In isosceles triangle \(BCD\), the equal base angles are \(x\), so its vertex angle is \(180^\circ-2x\). This equals the polygon interior angle \(135^\circ+x\). Thus \(180-2x=135+x\), giving \(x=15\). The exterior angle is \(180-150=30^\circ\), so the number of sides is \(360/30=12\).

(a) By the alternate segment theorem, \(x=35\). Opposite angles in cyclic quadrilateral \(BCDE\) sum to \(180^\circ\), so \(y=180-75=105\).

(b) The angle at the centre is twice the angle at the circumference standing on the same arc: \(2(75^\circ)=150^\circ\).

(c) The major sector angle is \(360-150=210^\circ\). Area \(=\frac{210}{360}\pi(6)^2=21\pi\text{ cm}^2\).

Angles on the straight line \(ADB\) give \(x=180-74=106\). In triangle \(ACD\), \(\angle ACD=180-90-74=16^\circ\). Thus \(y=180-16=164\).

The exterior angle is \(180^\circ-140^\circ=40^\circ\). Therefore \(n=360/40=\boxed{9}\).

The exterior angle is \(180-135=45^\circ\), so the number of sides is \(360/45=\boxed{8}\).

Opposite angles of a cyclic quadrilateral sum to \(180^\circ\), so \(\angle BAD=180^\circ-75^\circ=105^\circ\).

Angles of triangle ABD sum to \(180^\circ\), giving \(\angle ABD=180^\circ-105^\circ-20^\circ=55^\circ\).

By the alternate segment theorem, the angle between tangent AE and chord AD equals the angle in the opposite segment: \(\angle EAD=\angle ABD=\boxed{55^\circ}\).

(a) The line from the centre to a chord midpoint is perpendicular to the chord. Thus \(AM=3\text{ cm}\) and triangle OMA is right-angled. \(OA=\sqrt{3^2+4^2}=\boxed{5\text{ cm}}\).

(b) AC is a diameter, so ABC is an angle in a semicircle.

(a) Angles on the same chord AB are equal: \(\angle ACB=\angle ADB=\boxed{41^\circ}\).

(b) In triangle ABC, \(\angle BAC=180^\circ-86^\circ-41^\circ=53^\circ\). Angles on chord BC are equal, so \(\angle BDC=\angle BAC=\boxed{53^\circ}\).

Each exterior angle is \(360^\circ/12=30^\circ\), so each interior angle is \(150^\circ\). The diagonal through opposite vertices is a line of symmetry and bisects the endpoint interior angle. Therefore \(\boxed{x=75^\circ}\).

The radius OS is perpendicular to the tangent, so \(\angle PSO=90^\circ-78^\circ=12^\circ\). By the alternate segment theorem, \(\angle PQS=78^\circ\).

The central angle \(\angle SOQ=2\angle SPQ=36^\circ\). Since \(OS=OQ\), \(\angle SQO=(180^\circ-36^\circ)/2=72^\circ\). Therefore \(\angle PQO=78^\circ-72^\circ=6^\circ\). The ratio is \(12:6=\boxed{2:1}\).

The base angles at B and C are equal, so \(2x+30^\circ=180^\circ\). Hence \(\boxed{x=75^\circ}\).

Angles around the point total \(360^\circ\). There are \(3+4+5=12\) parts, each \(30^\circ\), so \(b=120^\circ,c=150^\circ\).

The exterior angles are \(180^\circ-120^\circ=60^\circ\) for B and \(180^\circ-150^\circ=30^\circ\) for C. Thus B has \(360/60=6\) sides and C has \(360/30=12\) sides. Since \(12=2(6)\), the result follows.

(a) Opposite angles of cyclic quadrilateral ABCD sum to \(180^\circ\), so \(\boxed{\angle BCD=70^\circ}\).

(b)(i) \(\angle BDC=\angle BEC=45^\circ\), since both subtend chord BC in the same segment. In BCD, \(\angle DBC=180^\circ-70^\circ-45^\circ=\boxed{65^\circ}\).

(b)(ii) By the alternate segment theorem, \(\angle DCG=\angle DBC=\boxed{65^\circ}\).

The exterior angle is \(180^\circ-150^\circ=30^\circ\), so the number of sides is \(360/30=\boxed{12}\).

(a) \(OA=OC\), so \(\angle AOC=180^\circ-2(35^\circ)=\boxed{110^\circ}\).

(b) The angle at the circumference is half the angle at the centre: \(\boxed{\angle ABC=55^\circ}\).

(c) The alternate segment theorem gives \(\boxed{\angle DAC=\angle ABC=55^\circ}\).

(d) \(\angle ACB=35^\circ+40^\circ=75^\circ\), hence \(\angle CAB=180^\circ-75^\circ-55^\circ=50^\circ\). Since \(\angle CAO=35^\circ\), \(\angle OAB=50^\circ-35^\circ=\boxed{15^\circ}\).

Corresponding angles give \(\boxed{w=158^\circ}\). Alternate angles between the parallel lines give \(\boxed{x=76^\circ}\). The angle adjacent to \(158^\circ\) is \(22^\circ\); the triangle then gives \(y=180^\circ-22^\circ-76^\circ=\boxed{82^\circ}\).

\(\angle QPR=90^\circ\), because the angle in a semicircle is a right angle. Triangle angles total \(180^\circ\), so \(\angle PRQ=180^\circ-90^\circ-74^\circ=\boxed{16^\circ}\).

Corresponding angles give the upper-right triangle angle \(65^\circ\); vertically opposite angles give the upper-left one \(45^\circ\). Thus \(\boxed{x=180^\circ-45^\circ-65^\circ=70^\circ}\). Alternate angles give \(\boxed{y=65^\circ}\).

(a) Opposite angles of a cyclic quadrilateral sum to \(180^\circ\), so \(\angle DCB=180^\circ-60^\circ=\boxed{120^\circ}\).

(b) By the alternate segment theorem, \(\angle ABD=\angle EAD=35^\circ\). Since AB is parallel to DC, alternate angles give \(\angle CDB=35^\circ\). In DCB, \(\angle DBC=180^\circ-120^\circ-35^\circ=\boxed{25^\circ}\).

The internal angle at D is \(180^\circ-104^\circ=76^\circ\). Quadrilateral angles total \(360^\circ\), so \(\boxed{x=360^\circ-76^\circ-130^\circ-70^\circ=84^\circ}\). In isosceles AFB, the angle at A is \(30^\circ\), hence each base angle is \((180^\circ-30^\circ)/2\), giving \(\boxed{y=75^\circ}\).

Exterior angle \(=180^\circ-172^\circ=8^\circ\). Number of sides \(=360/8=\boxed{45}\).

Let T be the intersection of EB and AC. In ECT, \(x=180^\circ-125^\circ-26^\circ=\boxed{29^\circ}\).

