Question 1 of 24
Convert 135 degrees to radians, giving an exact answer.
Show a hint
Multiply the degree measure by pi/180 and simplify.
Cambridge International AS Level · Mathematics 9709
This original practice pack covers Cambridge 9709 Pure Mathematics 1 section 1.4: the definition of a radian, conversion between degrees and radians, arc length, sector area, related triangles, segments and composite circular regions. It explicitly distinguishes an arc from a complete boundary and a sector from a segment. Every spatial problem fully describes its centre, radii, angle, chosen arc and required boundary or region, so it needs no diagram and makes no hidden geometrical assumption. It is not complete AS Mathematics coverage.
2026 / 2027 · Academic review not recorded · Published 2026-09-08
AI-assisted practice — not independently academically reviewed. Answers may contain errors; check important results against your course materials.
Guided tutorial
16 lesson sections
Explanation
For a circle with centre O and radius r, a central angle theta radians cuts off an arc of length s = r theta. Therefore theta = s/r. One radian is the central angle that subtends an arc whose length is exactly one radius. A radian is an angle unit, while r and s are lengths measured in the same unit, so their ratio has no physical unit.
A complete turn subtends the full circumference 2 pi r, so its angle is (2 pi r)/r = 2 pi radians. Hence 360 degrees = 2 pi radians and 180 degrees = pi radians. These identities connect the two angle systems without changing the geometrical angle itself.
The formulae s = r theta and A = (1/2)r^2 theta require theta in radians. If an angle is supplied in degrees, convert it first. Writing the unit beside an approximate angle prevents a calculator value such as 1.2 from being mistaken for 1.2 degrees.
Worked example
Convert 150 degrees to radians and 7 pi/12 radians to degrees. To change degrees to radians, multiply by pi/180: 150(pi/180) = 5 pi/6 radians after cancelling a factor of 30.
To change radians to degrees, multiply by 180/pi: (7 pi/12)(180/pi) = 7(15) = 105 degrees. The factor pi cancels because the given radian angle is an exact multiple of pi.
A scale check supports both results: 150 degrees is between 90 degrees and 180 degrees, so 5 pi/6 is between pi/2 and pi; 7 pi/12 is a little more than pi/2, so 105 degrees is a sensible conversion.
Explanation
Two radii from O to distinct points A and B identify two arcs. The minor arc AB corresponds to the smaller central angle, normally between 0 and pi radians. The major arc AB corresponds to the reflex angle 2 pi - theta when theta is the minor angle. Naming only the endpoints does not by itself identify which arc is intended.
For a semicircle, the central angle is pi radians and the curved arc has length pi r. For a full circle, the angle is 2 pi and the circumference is 2 pi r. A straight angle, a semicircular arc and a semicircular region are related but are not interchangeable quantities.
Always select the angle before using a formula. If theta is the stated minor angle, use theta for the minor arc or sector and 2 pi - theta for the major one. The two arc lengths sum to 2 pi r, and the two sector areas sum to pi r^2.
Worked example
A circle has centre O and radius 8 cm. Its minor arc AB has length 14 cm. Find the smaller angle AOB in radians and degrees. Because the named length is the curved minor arc, s = r theta gives theta = 14/8 = 7/4 = 1.75 radians.
For a degree value, multiply by 180/pi: theta = (7/4)(180/pi) = 315/pi degrees, approximately 100.27 degrees. This remains below 180 degrees, consistent with the statement that AB is the minor arc.
The units also check: 14 cm divided by 8 cm is dimensionless, leaving an angle in radians. Using 14 as a perimeter or adding radii would answer a different question because only the curved arc length was given.
Explanation
The formula s = r theta returns only the curved arc. A sector bounded by radii OA and OB and arc AB has perimeter r theta + 2r because both straight radii are also on its boundary. If a question requests only arc AB, adding the radii is incorrect.
A segment is bounded by an arc and its chord, not by two radii. Its perimeter is therefore arc length plus chord length. For a minor central angle theta, the chord has length 2r sin(theta/2), obtained by bisecting the isosceles triangle AOB.
Composite boundaries must be traced once from start to finish. Include every exposed arc and straight edge exactly once, and exclude internal construction lines. State whether an answer is an arc, a complete perimeter, or only one part of a path.
Worked example
A sector has centre O, radius 8 cm and central angle 1.2 radians. Its boundary consists of minor arc AB together with radii OA and OB. Find its perimeter. The arc length is s = r theta = 8(1.2) = 9.6 cm.
