Questions on radians, arcs and small angles

Question 1

Convert 60° to radians and 3π/4 radians to degrees, exactly.

Accepted Answer

Since 180° = π, dividing by 3 gives 60° = π/3. Going the other way, 3π/4 = 3 × 45° = 135°. B1 B1 for the two conversions. Exactly means the π stays; 1.047 is not an answer here.

Question 2

A sector has radius 5 cm and angle 0.8 radians. Find its arc length and area.

Accepted Answer

The arc is s = rθ = 5 × 0.8 = 4 cm, and the area is ½r²θ = ½ × 25 × 0.8 = 10 cm². B1 for the arc length, B1 for the area. Both formulae are only valid with θ in radians, so an angle handed over in degrees must be converted first.

Question 3

State what is meant by an angle of one radian, and find, to 1 decimal place, the number of degrees in one radian.

Accepted Answer

One radian is the angle subtended at the centre of a circle by an arc whose length equals the radius. B1 for the definition. Since a half-turn is π radians and also 180°, one radian is 180/π = 57.3°. B1 for 57.3°. The definition is what makes s = rθ true with no conversion factor attached: an angle of θ radians subtends θ arcs each of length r, so the total arc is rθ.

Question 4

Using the small-angle approximations, write down approximate values of tan 0.02 and cos 0.02, where the angles are in radians.

Accepted Answer

tan θ ≈ θ gives tan 0.02 ≈ 0.02. B1 for 0.02. cos θ ≈ 1 − θ²/2 gives cos 0.02 ≈ 1 − 0.0002 = 0.9998. B1 for 0.9998. The squared term is what makes the cosine approximation say anything: with it dropped the statement would be cos θ ≈ 1, which never changes. A calculator in radian mode agrees with both to 4 decimal places, and the whole claim depends on the angles being radians, since tan 0.02° is 0.000349.

Question 5

A sector has radius 9 cm and angle 1.4 radians. Find its perimeter and its area.

Accepted Answer

The arc is 9 × 1.4 = 12.6 cm, and the perimeter adds the two radii: 12.6 + 18 = 30.6 cm. The area is ½ × 81 × 1.4 = 56.7 cm². B1 for the arc, B1 for adding the two radii, B1 for the area.

Question 6

An arc of length 12 cm subtends an angle θ at the centre of a circle of radius 7.5 cm. Find θ, and the area of the sector.

Accepted Answer

Rearranging s = rθ gives θ = 12/7.5 = 1.6 radians, and then the area is ½ × 7.5² × 1.6 = 45 cm². M1 for rearranging s = rθ, A1 for 1.6 radians, B1 for the area. The θ that comes out is already in radians, so it drops straight into the area formula.

Question 7

Find the area of the minor segment cut off by a chord in a circle of radius 6 cm, where the chord subtends 1.2 radians at the centre. Give 3 significant figures.

Accepted Answer

A segment is the sector with the triangle taken out. ½ × 36 × 1.2 − ½ × 36 × sin 1.2 = 21.6 − 16.78 = 4.82 cm². The triangle uses ½r² sin θ, with the two radii as its sides and θ between them. M1 for sector minus triangle, A1 for the two areas, A1 for 4.82 cm². The calculator must be in radian mode, since sin 1.2 in degrees is 0.0209 and the answer collapses; and the two terms are close together, so carry full accuracy into the subtraction rather than rounding each part first.

Question 8

A sector of a circle has area 30 cm² and angle 0.75 radians. Find its radius and its arc length, each to 3 significant figures.

Accepted Answer

Rearrange A = ½r²θ for the radius: r² = 2A/θ = 60/0.75 = 80, so r = √80 = 8.94 cm. M1 for the rearrangement, A1 for 8.94 cm. Then the arc is s = rθ = √80 × 0.75 = 6.71 cm. B1 for the arc length. Carry the exact √80 into the second formula and round once, at the end. Substituting back is a quick check: ½ × 80 × 0.75 = 30, the area the question gave.

Question 9

A chord of a circle of radius 10 cm subtends an angle of 2 radians at the centre. Find the perimeter of the minor segment cut off by that chord, to 3 significant figures.

Accepted Answer

The boundary of a segment is its arc together with its chord. The arc is s = rθ = 10 × 2 = 20 cm. B1 for the arc. For the chord, drop the perpendicular from the centre to it: that perpendicular bisects both the chord and the angle, leaving a right-angled triangle with hypotenuse 10 and an angle of 1 radian at the centre. So half the chord is 10 sin 1 and the chord is 2 × 10 sin 1 = 16.83 cm. M1 for halving the angle in a right-angled triangle, A1 for 16.8 cm. The perimeter is 20 + 16.83 = 36.8 cm. A1 for 36.8 cm. The calculator must be in radian mode: sin 1 is 0.8415, while sin 1° is 0.0175 and would give a chord of 0.35 cm, shorter than the arc it subtends, which no chord ever is.

Question 10

The minute hand of a clock is 9 cm long. Find the distance travelled by the tip of the hand in 25 minutes, and the area swept out by the hand in that time, each to 3 significant figures.