Angles AEB and ACB subtend the same chord AB, so \(\boxed{y=52^\circ}\).

\(\angle AEC=x+y=81^\circ\). Triangle AEC gives \(\angle EAC=180^\circ-81^\circ-26^\circ=73^\circ\). Opposite angles of cyclic quadrilateral EADC sum to \(180^\circ\), so \(\boxed{z=107^\circ}\).

Exterior angle \(=360^\circ/20=18^\circ\). Interior angle \(=180^\circ-18^\circ=\boxed{162^\circ}\).

(a) In triangle CXD, \(\angle CDB=180^\circ-88^\circ-55^\circ=\boxed{37^\circ}\), because triangle angles sum to 180°. \(\angle ABD=\boxed{55^\circ}\), because angles in the same segment subtending chord AD are equal. \(\angle AED=180^\circ-55^\circ=\boxed{125^\circ}\), because opposite angles in cyclic quadrilateral ABDE sum to 180°.

(b)(i) Corresponding sides DX and AX give length ratio \(8/4=2\). Thus \(CX=2(BX)=2(2.7)=\boxed{5.4\text{ cm}}\). (b)(ii) Square the length ratio: \(\boxed{4:1}\).

Angles around C sum to \(360^\circ\), so \(\angle ACB=360^\circ-148^\circ-82^\circ=130^\circ\). Equal sides give equal base angles: \(\angle CAB=(180^\circ-130^\circ)/2=\boxed{25^\circ}\).

The 15-gon interior angle is \(180^\circ-360^\circ/15=156^\circ\). Hence the n-gon interior angle is \(156^\circ+14^\circ=170^\circ\). Its exterior angle is \(10^\circ\), so \(\boxed{n=360/10=36}\).

(a) Co-interior angles on transversal AB sum to \(180^\circ\): \(\angle ABC=180^\circ-75^\circ=\boxed{105^\circ}\).

(b) Corresponding angles on transversal DC give \(\angle AED=\boxed{80^\circ}\).

(c) Triangle AED is isosceles, so \(\angle DAE=(180^\circ-80^\circ)/2=50^\circ\). Hence \(\angle DAB=50^\circ+75^\circ=\boxed{125^\circ}\).

The two base angles are equal: \(x=180^\circ-25^\circ-25^\circ=\boxed{130^\circ}\).

A, O and C are collinear. Since a radius is perpendicular to its tangent, \(OB\perp EB\); the parallel diameter AC gives \(\angle AOB=90^\circ\). Therefore \(\angle DOB=44^\circ+90^\circ=134^\circ\).

Quadrilateral ODEB has right angles at D and B, so \(\angle DEB=360^\circ-90^\circ-90^\circ-134^\circ=46^\circ\). Right triangles EOD and EOB are congruent (common hypotenuse EO and equal radii), so EO bisects this angle. Thus \(\boxed{x=23^\circ}\).

Opposite angles p and m sum to 180°. Their five ratio parts total 180°, so each part is 36°: p=72°, m=108°. The angle opposite y is \(m+10=118^\circ\). Hence \(\boxed{y=180^\circ-118^\circ=62^\circ}\).

At F, the straight angle is split into \(12^\circ,60^\circ,x\). Hence \(x=180^\circ-12^\circ-60^\circ=\boxed{108^\circ}\). Since BC and AD are parallel, co-interior angles x and y sum to 180°, so \(\boxed{y=72^\circ}\).

The central angle AOB is twice the angle at the circumference: \(2(67^\circ)=134^\circ\). Radii OA and OB are perpendicular to their tangents, so in quadrilateral OATB, \(\angle ATB=360^\circ-90^\circ-90^\circ-134^\circ=\boxed{46^\circ}\).

The acute angle at the upper intersection is 55° by corresponding angles. x is its supplement on a straight line, so \(\boxed{x=180^\circ-55^\circ=125^\circ}\).

(a) \(\boxed{x=104^\circ}\), because the angle at the centre is twice the angle at the circumference subtended by the same chord AC. (b) By the alternate segment theorem, angle BAC=65°. Triangle AOC is isosceles, so angle OAC=(180−104)/2=38°. Hence \(\boxed{y=65^\circ-38^\circ=27^\circ}\).

(a) Interior angle \(=180^\circ-360^\circ/10=144^\circ\). The symmetry line bisects it, so \(\boxed{d=72^\circ}\). (b) \(\boxed{n=360/45=8}\).

Opposite angles sum to 180°: \(4m+5m=180\), so m=20. The other pair gives \(p+(4m+38)=180\). Hence \(\boxed{p=62}\).

The six equal angles at C fill a straight angle, so each is \(180^\circ/6=30^\circ\). The other two angles in a kite are equal. Their sum is \(360^\circ-40^\circ-30^\circ=290^\circ\), so \(\boxed{x=145^\circ}\).

(a) By the alternate segment theorem, \(\boxed{\angle PRT=50^\circ}\). (b) Opposite cyclic angles: PQR=180−48=132°. Since PQ=QR the base angles are equal: \(\angle QPR=(180-132)/2=\boxed{24^\circ}\).

\(y+11.5y=180\), so y=14.4. The exterior angle is y°, so the number of sides is \(360/14.4=\boxed{25}\).

(a) Supplementary and corresponding angles give \(\boxed{a=142,b=142}\). (b) Exterior=360/12=30°, interior=\(\boxed{150^\circ}\). (c) Alternate segment theorem gives \(\boxed{f=56}\). Radius OM is perpendicular to the tangent, so OMP=90−56=34°. OM=OP, so \(\boxed{g=34}\). (d) The triangle opposite k has third angle 180−24−27=129°. Opposite cyclic angles sum to 180°, so \(\boxed{k=51}\).

Exterior angle=360/15=24°. Interior angle \(=180-24=\boxed{156^\circ}\).

(a) Angles on a straight line give \(x=180-122=\boxed{58^\circ}\). (b) In ABC, angle C=180−90−34=56°. Thus \(y=56-17=\boxed{39^\circ}\). (c) The interior angle at D is 360−90−122−39=109°. The marked reflex angle is \(z=360-109=\boxed{251^\circ}\).

\(x+(x+132)=180\), so \(2x=48\) and \(x=24^\circ\). Number of sides \(=360/24=\boxed{15}\).

The central angle AOC=2×64=128°. Radii meet tangents at 90°, so in AODC, \(x=360-128-90-90=\boxed{52^\circ}\).

The two base angles are 43°, so \(x=180-2(43)=\boxed{94^\circ}\).

Adjacent angles sum to 180°: \(132-2x+15+5x=180\). Thus \(3x=33\), \(x=11\). The angles are 110° and 70°, so the smallest is \(\boxed{70^\circ}\).

(a)(i) Angles ADB and ACB stand on the same chord AB: \(\boxed{x=41^\circ}\). (a)(ii) By the alternate segment theorem for chord AD, \(\boxed{y=37^\circ}\). (b) The reflex central angle POR=2×115=230°. Thus \(w=360-230=\boxed{130^\circ}\).

