Additional Mathematics | How to Find the Central Angle and Arc Length from a Chord

If a circle’s radius and chord length are known, the central angle of the minor arc can be found using an isosceles triangle. For radius r and chord length c, the minor central angle θ satisfies c = 2r sin(θ/2), so θ = 2arcsin[c/(2r)] when 0 < c ≤ 2r. Use radians for the arc-length formula s = rθ.

For example, a circle of radius 10 with a chord of length 10 has minor central angle θ = π/3 radians, or 60°. Its minor arc is 10π/3 units long, whereas the chord itself is only 10 units long. This guide derives the relationship, explains the choice between minor and major arcs, calculates sector and segment areas, and shows how to check the result using the chord’s distance from the centre.

A chord and an arc connect the same two points

A chord is a straight line segment joining two points on a circle. An arc is the curved portion of the circumference joining those points. Except in the degenerate case of coincident endpoints, the shorter arc is longer than its chord.

The two radii drawn from the centre to the chord endpoints form an isosceles triangle. Its equal sides both have length r, while the third side is the chord with length c.

The central angle is the angle at the circle’s centre between those radii. For distinct non-diametrically opposite points, there is a smaller central angle and a larger angle representing the other arc.

These distinctions matter because knowing a chord’s length does not by itself say whether the problem asks for the minor arc or the major arc. Begin by naming the requested arc and interpreting its central angle before using trigonometric formulas.

The perpendicular from the centre bisects the chord

Let O be the centre, A and B the chord endpoints, and M the midpoint of AB. The perpendicular from O to the line AB meets the chord at M.

Triangles OMA and OMB are congruent right triangles because OA = OB = r and OM is common. Therefore AM = MB = c/2.

The line OM also bisects the smaller central angle AOB. If that angle is θ, the angle AOM is θ/2.

This construction creates a right triangle containing the radius r as hypotenuse, the half-chord c/2 opposite θ/2, and the perpendicular centre-to-chord distance d adjacent to θ/2. From this single triangle we can derive both the sine relation and the Pythagorean chord-distance formula.

Derive the chord formula using sine

In right triangle OMA, sin(θ/2) = opposite/hypotenuse = (c/2)/r.

Multiply by 2r to obtain c = 2r sin(θ/2). This formula describes chord length for a central angle θ corresponding to the arc selected between the endpoints.

For the minor arc, choose θ between zero and π radians, inclusive at the diameter case. Then θ/2 lies between zero and π/2, where sine is increasing and the principal inverse sine selects the correct half-angle.

Therefore θ = 2arcsin[c/(2r)]. The ratio c/(2r) is dimensionless and must lie between zero and one for a real chord. The formula becomes invalid if the stated chord is longer than the diameter.

Worked example: radius ten and chord ten

Let r = 10 and c = 10. The chord formula becomes 10 = 20 sin(θ/2).

Divide by 20: sin(θ/2) = 1/2. For the minor central angle, 0 < θ/2 ≤ π/2, so θ/2 = π/6.

Thus θ = π/3 radians, equivalent to 60°. The minor arc length is s = rθ = 10(π/3) = 10π/3.

The chord measures 10 units, while the minor arc measures about 10.472 units. The arc is a little longer because it curves outward rather than taking the straight shortest path between the endpoints. Exact π notation should be kept when the question requests an exact answer.

Why the minor-angle restriction matters

The equation sin(θ/2) = 1/2 also has solutions with θ/2 = 5π/6, which gives θ = 5π/3. This is the major central angle between the same two radii.

The principal inverse sine returns the angle in its restricted range. For a chord of length 10 in a circle of radius 10, the calculation θ = 2arcsin(1/2) gives the minor angle π/3, not the major angle 5π/3.

That is correct when the question asks for the smaller sector or minor arc. If the task refers to the major arc, use the complementary central angle 2π − π/3 = 5π/3.

The same chord therefore corresponds to two arcs of different lengths. A complete solution has to follow the wording, not merely report whichever angle the calculator displays first.

Find the major arc as a second answer

The full circumference of a radius-10 circle is 2πr = 20π. The minor arc has length 10π/3.

The major arc is the remaining part of the circumference: 20π − 10π/3 = 60π/3 − 10π/3 = 50π/3.

Alternatively, use its central angle 5π/3 directly: s_major = r(5π/3) = 50π/3.

The two arc lengths add to 20π, as required. This sum is a strong independent check, and it distinguishes the major arc from the chord or the minor arc even if a sketch is not drawn to scale.

Radians are required in s = rθ

The formula s = rθ assumes θ is measured in radians. A radian is defined by the ratio of arc length to radius, which is why the formula has such a simple form.

If the central angle is 60°, convert it first: θ = 60π/180 = π/3 radians. Then s = 10π/3.

Substituting 60 directly into s = 10θ would produce 600, an incorrect arc length. In degrees, the equivalent formula is s = (θ_deg/360)2πr.