Accepted Answer

In 25 minutes the hand turns through 25/60 of a full turn, and a full turn is 2π radians, so θ = (25/60) × 2π = 5π/6 radians. M1 for θ = 5π/6. The tip runs along an arc of radius 9: s = rθ = 9 × 5π/6 = 7.5π = 23.6 cm. A1 for 23.6 cm. The hand sweeps out a sector of the same radius and angle: A = ½r²θ = ½ × 81 × 5π/6 = 33.75π = 106 cm². M1 for the sector area formula, A1 for 106 cm².

Question 11

A satellite 8 m across is directly overhead, at a distance of 500 km from an observer. Using the small-angle approximation, find the angle in radians that the satellite subtends at the observer, and convert that angle to degrees. Give each answer to 2 significant figures.

Accepted Answer

For a very small angle the arc and the straight line across it are nearly the same length, so the satellite's 8 m width may be treated as an arc of a circle of radius 500 km. Both lengths must be in the same unit: 500 km is 500 000 m. Then s = rθ gives θ ≈ 8 ÷ 500 000 = 1.6 × 10⁻⁵ radians. M1 for s = rθ with the lengths in matching units, A1 for 1.6 × 10⁻⁵. To convert, multiply by 180/π = 57.2958: 1.6 × 10⁻⁵ × 57.2958 = 9.2 × 10⁻⁴ degrees. M1 for multiplying by 180/π, A1 for 9.2 × 10⁻⁴. Leaving the distance in kilometres puts the answer out by a factor of a thousand. At this size the approximation is free: sin θ and θ agree to nine significant figures.

Question 12

Using the small-angle approximations, find the approximate value of (1 − cos 2θ)/(θ sin θ) when θ is small and in radians.

Accepted Answer

cos 2θ ≈ 1 − (2θ)²/2 = 1 − 2θ², so the top is approximately 2θ². The bottom is θ × sin θ ≈ θ × θ = θ². The ratio is 2θ²/θ² = 2, independent of θ. M1 for the expansion of cos 2θ, A1 for the numerator, M1 for sin θ ≈ θ below, A1 for the value 2. The replacement for cos must use the angle actually present, 2θ, squared whole: (2θ)² = 4θ², not 2θ².

Question 13

The approximation sin θ ≈ θ is stated for θ in radians. Show it performs well at θ = 0.1 radians, and explain why the same statement fails completely if θ is measured in degrees.

Accepted Answer

sin 0.1 = 0.0998…, within a fifth of a percent of 0.1. In radians a small angle's arc and sine are nearly the same length, which is what the approximation says. In degrees the claim reads sin 0.1° ≈ 0.1, but sin 0.1° = 0.0017…: the number 0.1 no longer measures the arc, so the geometry behind the approximation is gone. B1 for sin 0.1 = 0.0998, B1 for why arc and sine nearly agree, B1 for sin 0.1° = 0.0017, B1 for why the degree version fails.

Question 14

Three circles, each of radius 5 cm, are placed so that each one touches the other two. Find the exact area of the region enclosed between the three circles, and give its value to 3 significant figures.

Accepted Answer

Join the three centres. Each side of that triangle joins the centres of two circles that touch, so each side is 5 + 5 = 10 cm and the triangle is equilateral, with every angle π/3 radians. B1 for the equilateral triangle of side 10. The region between the circles is that triangle with three sectors cut away, one at each corner. Each sector has radius 5 and angle π/3, so its area is ½ × 25 × π/3 = 25π/6 cm². M1 for a sector of radius 5 and angle π/3, A1 for 25π/6. The triangle's area is ½ × 10 × 10 × sin (π/3) = 25√3 cm². M1 for the triangle's area. Taking the three sectors away leaves 25√3 − 3 × 25π/6 = 25√3 − 25π/2 cm². A1 for the exact area. Numerically that is 43.301 − 39.270 = 4.03 cm². A1 for 4.03 cm². The three sectors have the same radius and their angles add to π, the angle sum of the triangle, so together they make exactly half a circle. The two numbers being subtracted at the end are close, so keep full accuracy until the last line.

Question 15

A sector of a circle has radius r cm, angle θ radians and perimeter 20 cm. Show that its area A cm² satisfies A = 10r − r², and hence find the greatest possible area, the radius at which it occurs and the angle of that sector.

Accepted Answer

The perimeter of a sector is its arc plus its two radii, so rθ + 2r = 20 and rθ = 20 − 2r. M1 for the perimeter equation. Now A = ½r²θ = ½r(rθ) = ½r(20 − 2r) = 10r − r², as required. A1 for grouping rθ and substituting for it. Completing the square, A = 25 − (r − 5)², and a square is never negative, so A is greatest when the square is zero. M1 for completing the square. The greatest area is 25 cm², at r = 5 cm. A1 for both. Then rθ = 20 − 10 = 10, so θ = 2 radians. A1 for θ = 2. The answer for θ does not depend on the perimeter: the same working with perimeter P gives r = P/4 and θ = 2 every time. Two radians is a little under 115°.