Angles around a point total 360°: \(w=360-90-80=\boxed{190^\circ}\).

(a) \(w=360\times3/8=\boxed{135^\circ}\). (b)(i) ABD=ACD=49° (same segment). In ABM, \(BAM=180-105-49=\boxed{26^\circ}\). (ii)(a) By the alternate segment theorem, BDA=PAB=45°. Then \(BAD=180-49-45=\boxed{86^\circ}\). (ii)(b) A diameter would subtend 90° at A, but BAD=86°.

(c)(i) OA⊥TA and OT bisects AOC. \(AOC=2\cos^{-1}(6.75/11.5)=108.117\ldots^\circ\), rounding to 108.12°. (ii) \(L=\frac{AOC}{360}(2\pi\times6.75)=\boxed{12.7\text{ cm}}\). (iii) \(A=\frac{360-AOC}{360}\pi(6.75)^2=\boxed{100\text{ cm}^2}\).

The two angles sum to \(180^\circ\). Exterior angle \(=180/12=15^\circ\). Number of sides \(=360/15=24\).

(a) \(\angle CBD=42^\circ\) (angles in the same segment, chord CD).

(b) \(\angle ACB=55^\circ\) (alternate segment theorem, chord AB).

(c) \(\angle ABC=95^\circ\), so \(\angle ADC=180^\circ-95^\circ=85^\circ\).

(d) \(\angle ACD=180^\circ-42^\circ-85^\circ=53^\circ\). Hence \(\angle BCD=55^\circ+53^\circ=108^\circ\).

(e) \(\angle PAD=\angle ACD=53^\circ\) (alternate segment theorem).

Angles of a quadrilateral sum to \(360^\circ\): \(\angle ADC=360-104-56-71=129^\circ\). Hence \(\angle ADE=180-129=51^\circ\).

(a) SSS: \(AO=CO\), \(AB=CB\), and \(OB\) is common.

(b)(i) \(\angle QSR=42^\circ\) by the alternate segment theorem.

(ii) \(\angle PQS=42^\circ\) by alternate angles, since \(PQ\parallel SR\).

(iii) The angle at the centre is twice that at the circumference subtending chord \(PS\): \(\angle POS=84^\circ\).

\(a=58^\circ\) because vertically opposite angles are equal. \(b=180^\circ-58^\circ=122^\circ\) because co-interior angles between parallel lines sum to \(180^\circ\).

\(\angle CAB=32^\circ\) by alternate angles, and \(\angle ABD=32^\circ\) because it subtends the same chord \(AD\) as \(\angle ACD\).

In triangle \(AEB\), \(x=180-32-32=116^\circ\).

(a) Each interior base angle is \((180-42)/2=69\). The exterior angle \(x=180-69=111^\circ\).

(b) \(c=360(5/12)=150^\circ\). (c) \(d=360/5=72^\circ, h=180-360/6=120^\circ\). Hence \(d/h=72/120=3/5\).

(d) \(x+(2x-5)+(x+20)+(3x-40)=360\Rightarrow7x=385\Rightarrow x=55\). Opposite angles \(S+Q=55+125=180^\circ\) and \(P+R=105+75=180^\circ\). Therefore the quadrilateral is cyclic.

(e) Major arc angle \(360-50=310^\circ\). Length \(\frac{310}{360}(2\pi\times9)=48.7\text{ cm}\).

(a) \(\angle BAC=180-114=66^\circ\). Since \(AC=BC\), \(\angle ABC=66^\circ\) too. From \(C\), \(B\) is 66° west of south, so its bearing is \(180+66=246^\circ\).

(b)(i) \(\angle PRS=58^\circ\) by the alternate segment theorem. (ii) Opposite angles in a cyclic quadrilateral sum to 180°: \(\angle PQR=180-74=106^\circ\).

(iii) \(\angle PRQ=27^\circ\) by the alternate segment theorem. Triangle \(PQR\) gives \(\angle RPQ=180-106-27=47^\circ\).

(c) A radius is perpendicular to a tangent, so \(\angle OAT=\angle OBT=90^\circ\). Tangents from the same external point are equal, so \(AT=BT\). The hypotenuse \(OT\) is common, hence the triangles are congruent by RHS.

\(\angle AOT=\angle BOT=180-90-34=56^\circ\). Since \(A,O,C\) lie on a straight line, \(\angle BOC=180-56-56=68^\circ\).

In isosceles triangle \(BOC\), \(OB=OC\) radii, so \(\angle OBC=(180-68)/2=56^\circ\). Thus \(\angle OBC=\angle BOT\) are equal alternate angles, giving \(CB\parallel OT\).

The two base angles are equal: \(x=180-2(41)=98^\circ\).

Since \(OA=OC\), \(\angle OAC=20^\circ\), so \(\angle AOC=140^\circ\). Thus \(\angle ABC=70^\circ\) (angle at centre twice angle at circumference).

Radius \(OC\) is perpendicular to the tangent, so \(\angle OCB=90-53=37^\circ\). \(OB=OC\) gives \(\angle OBC=37^\circ\). Hence \(x=70-37=33^\circ\).

(a) \(OA=OB\), so \(\angle AOB=180-2(28)=124^\circ\). The angle at circumference is half: \(\angle ACB=62^\circ\).

(b) Alternate segment theorem gives \(\angle PQR=47^\circ\). Hence \(\angle RPQ=180-52-47=81^\circ\).

The other base angle of the triangle is \(180-114=66^\circ\). The first base angle is 50° by parallel-line and vertically opposite angles. Hence \(x=180-66-50=64^\circ\).

(a) \(180(10-2)/10=144\) degrees.

(b) The tangent–chord theorem for chord \(ED\) gives \(w=x=20\) degrees. Opposite cyclic angles give \(\angle AED=180-75=105^\circ\). In \(AED\), \(\angle ADE=180-105-20=55^\circ\).

Parallel lines give \(\angle BDC=w=20^\circ\). Angles above tangent \(FG\) sum to 180°: \(20+55+y+20+25=180\), so \(y=60\) degrees. Also \(\angle EDC=180-20-25=135^\circ\). Co-interior angles between the parallel lines give \(z=180-135=45^\circ\).

Centre a protractor on the vertex and align its zero with the horizontal ray. Read 40°.

(a) Exterior angle \(180-156=24\)°, so \(360/24=15\) sides. (b) \(OA=OB\) gives \(\angle AOB=180-2(52)=76^\circ\). The angle at the circumference is half this: 38°.

(c) \(\angle PSR=180-112=68^\circ\). Angles \(PSR,PTR\) stand on the same chord \(PR\) in the same segment, hence \(\angle PTR=68^\circ\).