A student should label the angle’s unit next to the result. The numerical distinction between 60 and π/3 is large, but both represent the same rotation. Consistent units are vital in any trigonometry question connecting angles, arcs and sectors.

The sector area is one-half r-squared theta

A sector is the region bounded by two radii and the arc between them. When θ is in radians, its area is A_sector = (1/2)r²θ.

For the radius-10, 60° minor sector, A_sector = (1/2)(100)(π/3) = 50π/3 square units.

The same result follows by taking the fraction θ/(2π) = (π/3)/(2π) = 1/6 of the full circle area 100π.

This is a useful connection: a central angle determines both a fraction of the circumference and an equal fraction of the circle’s area. The arc length and sector area are different physical quantities, so report linear units for the arc and square units for the area.

Calculate the isosceles triangle’s area

The two radii and chord form an isosceles triangle with sides r and r enclosing central angle θ. Its area is A_triangle = (1/2)r² sin θ.

For r = 10 and θ = π/3, this gives (1/2)(100)sin(π/3) = 50(√3/2) = 25√3 square units.

This formula is the familiar triangle area rule (1/2)ab sin C with a = b = r and C the included central angle.

The triangle lies within the minor sector for 0 < θ ≤ π. Its area will be subtracted from the sector's area to obtain the area of the minor circular segment bounded by the chord and the minor arc.

Find the minor circular segment area

The minor segment is the region between the chord and the minor arc. It equals the minor sector area minus the isosceles triangle area.

Therefore A_segment = (1/2)r²(θ − sin θ), with θ in radians for the minor central angle.

For the example, A_segment = 50(π/3 − √3/2) = 50π/3 − 25√3 square units. This is approximately 9.06 square units.

The area is positive, as expected: a curved sector contains the triangle with the same two radii but extends beyond the straight chord. Be careful not to subtract the sector from the triangle, which would yield an impossible negative area for this minor segment.

Derive the centre-to-chord distance

The same right triangle OMA gives d² + (c/2)² = r². Hence d = √[r² − (c/2)²], taking the non-negative square root because d is a distance.

For r = 10 and c = 10, d = √(100 − 25) = √75 = 5√3.

The cosine relation also gives cos(θ/2) = d/r = (5√3)/10 = √3/2. Thus θ/2 = π/6, confirming θ = π/3.

This is a second way to reach the angle. It connects chord geometry to the centre-to-line distance technique, and it is particularly helpful when the question supplies or requests d as well as c and r.

An example with radius five and chord eight

Let r = 5 and c = 8. The half-chord length is 4, so the right triangle has hypotenuse 5 and one leg 4. Pythagoras gives d = √(25 − 16) = 3.

The half-angle satisfies sin(θ/2) = 4/5, so θ = 2arcsin(4/5), approximately 1.85459 radians.

The minor arc length is s = 5θ = 10arcsin(4/5), approximately 9.273 units. This exceeds the straight chord length 8, as it should.

The triangle’s familiar 3–4–5 dimensions make this a convenient check. The minor central angle is larger than 90°, which is plausible because a chord occupying eight of the circle’s ten diameter units cuts a fairly wide arc.

The segment area for the three-four-five geometry

In the radius-5, chord-8 example, sin(θ/2) = 4/5 and cos(θ/2) = 3/5.

Using the double-angle identity, sin θ = 2(4/5)(3/5) = 24/25. The isosceles triangle area is (1/2)(25)(24/25) = 12 square units.

The sector area is (25/2)θ, approximately 23.182 square units. Therefore the minor segment area is (25/2)θ − 12, approximately 11.182 square units.

This calculation shows how geometry and trigonometry connect. The triangle area can be found exactly from a 3–4–5 right triangle even though θ itself is not a simple rational multiple of π.

What happens when the chord is a diameter?

The longest possible chord has length c = 2r and passes through the centre. Then c/(2r) = 1, so θ = 2arcsin 1 = π radians.

The chord divides the circle into two semicircles. The smaller and larger arc lengths are equal, both πr, because the two central angles are both π.

The associated sector area is half the full circle area, (1/2)πr². The triangle formed by the two radii has zero area because its included angle is 180° and sin π = 0.

The chord-distance formula gives d = 0, confirming that the line of the diameter passes through the centre. These endpoint cases are useful for checking that the general formulas behave sensibly.

A chord longer than the diameter is impossible

Suppose a question claims that a circle has radius 5 and a chord of length 12. Its diameter is only 10, so no such chord exists.

The sine formula makes the contradiction explicit: c/(2r) = 12/10 = 1.2, while the sine of a real angle cannot exceed 1.

Likewise, the chord-distance expression would require d² = 25 − 36 = −11, impossible for a real distance.

This should be reported as inconsistent data, not handled by pressing inverse sine on 1.2 or inventing a complex angle for a real-plane geometry problem. Check the simple bound 0 ≤ c ≤ 2r before undertaking longer calculations.

