=r\omega all demand radians (and radians/time). RPM and degrees must be converted before they enter these formulas, never after.

Formulas, Proofs & Tips

Tip
Radians, arc length and sector area
π rad=180∘,s=rθ,A=12r2θ(θ in radians)\pi \text{ rad}=180^\circ,\qquad s=r\theta,\qquad A=\tfrac12 r^{2}\theta \quad (\theta \text{ in radians})

What it means. A radian is the angle that cuts an arc as long as the radius.

Example. 90∘=π290^\circ=\tfrac{\pi}{2} rad; with r=6r=6, θ=π2\theta=\tfrac\pi2, arc length s=rθ=3πs=r\theta=3\pi.

Why it works. A full circle has circumference 2πr2\pi r, which is 2π2\pi radius-lengths — so a full turn is 2π2\pi radians and half a turn is π=180∘\pi=180^\circ. Since θ\theta radians is θ2π\tfrac{\theta}{2\pi} of the circle, the arc is θ2π⋅2πr=rθ\tfrac{\theta}{2\pi}\cdot 2\pi r=r\theta.

Tip. s=rθs=r\theta and A=12r2θA=\tfrac12r^2\theta only work in radians. Convert first, or use the θ360\tfrac{\theta}{360} versions instead.

Tip
Coterminal and reference angles
θ±360∘k (coterminal);θref=acute angle to the x-axis\theta \pm 360^\circ k \ (\text{coterminal}); \qquad \theta_{\text{ref}} = \text{acute angle to the } x\text{-axis}

What it means. Angles that land in the same place, and the acute angle that carries the trig value.

Example. 400∘400^\circ is coterminal with 40∘40^\circ; the reference angle of 150∘150^\circ is 30∘30^\circ.

Why it works. A full turn returns to the same ray, so adding 360∘360^\circ changes nothing about the terminal side — and therefore nothing about sine or cosine. The reference angle forms a congruent right triangle, so the ratios match up to sign.

Tip. Find the reference angle, take the trig value there, then attach the sign from the quadrant (ASTC).