(d) \(\angle GMK=90^\circ\) because \(KG\) is a diameter. \(\angle OGH=90^\circ\) because the tangent is perpendicular to the radius. \(\angle KGM=\angle HOG\) since \(GM\) is parallel to \(OH\) and \(GK,GO\) are on the same line. Therefore the triangles are similar by AA.

Angle sum \(180(5-2)=540\)°. Ratio sum 30. Largest \(540(9/30)=162\)°.

Tangent–chord theorem gives \(x=38\)°. Parallel lines then give \(\angle ACB=38^\circ\). Opposite cyclic angles give \(\angle ABC=180-60=120^\circ\). In \(ABC\), \(y=180-120-38=22\)°.

In triangle \(DEF\), the third angle is \(180-8-82=90\)°. It stands on \(DE\), so \(DE\) is a diameter (converse of the angle-in-a-semicircle theorem).

Interior angle at \(S\) is \(360-75-129-73=83\)°. Its exterior angle \(\angle PST=180-83=97^\circ\).

(a) \(\angle PQR=90^\circ\) (angle in a semicircle). \(\angle PRQ=180-90-29=61^\circ\) (triangle angle sum). \(\angle PSQ=61^\circ\) (angles subtended by chord PQ in the same segment).

(b) \(\angle ABT=98^\circ\) by the alternate segment theorem. Since AB = BT, \(\angle BAT=\angle ATB=(180-98)/2=41^\circ\). Hence \(\angle BTC=41^\circ\) by the alternate segment theorem, and \(\angle CBT=180-98=82^\circ\). In triangle CBT, \(\angle ACT=\angle BCT=180-82-41=57^\circ\).

For this working, call the left, top-left, top-right, bottom-right and bottom vertices L, A, B, C and D, and the crossing of AC and LB X. Angles subtended by chord LA give \(u=20\)°. In triangle AXB, \(\angle AXB=180-56=124^\circ\) and \(\angle XAB=180-124-20=36^\circ\). Angles subtended by BC give \(x=36\)°. Triangle LBC gives \(\angle LCB=180-72-36=72^\circ\). Thus \(\angle ACB=72-20=52^\circ\), and angles subtended by AB give \(v=52\)°. Opposite angles in cyclic LBCD give \(w=180-72=108\)°.

Corresponding and vertically opposite angles give 55° on the top straight line. \(x=180-71-55=54\)°.

Equal base angles are both 34°. \(x=180-2(34)=112\)°.

Opposite angles add to 180°: \(101+79=180\) and \(95+85=180\).

\(7x+44+x+8=180\) gives \(8x=128,x=16\). Exterior angle \(24\)°. Number of sides \(360/24=15\).

(a) Cosine rule: \(\cos B=\frac{9.5^2+7.7^2-10^2}{2(9.5)(7.7)}=\frac{2477}{7315}\). Thus \(B=70.2072248\ldots^\circ\), which rounds to 70.2°. (b)(i) Centre angle is twice the circumference angle: 140.4°. (ii) OA = OC, so \((180-140.4)/2=19.8^\circ\). (iii) Radius is perpendicular to tangent: \(90-19.8=70.2^\circ\).

(c) Bisect AC; half the chord is 5 cm. \(r=5/\sin B=5.31\) cm. (d) Use unrounded values: \(100\frac{\frac12(9.5)(7.7)\sin B}{\pi r^2}\approx38.8\)%.

Alternate segment theorem gives angle EHF = 47°. Angles subtended by chord FG give angle FHG = angle FEG = 25°. Thus angle EHG = 72°. Since EG = EH, angle EGH is also 72°, so \(x=180-72-72=36^\circ\).

(a) Vertically opposite angles give a = 126°. The angle supplementary to 126° is 54°, which gives b = 54° using the parallel lines. Co-interior angles give \(c=180-63=117^\circ\). (b) Angle PRQ = 90° (angle in a semicircle). Triangle RPQ gives \(\angle RQP=180-90-58=32^\circ\). Alternate angles between RQ and PS give x = 32°. (c) B lies on the minor arc AC, so its angle subtends the major arc. \(y=(360-142)/2=109^\circ\).

Exterior angle \(180-175=5\)°. Number of sides \(360/5=72\).

Vertically opposite angles give \(a=59^\circ\). Corresponding angles give \(b=37^\circ\). The triangle angle sum gives \(c=180-59-37=84^\circ\).

Radii meet tangents at 90°. In quadrilateral AOCD, \(\angle AOC=360-90-90-44=136^\circ\). Angle at the circumference on the same arc is half the central angle: \(x=136/2=68^\circ\).

(a) Alternate segment theorem gives angle PQT = 50°. Since PT = PQ, angle PTQ is also 50°. Angle TPQ \(180-50-50=80\)°. (b) An exterior angle of a cyclic quadrilateral equals its opposite interior angle, so w = 68°. The other opposite angles give \(3x+2x=180\), hence x = 36°.

Corresponding angles on parallel lines are equal.

\(2y+100=360\), hence \(y=130\)°.

Exterior angle \(180-174=6\)°. Number of sides \(360/6=60\).

Exterior angle \(180-156=24\)°. Number of sides \(360/24=15\).

(a) \(x=55^\circ\) by the alternate segment theorem: the tangent–chord angle equals the angle in the opposite segment. (b) \(SV=SR\) because tangents from the same external point have equal length.

\(\angle DCA=58^\circ\), so \(\angle DAC=180-2(58)=64^\circ\). Then \(\angle CAB=86-64=22^\circ\). The other two angles of ABC are equal: \(x=(180-22)/2=79^\circ\).

Exterior angle \(1.5\)°. Number of sides \(360/1.5=240\).

Corresponding angles and angles on a straight line give \(x=180-132=48^\circ\).

(a)(i) \(\angle CBD=124/2=62^\circ\) because the angle at the centre is twice the angle at the circumference standing on the same arc. (ii) OC = OD, so COD is isosceles: \(\angle OCD=(180-124)/2=28^\circ\). Thus \(\angle BCD=35+28=63^\circ\). Opposite angles of a cyclic quadrilateral sum to 180°: \(\angle BAD=180-63=117^\circ\). (b) \(\angle PQS=42^\circ\) (same segment) and \(\angle QPS=90^\circ\) (diameter). Hence \(QS=5.9/\cos42^\circ\) and circumference \(\pi QS=24.9418\ldots\approx24.9\) cm.

(a) OA = OB, so \(\angle AOB=180-2(38)=104^\circ\). Angle at circumference \(\angle APB=104/2=52^\circ\). (b)(i) Alternate segment theorem gives FEC = 50°. Since FC = FE, FCE = 50° as well. Hence \(\angle EFC=180-50-50=80^\circ\). (ii) Opposite cyclic angles sum to 180°: \(\angle CDE=180-80=100^\circ\).

\(180^\circ-360^\circ/24=165^\circ\).