Work backwards from the central angle to the chord

Suppose a circle has radius 6 and a minor central angle of 2π/3 radians. The chord formula gives c = 2(6)sin[(2π/3)/2] = 12sin(π/3).

Since sin(π/3) = √3/2, the chord length is 6√3 units.

The minor arc length is s = rθ = 6(2π/3) = 4π units. The minor sector area is (1/2)(36)(2π/3) = 12π square units.

The triangle area is (1/2)(36)sin(2π/3) = 9√3 square units, so the minor segment area is 12π − 9√3. This reverse example tests whether the learner knows which half-angle belongs in the chord formula.

Recover the radius from chord and central angle

If the chord length c and minor central angle θ are known, rearrange c = 2r sin(θ/2) to obtain r = c/[2sin(θ/2)], provided θ is non-zero and within the valid minor-angle range.

For c = 10 and θ = π/3, r = 10/[2sin(π/6)] = 10/(2 × 1/2) = 10.

This agrees with the worked example and allows the arc length or sector area to be calculated afterwards.

A student should not divide by sin θ instead of sin(θ/2). The geometry uses half the central angle because the perpendicular to the chord bisects the isosceles triangle.

Recover chord length from arc length

Suppose a circle has radius 10 and a minor arc length of 5π. Use s = rθ to find θ = (5π)/10 = π/2 radians.

Then c = 2r sin(θ/2) = 20sin(π/4) = 20(√2/2) = 10√2.

The chord is about 14.142 units long, shorter than the curved arc length 5π, approximately 15.708 units.

This is a two-stage problem: arc length determines central angle, and central angle determines chord length. Understanding the link is more transferable than memorising a separate direct formula for every pair of measurements.

The arc must exceed the positive chord

For a minor angle θ with 0 < θ ≤ π, the chord is c = 2r sin(θ/2), while the minor arc is s = rθ.

Let z = θ/2. Then s = 2rz and c = 2r sin z. For positive z, sin z < z, so c < s.

Thus a non-degenerate minor arc is longer than its chord. The difference can be small for a very short arc, but it remains positive.

This is a strong answer check: if a computed minor arc length is shorter than the chord, examine the angle unit, chosen arc, radius or missing half-angle. Geometry gives a reliable bound before any calculator approximation is trusted.

Do not confuse central and inscribed angles

A central angle is measured at the centre of the circle, between two radii. An inscribed angle has its vertex on the circumference and subtends an arc.

The inscribed angle theorem says that an inscribed angle subtending a particular arc is half the corresponding central angle, under its usual geometric conditions.

This is distinct from the half-angle θ/2 used in the chord derivation. That half-angle appears because a perpendicular radius bisects the isosceles central triangle.

Students sometimes see the word “half” in both facts and mix their meanings. Label the actual angle and its vertex on a sketch before selecting the theorem or ratio.

A quick solution structure for examinations

First write the radius and chord length, and check that the chord does not exceed the diameter. Second draw the isosceles triangle and its perpendicular bisector to obtain sin(θ/2) = c/(2r).

Third calculate the minor central angle θ in degrees or radians as requested. Convert to radians before using s = rθ or A_sector = r²θ/2.

Fourth, if the major arc is requested, subtract the minor angle from 2π or the minor arc from the full circumference. Fifth, for a segment area, subtract the triangle area from the sector area.

Finally check units and bounds: chord no longer than diameter, arc at least as long as chord, and segment area positive. These checks catch common mistakes without repeating the entire calculation.

Practice tasks with checked answers

Question one: radius 10, chord 10. Minor central angle: π/3. Minor arc: 10π/3.

Question two: same circle and chord. Major central angle: 5π/3. Major arc: 50π/3.

Question three: radius 5, chord 8. Centre-to-chord distance: 3. Minor angle: 2arcsin(4/5); minor arc: 10arcsin(4/5).

Question four: radius 6 and minor central angle 2π/3. Chord: 6√3. Minor arc: 4π.

Question five: radius 10 and minor arc 5π. Chord: 10√2.

Question six: a circle of radius 5 is claimed to contain a chord of length 12. Answer: impossible, because the diameter is only 10.

For each, identify whether the given quantities belong to the triangle, the chord, the arc or the sector before choosing a formula.

Chord geometry connects right-triangle trigonometry, radians, arcs and sectors. Some segment-area extensions may be enrichment depending on the student’s assessed syllabus and school sequence.

For Singapore, consult the appropriate SEAB references: 2026 O-Level Additional Mathematics 4049, 2027 SEC G3 K341 or 2027 SEC G2 K232, as applicable.

A strong learner should be able to derive the chord relationship from a right triangle and decide whether a question wants a minor arc, major arc, sector or segment before calculating. Properly taught kids shine a bright light into the future.