The complete sector boundary also contains two radii, each 8 cm. Hence the perimeter is 9.6 + 8 + 8 = 25.6 cm. Reporting 9.6 cm alone would give the arc length, not the requested perimeter.
No extra circular edge is present: the boundary was explicitly defined as one arc and two radii. A quick magnitude check is useful because this minor arc is shorter than a semicircle of radius 8, whose arc length is 8 pi, while the full perimeter must exceed 16 cm.
Explanation
A sector with radius r and central angle theta radians occupies the fraction theta/(2 pi) of a complete circle. Multiplying this fraction by pi r^2 gives A = (1/2)r^2 theta. Area units are squared, unlike arc length and perimeter units.
The formula can be rearranged. If A and r are known, theta = 2A/r^2. If A and theta are known, r = sqrt(2A/theta), taking the positive root because radius is a positive length. Any degree angle must be converted before substitution.
Sector means the region enclosed by two radii and their connecting arc. It is not the same as the segment enclosed by a chord and arc. The formula (1/2)r^2 theta gives the whole sector, including the central triangle within it.
Worked example
A sector has centre O, radius 6 cm and area 27 cm^2. Its angle theta is between 0 and pi radians. Find theta and the sector's curved arc length. From 27 = (1/2)(6^2)theta, we get 27 = 18 theta, so theta = 1.5 radians.
Now use s = r theta with the same central angle: s = 6(1.5) = 9 cm. The requested curved length excludes the two radii; the complete sector perimeter would instead be 9 + 6 + 6 = 21 cm.
The condition 0 < theta < pi identifies a minor sector and is satisfied because 1.5 < pi. The area calculation uses cm^2, while the resulting arc uses cm, which checks the dimensions of the two answers.
Explanation
Let A and B lie on a circle with centre O and radius r, and let the minor angle AOB be theta radians. The minor sector AOB is bounded by OA, OB and the minor arc AB. The minor segment cut off by chord AB is bounded only by chord AB and that minor arc.
For 0 < theta < pi, triangle AOB lies inside the minor sector. Its area is (1/2)r^2 sin theta because two sides have length r and their included angle is theta. Therefore minor segment area = (1/2)r^2(theta - sin theta).
This subtraction is valid only after identifying the minor configuration. A major segment is most safely found as circle area minus minor segment area. Calculator trigonometric mode must be radians when evaluating sin theta if theta is in radians.
Worked example
In a circle with centre O and radius 10 cm, points A and B subtend the minor angle AOB = 1.2 radians. Find the area of the minor segment bounded by chord AB and the minor arc AB. The minor sector area is (1/2)(10^2)(1.2) = 60 cm^2.
Triangle AOB has area (1/2)(10)(10)sin(1.2) = 50 sin(1.2) cm^2. Hence the minor segment area is 60 - 50 sin(1.2), approximately 13.398 cm^2, or 13.4 cm^2 to 3 significant figures.
The answer is smaller than the sector area because the triangle is removed. The statement names a chord-and-minor-arc boundary, so the requested region is a segment; no perimeter is being calculated and no unseen construction is required.
Explanation
If theta is the minor central angle, the major sector angle is 2 pi - theta. Its area is (1/2)r^2(2 pi - theta), and its arc length is r(2 pi - theta). Using theta unchanged would calculate the minor quantity instead.
The major segment bounded by chord AB and the major arc AB contains most of the circle. It equals the full circle area minus the minor segment area. Equivalently, it is the major sector plus triangle AOB; the triangle is outside the major sector but inside the major segment.
These conditions explain the sign choice: minor segment = minor sector - triangle, while major segment = major sector + triangle. The minor and major segments share chord AB and together fill the circle, so their areas must sum to pi r^2.
Worked example
A circle has centre O and radius 4 cm. Chord AB subtends the minor angle pi/3 at O. Find the exact area of the major segment bounded by chord AB and the major arc AB. First find the minor segment so the chosen region is unambiguous.
The minor sector area is (1/2)(4^2)(pi/3) = 8 pi/3. Triangle AOB has area (1/2)(4^2)sin(pi/3) = 8(sqrt(3)/2) = 4 sqrt(3). Thus the minor segment area is 8 pi/3 - 4 sqrt(3).
Subtract from the circle area 16 pi: major segment area = 16 pi - (8 pi/3 - 4 sqrt(3)) = 40 pi/3 + 4 sqrt(3) cm^2. Adding the two segment expressions recovers 16 pi, independently checking the result.