The alternate segment theorem gives angle ADC = 64°. Opposite angles of a cyclic quadrilateral sum to 180°, so \(\angle ABC=180^\circ-64^\circ=116^\circ\).

Base angles ABC and ACB are both 32°. \(\angle BAC=180^\circ-2(32^\circ)=116^\circ\).

The angle at the circumference subtending the same arc as the 110° central angle is half as large, so x = 55°. Angles in the same segment are equal, so y = 24°.

QR bisects angle PQB, so angle PQR = angle RQB = x°. The alternate segment theorem gives angle QPR = angle RQB = x°. Equal angles PQR and QPR have equal opposite sides, PR = QR; hence PQR is isosceles.

Angles on the same side of a transversal between parallel lines sum to 180°. \((97-3x)+(69+5x)=180\), so \(166+2x=180\) and x = 7.

(a) Opposite angles in a cyclic quadrilateral give \(180^\circ-131^\circ=49^\circ\). (b) The central angle is twice the angle on the same arc: \(2(49^\circ)=98^\circ\). (c) Alternate segment theorem gives 20°. (d) Radius OA is perpendicular to AT, so \(90^\circ-20^\circ=70^\circ\).

Angle ABC = 180° − 50° = 130°. Equal base angles give \(x=(180^\circ-130^\circ)/2=25^\circ\).

(a) Exterior angle30°, so \(n=360/30=12\). (b)(i) Angle KLM=90° since KM is a diameter. Equal KL and LM give angles LKM and KML45°. Radius KM is perpendicular to tangent KS, so z=90°−45°=45°.

(ii) The alternate segment theorem gives58° at the left-hand vertex of the triangle containing the marked27° angle at A. The marked x is its exterior angle, so \(x=27^\circ+58^\circ=85^\circ\).

(iii) Angle GFJ=180°−108°=72°. Since FG is parallel to HJ, angle HJF=108°. Opposite angles in the cyclic quadrilateral sum to180°, so \(y=180^\circ-108^\circ=72^\circ\).

(c) OA=OB=OC=OD, since all are radii. Chords at equal distances from the centre have equal lengths, so AB=CD. Therefore the triangles are congruent by SSS.

Exterior angle \(180-156=24\) degrees. \(n=360/24=15\).

\(180-360/40=171\) degrees.

Opposite angles give \(2x+(x+60)=180\), so x=40. The opposite pair \((4x-87)+y=180\) gives \(y=180-(160-87)=107\) degrees.

The right-hand interior base angle is180°−120°=60°, so x=60° by alternate angles. The left-hand interior base angle is180°−140°=40°. The triangle’s third angle is \(y=180-60-40=80\) degrees.

(a) v=40° by the alternate segment theorem with angle ABE. w=80° because EC=EB makes triangle EBC isosceles. x=40° by angles subtended by chord EA. y=180°−80°=100° by opposite angles in cyclic EABC. Angle ABC=40°+80°=120°, so z=180°−120°=60° in cyclic ABCD.

(b) The minor central angle is twice KML: \(360-11x=2(2x)\Rightarrow15x=360\Rightarrow x=24\).

(c)(i) \(\angle ADX=\angle BCX\) (same chord AB), \(\angle DAX=\angle CBX\) (same chord DC), and \(\angle AXD=\angle BXC\) (vertically opposite). Corresponding angles are equal, giving similarity. (ii)(a) \(DX/CX=AD/BC=10/8\Rightarrow DX=7(10/8)=8.75\) cm. (b) In BXC, \(\cos\angle BXC=\frac{5^2+7^2-8^2}{2(5)(7)}=\frac17\), so angle BXC=81.8°.

(a)(i) x=162°/2=81°: the angle at the centre is twice the angle at the circumference subtending CD. (ii) y=81° by the alternate segment theorem for chord CD. (iii) \(\angle ADC=180^\circ-42^\circ-81^\circ=57^\circ\) since EDF is straight. Opposite angles of cyclic ABCD sum to180°, so z=180°−57°=123°.

(b)(i) Angles PTU and PRQ are equal, and PUT and PQR are equal (corresponding angles on parallel lines); angle RPQ is common. Equal corresponding angles give similarity. (ii)(a) PT :PR=4 :7, so PU :PQ=4 :7. (b) Area PQR=\(20(7/4)^2=61.25\) cm². Subtract PUT: \(61.25-20=41.25\) cm².

OA=OB, so angle ABO=angle OAB=21°. Thus angle AOB=180°−42°=138°, and angle BOC=138°−104°=34°. Since OB=OC, \(w=(180-34)/2=73\)°.

The reflex angle AOC subtended by the same arc as angle ABC is \(2(130)=260\)°. The obtuse angle is \(360-260=100\)°.

(a) Trapezium. (b) Obtuse.

Angle MKJ=22° (same chord MJ). Angle JKL=90° (diameter JL), so angle NKL=90°−22°=68°. Angle KNL=180°−104°=76°. In triangle KNL, \(d=180-68-76=36\)°.

(a) Base angles ABC and ACB=(180°−26°)/2=77°. Angle CBD=77° by alternate angles, hence x=180°−77°=103°. (b) Let L be the unlabelled left point on the circle, R the right point, and K the intersection of LR and PQ. Angle PLR=58° by the tangent-chord theorem. In triangle PLK, angle LPK=17°; y is its exterior angle, so y=17°+58°=75°.

Exterior angle \(180-176=4\)°. Therefore \(n=360/4=90\).

(a) OA=OB, so both base angles in AOB are 2x°. Hence \(\angle AOB=(180-4x)^\circ\). (b) The angle at the circumference is half the angle at the centre: \(\angle ACB=(90-2x)^\circ\). (c) \(\angle DCB=x+(90-2x)=90-x\) degrees. Opposite angles of cyclic quadrilateral DABC sum to 180°, giving \(\angle DAB=(90+x)^\circ\).

(a) Straight line: p=180−48=132°. The triangle angle at the intersection is 48° (vertically opposite), so q=180−55−48=77°. (b) \(x+(x+5)+(2x-25)+(x+10)=360\Rightarrow5x=370\Rightarrow x=74\). (c) \(180-\frac{360}{72}=175^\circ\).

(d) Radius OA is perpendicular to the tangent, so u=90−60=30°. Alternate segment theorem gives v=60° and w=60°. The central angle x=2v=120°. Angle DAB=u+20=50°, so central angle DOB=100°. Since OD=OB, \(y=(180-100)/2=40\)°.

(e) ABC subtends the reflex angle AOC. \(2(5x)=360-(3x+22)\Rightarrow13x=338\Rightarrow x=26\).

(a)(i) Opposite cyclic angles sum to 180°, giving ABC=180−128=52°. The central angle subtending the same arc AC is twice this: AOC=104°. (ii) OB=OC, so OBC=BCO=30°. Hence ABO=ABC−OBC=52−30=22°. OA=OB gives BAO=22°. (iii) ABD=ACD=28° (angles in the same segment). (iv) Sector perimeter \(2(9.6)+\frac{104}{360}(2\pi9.6)=36.6\) cm.

(b)(i) Length scale factor \(\sqrt[3]{2187/648}=1.5\). Area \(36(1.5)^2=81\) cm². (ii) \(r=\sqrt[3]{\frac{3(2187)}{4\pi}}=8.05\) cm.

a=63° because corresponding angles on parallel lines are equal. Then b=180−63−58=59° because the angles in a triangle sum to 180°.

Angle ACB=43° by alternate angles. AB=AC, so angle ABC=43°. Thus \(x=180-43-43=94\)°.

w=180−85=95° (opposite angles of cyclic quadrilateral ABCE). x=85° (angles ADC and AEC subtend chord AC). In triangle formed by DC and the intersecting chords AD and CE, y=180−85−47=48°.

(a) 9x=180, so x=20. Sides=360/20=18. (b) Angle DBC=58° (same segment as DAC), DCB=90° (diameter DB). Thus CDB=180−58−90=32°. (c)(i) Angle PRQ=24° (half central POQ); parallel PO and QR give OPR=24°. (ii) Major central angle=360−48=312°. Arc=\(\frac{312}{360}(2\pi\cdot5.4)=29.405\ldots\approx29.4\) cm.

(a)(i)(a) OE=OD, so OED=28° and EOD=180−56=124°. EBD=124/2=62° because a central angle is twice the angle at the circumference on the same arc. (b) EAD=62° (same segment as EBD). (ii) AED=180−109=71° (opposite angles of cyclic quadrilateral ACDE). AEO=71−28=43°, so BEO=43−35=8°. (b)(i) 12 exterior angles=180°, so exterior angle=15° and number of sides=360/15=24. (ii) (24−2)×180=3960°.

Angle BAC=56° by alternate angles. Equal base angles give x=(180−56)/2=62°. At C, angle NCM is supplementary to MCB, which equals ABC by corresponding angles; y=180−62=118°.

ON=5 cm: equal chords are equidistant from the centre.

(a) OA⊥EA, so BAO=90−61=29°. (b)OAC=55−29=26°; OA=OC implies AOC=180−2×26=128°. (c)ABC=128/2=64° by angle at centre. (d) Opposite cyclic angles sum 180°: CDA=116°.

(a)(i) x=27° by alternate angles. (ii) Similar triangles EHZ and GFZ give \(\frac{EZ}7=\frac59\Rightarrow EZ=\frac{35}9\approx3.89\) cm. (b) OA=OB, so angle OBA=52°. Equal chords AB,BC give angle OBC=52°. Hence angle ABC=104°, so opposite cyclic angle ADC=180−104=76°. (c)(i)90°, angles in a semicircle. (ii)27°, since OQ is perpendicular to the tangent and \(90-63=27\). (iii) A rectangle: angle QSP=63° in triangle PQS, and angle PSR=63+27=90°, giving four right angles.

Octagon interior angle \(180-360/8=135^\circ\). Triangle angle is 60°. Angles around a point sum to 360°, so \(x=360-135-60=165^\circ\).

(a) 5x+6x+7x=180, so x=10. (b) 7x=70°.

(a) Angles subtended by chord DC give \(u=\angle DBC=\angle DAC=35^\circ\). Parallel lines give \(\angle ACB=35^\circ\). In BXC the angle at X is 110°, and the vertically opposite angle v is 110°. (b) Minor central angle FOG=360−210=150°. Angle FHG subtends the minor arc FG, so p=150/2=75°.

Opposite angles in cyclic ABDE give w=180−140=40°. Angle ABC=180−95=85°, so opposite angle x=ADC=180−85=95°. Angle DBC=85−40=45°; the same chord DC gives y=DAC=45°.

Angles on a straight line give \(p=180-70=110^\circ\). Alternate interior angles give \(q=70^\circ\).

(a) Vertically opposite angles give x=38°. Triangle ACE gives y=180−24−38=118°. Using parallel lines and a straight line gives z=180−118=62°. (b) AB=AD gives ADB=42°. Angles subtending AB give ACB=42°. Since AC=BC, CAB=CBA=(180−42)/2=69°. (c) QRS subtends the major arc QS, so \((360-146)/2=107^\circ\).

\((12-2)180/12=150^\circ\).

\(x=360-83-104-72=101^\circ\).

Interior angle=\((5-2)180/5=108^\circ\). The symmetry line bisects it, so \(d=108/2=54^\circ\).

\(30x=180\Rightarrow x=6^\circ\). Number of sides=\(360/6=60\).

Angle at the circumference on arc AC is \(w=108/2=54^\circ\). Opposite cyclic angles give \(x=180-54=126^\circ\). The exterior angle of a cyclic quadrilateral equals its opposite interior angle, so \(y=60^\circ\).

\(x=180-120=60^\circ,\quad y=180-80-60=40^\circ\).

(a) Same chord BC: BEC=19°. (b) AEB=ADB=23°; EAB=180−67−23=90°. BAC=BEC=19°, so \(ABC=180-19-23=138^\circ\), using ACB=ADB=23°. (c) CBE=ABC−ABE=138−67=71°; \(BCE=180-71-19=90^\circ\).

(a) Interior sum=\((8-2)180=1080^\circ\). Each reflex interior angle=\((1080-4(32))/4=238^\circ\). Therefore \(x=360-238=122^\circ\).

(b) The reflex angle at O subtended by PQR is twice y: \(360-(2y-60)=2y\Rightarrow420=4y\Rightarrow y=105^\circ\).

(a) Alternate angles between parallel lines are equal, so \(x=47^\circ\). (b) The fourth interior angle is \(360^\circ-115^\circ-85^\circ-97^\circ=63^\circ\). Thus \(y=180^\circ-63^\circ=117^\circ\). (c) The non-reflex central angle is \(2(58^\circ)=116^\circ\). Hence the reflex angle is \(z=360^\circ-116^\circ=244^\circ\).

Each exterior angle is \(180^\circ-172^\circ=8^\circ\). The number of sides is \(360/8=45\).

(a) The angle at the centre is twice the angle at the circumference: \(x=2(56^\circ)=112^\circ\). (b) Angles in the same segment are equal, so \(y=56^\circ\).

The sum of interior angles is \((6-2)180^\circ=720^\circ\). The sixth angle is \(720^\circ-5(115^\circ)=145^\circ\).

The base angles are equal: \(a=\frac{180^\circ-40^\circ}{2}=70^\circ\). Alternate angles between AC and BD give \(b=40^\circ\).

OA = OB, since both are radii. Thus triangle AOB is isosceles and \(\angle OAB=28^\circ\). The angles of a triangle sum to \(180^\circ\), so \(\angle AOB=180^\circ-28^\circ-28^\circ=124^\circ\). The angle at the circumference is half the angle at the centre standing on the same arc AB, hence \(\angle ACB=62^\circ\).