Explanation
An annular sector lies between two concentric circles. If its outer radius is R, inner radius is r and common angle is theta radians, its area is the difference of two sectors: (1/2)(R^2 - r^2)theta. The radii must share the same centre and bounding rays for this formula to apply.
Its complete perimeter has four parts: outer arc R theta, inner arc r theta, and two straight radial edges each of length R - r. Thus P = R theta + r theta + 2(R - r). The inner arc remains part of the boundary even though it curves in the opposite visual direction.
For any composite shape, separate area from boundary length. Subtract regions when computing area, but trace exposed edges when computing perimeter. An internal radius used only to divide a shape does not belong in the external perimeter.
Worked example
Two concentric circles have centre O, outer radius 7 cm and inner radius 4 cm. Two common bounding rays form an angle of 1.5 radians. Find the area and complete perimeter of the region between the circles and between the rays.
Its area is (1/2)(7^2 - 4^2)(1.5) = (1/2)(33)(1.5) = 24.75 cm^2. Its outer and inner arcs have lengths 7(1.5) = 10.5 cm and 4(1.5) = 6 cm.
Each straight radial boundary has length 7 - 4 = 3 cm. Therefore the complete perimeter is 10.5 + 6 + 3 + 3 = 22.5 cm. This includes both exposed arcs and both connectors, but not either full radius from O.
Explanation
For endpoints A and B on a circle of radius r, triangle AOB is isosceles. If its included angle theta is known, its area is (1/2)r^2 sin theta and its chord length is 2r sin(theta/2). Both trigonometric evaluations use the same angle unit selected on the calculator.
The cosine rule gives the equivalent chord relation AB^2 = r^2 + r^2 - 2r^2 cos theta. Conversely, if r and chord AB are known, this equation can determine theta, with the required minor or major condition deciding the appropriate angle.
A prompt must say whether a chord, arc, sector boundary or segment boundary is required. Chord AB is the straight distance between A and B; arc AB follows the circumference. They coincide only in the limiting sense as a very small angle approaches zero, not in an ordinary calculation.
Worked example
A circle has centre O and radius 9 cm. Points A and B subtend the minor angle AOB = 0.8 radians. Find chord AB, the area of triangle AOB, and the area of the minor segment bounded by chord AB and the minor arc AB.
Chord AB = 2(9)sin(0.8/2) = 18 sin(0.4), approximately 7.0095 cm. Triangle area = (1/2)(9^2)sin(0.8) = 40.5 sin(0.8), approximately 29.053 cm^2.
The minor sector area is (1/2)(9^2)(0.8) = 32.4 cm^2, so the minor segment area is 32.4 - 40.5 sin(0.8), approximately 3.347 cm^2. Each requested object was identified explicitly, and sector minus triangle gives the stated chord-and-arc region.
Work at your own pace
Question 1 of 24
Multiply the degree measure by pi/180 and simplify.
Quick recall
Card 1 of 14
Front
Review the essentials
First identify the requested object: angle, arc, chord, sector, segment, area or complete perimeter. Next write the relevant central angle and mark it as degrees or radians. Convert degrees before using s = r theta or A = (1/2)r^2 theta.
For a major object, replace a stated minor angle theta by 2 pi - theta. For a sector perimeter, add two radii to the arc. For a segment perimeter, add the chord to the arc. For a minor segment area, subtract the central triangle; for a major segment area, subtract the minor segment from the circle.
Finally check units, scale and conditions. Length answers use linear units, areas use square units, a minor angle is below pi, and complementary minor-plus-major quantities recover the whole circle. Do not infer a tangent, diameter, right angle, midpoint or symmetry unless the text states or logically establishes it.
This pack is limited to Cambridge 9709 Paper 1 section 1.4. It uses elementary triangle trigonometry where circular regions require triangle lengths, angles or areas. Broader trigonometric equations, identities, calculus and coordinate-circle methods belong to other syllabus sections and are excluded scope here.
Throughout, pi denotes the exact circle constant, theta denotes a central angle, and any decimal instruction is applied only after retaining sufficient intermediate precision. Exact forms involving pi, roots or sine are preserved unless a rounded value is requested.
Every exercise is standalone: the centre, radius or radii, endpoints, minor or major choice, included angle and requested region or boundary are stated in words. No answer depends on an omitted picture, a measured sketch, a conventional-looking orientation or an unstated assumption.
Authorship: original ai assisted.
Original practice, not an official examination paper. Readnary is not affiliated with the awarding body. Prepared with AI assistance.