Opposite angles in a cyclic quadrilateral sum to \(180^\circ\), giving \(\angle AEC=135^\circ\). In triangle ACE, \(\angle CAE=180^\circ-135^\circ-20^\circ=25^\circ\). AED is straight, so \(\angle CAD=25^\circ\). Since \(AC=CD\), triangle ACD is isosceles and \(\angle ADC=25^\circ\). Thus \(\angle ACD=130^\circ\) and \(\angle ECD=130^\circ-20^\circ=110^\circ\).

(a) The five equal angles meet at a point, so \(x=360^\circ/5=72^\circ\). (b) In one kite, the angles are \(72^\circ,42^\circ,y,y\). Hence \(2y=360^\circ-72^\circ-42^\circ=246^\circ\), giving \(y=123^\circ\).

By the alternate segment theorem, the angle between tangent PT and chord PQ equals angle PWQ. Thus \(\angle PWQ=42^\circ\).

Angle x at the circumference subtends the same arc as the central angle 110°, so \(x=110/2=55^\circ\). Opposite angles of a cyclic quadrilateral sum to 180°, giving \(y=180-55=125^\circ\).

The 67° exterior angle equals the sum of the two opposite interior angles. The marked parallel lines make the right-hand interior angle 42°. Thus \(a=67-42=25^\circ\).

(a) The two equal base angles are \((180-44)/2=68^\circ\). (b) Exterior angles total 360°, so \(360/40=9\) sides.

B. Its opposite angles add to 180°: \(113+67=180\) and \(99+81=180\).

(a) The exterior angle of a cyclic quadrilateral equals the opposite interior angle: \(u=80^\circ\). The central angle on the same arc is twice this, so \(v=160^\circ\). (b) The perpendicular from the centre bisects the chord. Distance \(=\sqrt{8^2-5^2}=\sqrt{39}\approx6.24\) cm. (c) Triangles XMN and XLK are similar: corresponding angles are equal, including the angles subtended by the same chords. \(\frac{MN}{KL}=\frac{XN}{XK}=\frac{2.5}{4.8}\). Thus \(MN=9.7(2.5/4.8)=5.05208\ldots\approx5.05\) cm. (d) Triangles YRS and YQP are similar. Their length ratio is \(YS/YP=1/2\) and their area ratio is 1/4. \(\frac5{12}x(x-1)=\frac14x(x+1)\). Since x is positive, divide by x and multiply by 12: \(5(x-1)=3(x+1)\). Hence \(2x=8\Rightarrow x=4\).

(a) Angle VWX = 57° by alternate angles. Hence \(\angle WVX=180-88-57=35^\circ\). (b) Similar triangles give \(AC/AQ=BC/PQ=12.6/8.4=1.5\), so \(AC=7.2(1.5)=10.8\) cm.

The angle subtended by a diameter is 90°. The angle subtended by the same chord as the 53° angle is 53°. In the resulting triangle, \(x=180-90-53=37^\circ\).

(a) The equal base angles are \((180-44)/2=68^\circ\). (b) \(n=360/24=15\) sides.

(a)(i) By the alternate segment theorem, angle ACB = 48°. Angle ABC = 90° because AC is a diameter. Thus angle BAC = 42°, and angle BDC subtending the same chord BC gives \(x=42^\circ\). (ii) Let E be the intersection of AC and BD. In ABE, angles at A and B are 42° and 27°, so angle AEB = \(180-42-27=111^\circ\). Vertically opposite angle CED gives \(y=111^\circ\). (b)(i) Sum the three semicircular arcs: \(\pi(6+4+2)=12\pi\approx37.7\) cm. (ii) Shaded area \(\tfrac12\pi(6^2+4^2-2^2)=24\pi\) cm² = \(2400\pi\) mm². Volume \(2400\pi(1.6)=3840\pi\approx12100\) mm³ (3 significant figures).

(a)(i) AB = AD (equal tangents), so \(\angle ABD=(180-48)/2=66^\circ\). (ii) OB is perpendicular to AB, giving \(90-66=24^\circ\). (iii) \(\angle BOD=180-2(24)=132^\circ\), so \(\angle BCD=132/2=66^\circ\). (iv) Opposite angles of BCDE sum to 180°, so \(\angle BED=180-66=114^\circ\). (b) \(A=\tfrac12(15)(15)\sin132^\circ=83.608\ldots\approx83.6\) cm². (c) \(\angle ABO=\angle ADO=90^\circ\), so these opposite angles sum to 180°.

(a) The parallel sides give angle FAB = 140° and angle BCD = 130°. A hexagon’s interior angles total \((6-2)180=720^\circ\). Thus \(\angle EFA=720-(140+90+130+140+120)=100^\circ\). (b)(i) Angle BAC equals BDC = 30° (same chord BC). Hence \(\angle ACB=180-100-30=50^\circ\). (ii) In BXC, \(\angle XBC=180-89-50=41^\circ\). Angle CAD equals CBD (same chord CD), so 41°. (iii) Similar: a pair of vertically opposite angles at X and equal angles subtending the same chord. (c) Triangles PYS and QYR are similar; corresponding sides PS and QR have ratio 11 : 10. Area ratio is its square, so \(A_{PYS}=23(11/10)^2=27.83\) cm². (d)(i) Exterior angles total 360°, so \(n(n/10)=360\Rightarrow n^2=3600\Rightarrow n=60\). (ii) \(180-60/10=174^\circ\).

\((25-2)180=4140^\circ\).

(a) 37°, because angles in the same segment subtending chord BD are equal. (b) 74°, because the central angle is twice the angle at the circumference subtending the same arc. (c) 143°, because opposite angles of cyclic quadrilateral ABCD sum to 180°.

(a)(i) 43°, angles in the same segment. (ii) OY = OZ, so OYZ is isosceles; angle OZY = 28°. Therefore \(\angle YOZ=180^\circ-56^\circ=124^\circ\). The angle at the circumference is half the central angle, so \(w=62^\circ\). (iii) Opposite angles of a cyclic quadrilateral sum to 180°: \(p+5p=180\Rightarrow p=30\). (b)(i) 1 : 2, because the perpendicular from the centre bisects a chord. (ii) OQ is common; MQ = NQ; OM = ON, because equal chords are equidistant from the centre O. The triangles are congruent by SSS.

(a)(i) Angle BAD = angle CAB because D lies on AC. With angle ADB = angle ABC, the remaining angles of the triangles are equal. (ii) Similar (AA). (iii) Corresponding side ratio \(AB/AC=BD/BC\), so \(BD=11(12/16)=8.25\) cm. (b)(i) u = 38°, same chord BC. (ii) v = 38°, equal chords BC and CD subtend equal angles. (iii) w = 78°, opposite angle AED in cyclic quadrilateral ABDE. (iv) Angle DBC = v = 38°, so angle ABC = 78° + 38° = 116°. Thus \(x=180-38-116=26^\circ\). (c) OP = OQ implies angle PQO = m°. Since PQ is parallel to OR, angle QOR = m° (alternate angles). OQ = OR implies angle OQR = angle QRO = 2m°. In triangle QOR, \(m+2m+2m=180\Rightarrow m=36\).

(a) Angle AOC = 116° (central angle is twice the circumference angle). OA = OC, so \(\angle OCA=(180-116)/2=32^\circ\). (b) Angle ADC = 122° (opposite ABC in a cyclic quadrilateral). Thus \(\angle DCA=180-122-23=35^\circ\).

(a)(i) x = 21°, same chord RQ. Angle SRT = 42°, so y = 42°, same chord SP. (ii) Triangles SRT and PQT are similar by equal angles; \(TQ/RT=PQ/SR\), so \(TQ=3.31(9.43)/8.23\approx3.79\) cm. (b) Diameter EF is perpendicular to tangent EK, so angle GEF = 115° − 90° = 25°. Angle EGF = 90° (semicircle), hence EFG = 65°. Cyclic opposite angle EHG = 115°. Parallel lines give EGH = GEF = 25°, so GEH = 180° − 115° − 25° = 40°. (c) OA = OC, giving angle AOC = 180° − 28° = 152°. D lies on the minor arc AC, so angle ADC = (360° − 152°)/2 = 104°. AD = DC makes triangle ADC isosceles: angle ACD = (180° − 104°)/2 = 38°.

(a)(i) p = 180° − 32° = 148°. (ii) Parallel lines give angle FYA = 32°. Triangle YFA then gives YAF = 58°, so q = 122°. (iii) Parallel sides give t = 148°. (iv) Interior angle sum of a hexagon = 720°: \(148+122+90+148+2x=720\Rightarrow x=106^\circ\). (b)(i) The tangent–chord theorem gives x = 63°. (ii) Angle SPQ = 63°, since it subtends chord SQ, equal to PS. Triangle PSQ is isosceles, so \(y=180^\circ-2(63^\circ)=54^\circ\).

\(9x+90=360\Rightarrow x=30\).

\(180^\circ-360^\circ/18=160^\circ\).

(a) 108°: angle at the centre is twice the angle at the circumference on the same arc. (b)(i) \(-4/3\). (ii) The tangent is perpendicular to OC, with gradient ¾. \((k+4)/(7-3)=3/4\Rightarrow k=-1\).

(a)(i) 52°, angles subtending chord DC in the same segment. (ii) 104°, centre angle is twice the circumference angle. (iii) 34°, because OC is perpendicular to tangent CE, giving 90° − 56°. (b)(i) \(BE=\sqrt{8.9^2+7^2-2(8.9)(7)\cos56^\circ}=7.650\ldots\approx7.65\) cm. (ii) \(\sin E=7\sin56^\circ/BE\Rightarrow E=49.335\ldots^\circ\approx49.3^\circ\). The obtuse alternative cannot fit with the 56° angle and side lengths.

The vertically opposite angle is 55°. The triangle’s third angle is \(180-55-50=75^\circ\). The adjacent exterior angle is \(a=180-75=105^\circ\).

(a) Opposite angles in cyclic quadrilateral ABDE give \(\angle BDE=180-85=95^\circ\). (b) \(\angle BAN=\angle BAC=19^\circ\) (angles subtended by BC). In triangle ABN, \(\angle ANB=180-58-19=103^\circ\). Therefore \(\angle AND=180-103=77^\circ\).

(a)(i) \(110+110+84+2x=540\Rightarrow x=118\). (ii) In isosceles BCD, \(\angle CBD=(180-118)/2=31^\circ\). (iii) By symmetry, EDA = BDC = 31°, so \(\angle ADB=84-31-31=22^\circ\). (b) The reflex central angle is twice ABC: \(360-3y=2(4y+4)\Rightarrow y=32\). (c)(i) Opposite angles of a cyclic quadrilateral sum to 180°. (ii) \(\angle PRS=(2/3)(102)=68^\circ\). PQS subtends the same chord PS, so PQS = 68°. (d) \(l=6.9\sqrt{5/7.2}=5.75\) cm.

(a) \(\pi(8.5)^2=226.9801\ldots\approx227\) cm². (b) A perpendicular from the centre bisects a chord, so \(MB=10.7/2=5.35\) cm. (c) \(\sin\angle MOB=5.35/8.5\Rightarrow\angle MOB=39.0067\ldots^\circ\approx39^\circ\). (d) Minor AOC \(=4(39)=156^\circ\); major angle = 204°. Arc \((204/360)2\pi(8.5)=30.2640\ldots\approx30.3\) cm. (e) AB = BC (given), TA = TC (tangents from one point), and TB is common. SSS proves congruence.

\(n=360/36=10\). It is a decagon.

(a) \(\angle ABC=42+28=70^\circ\). Opposite cyclic angles give \(\angle ADC=110^\circ\). (b) Let X be the intersection of AC and BD. \(\angle AXB=180-59=121^\circ\), so in ABX, \(\angle BAX=180-121-28=31^\circ\). Then in ABC, \(\angle ACB=180-70-31=79^\circ\). ADB subtends the same chord AB, so ADB = 79°.

The corresponding acute angle is 55°. It and p lie on a straight line, so \(p=180-55=125^\circ\).

OAB = 38°. OA = OB, so OBA = 38°, giving \(\angle AOB=180-2(38)=104^\circ\). The angle at the circumference is half this: \(\angle ACB=52^\circ\).

OC makes 42° with the horizontal, and OM is vertical, so COM = 48°. By symmetry DOM = 48°. (a) CAM subtends CM and is half COM: \(48/2=24^\circ\). (b) DCM subtends DM and is half DOM: 24°.

(a) If 48° is a base angle, the others are 48°, 84°. If it is the vertex angle, the others are 66°, 66°. (b) \((5-2)180=540^\circ\). (c) The boundary has 7 vertices, so its interior angles sum to \((7-2)180=900^\circ\). Each marked outer angle plus its inner angle is 360°; sum \(7(360)-900=1620^\circ\). (d)(i) \((2x+5)+(3y-20)+(4x-5)+(x+y-10)=360\Rightarrow7x+4y=390\). (ii) Co-interior angles A and B sum to 180°: \(2x+5+3y-20=180\Rightarrow2x+3y=195\). (iii) Multiply the first equation by 3 and the second by 4: \(21x+12y=1170,\ 8x+12y=780\Rightarrow13x=390\Rightarrow x=30,\ y=45\). (iv) In order A, B, C, D: 65°, 115°, 115°, 65°.

\(\angle ADC=360-73-120-82=85^\circ;\ x=180-85=95^\circ\).