Circle theorems and angle chains

(a) $74$°

1. $$3x+14+5x+6=180$$
2. $$8x=160$$
3. Opposite angles in a cyclic quadrilateral add to $180^\circ.$
4. 3x + 14 + 5x + 6 = 180, so 8x = 160 and x = 20.
5. Angle BAD = 3 $\times$ 20 + 14 = $74^\circ.$

• P1 Establishing $3x+14+5x+6=180$ or an equivalent valid method.
• P1 Establishing $8x=160$ or an equivalent valid method.
• A1 Correct answer: $74$°

(a) $12$ cm

1. $$13^{2}-5^{2}$$
2. $$\sqrt{144}$$
3. The radius OT is perpendicular to the tangent PT, so triangle OTP is right-angled at T.
4. By Pythagoras, $PT^{2}$ = $OP^{2}$ $-$ $OT^{2}$ = $13^{2}$ $-$ $5^{2}$ = 144.
5. PT = 12 cm.

• P1 Establishing $13^{2}-5^{2}$ or an equivalent valid method.
• P1 Establishing $\sqrt{144}$ or an equivalent valid method.
• A1 Correct answer: $12$ cm

(a) $70^\circ$

1. Angle $OAP =$ angle $OBP = 90^\circ$: a tangent is perpendicular to the radius at the point of contact.
2. Angles in quadrilateral $OAPB$ add up to $360^\circ$.
   $$\angle AOB = 360 - 90 - 90 - 40 = 140$$
3. The angle at the centre is twice the angle at the circumference.
   $$\angle ACB = 140 \div 2 = 70$$

• P1 Both right angles at $A$ and $B$.
• P1 Angle $AOB = 140^\circ$.
• A1 Correct answer: $70^\circ$.
• C1 Reasons: tangent perpendicular to radius; angles in a quadrilateral add to $360^\circ$; angle at centre is twice the angle at the circumference.

(a) $120^\circ$

1. Opposite angles of a cyclic quadrilateral add up to $180^\circ$.
   $$3x + x + 20 = 180$$
2. $$4x = 160 \;\Rightarrow\; x = 40$$
3. Angle $ABC = 3 \times 40 = 120^\circ$.

• P1 The equation $3x + x + 20 = 180$.
• P1 Finding $x = 40$.
• A1 Correct answer: $120^\circ$.
• C1 Opposite angles of a cyclic quadrilateral add up to $180^\circ$.

(b) $100^\circ$

1. $2y + y + 30 = 180$, so $y = 50$ and angle $BAD = 100^\circ$.

• M1 $3y + 30 = 180$.
• A1 Correct answer: $100^\circ$.

(a) $110$°

1. $$180-2(3x-4)+4x+18=180$$
2. $$x=13$$
3. Triangle ABD is isosceles, so angle ADB = angle ABD = (3x $-$ 4)$^\circ.$
4. Its angles sum to $180^\circ$, giving angle BAD = 180 $-$ 2(3x $-$ 4) = (188 $-$ 6x)$^\circ.$
5. Opposite angles of cyclic quadrilateral ABCD add to $180^\circ$: 188 $-$ 6x + 4x + 18 = 180.
6. This gives x = 13, so angle BAD = 188 $-$ 6 $\times$ 13 = $110^\circ.$

• P1 Establishing $180-2(3x-4)+4x+18=180$ or an equivalent valid method.
• P1 Establishing $x=13$ or an equivalent valid method.
• A1 Correct answer: $110$°

(a) $7.79$ cm

1. $$\frac{360-90-90-52}{2}$$
2. $$7/\sin(64)$$
3. OT and OU are perpendicular to their tangents. Angles in quadrilateral OTPU total $360^\circ$, so angle TOU = 360 $-$ 90 $-$ 90 $-$ 52 = $128^\circ.$
4. The perpendicular from O to TU bisects the chord and, in isosceles triangle OTU, bisects angle TOU.
5. The resulting right triangle has opposite side 7 cm, angle $64^\circ$ and hypotenuse r, so sin $64^\circ$ = 7/r.
6. r = 7/sin $64^\circ$ = 7.788213... cm, which is 7.79 cm to 3 significant figures.

• P1 Establishing $\frac{360-90-90-52}{2}$ or an equivalent valid method.
• P1 Establishing $7/\sin(64)$ or an equivalent valid method.
• A1 Correct answer: $7.79$ cm

(a) $74$°

1. The central angle is twice the angle at the circumference on the same arc.
   $$2\times 37$$
2. Therefore $74$°.

• M1 The central angle is twice the angle at the circumference on the same arc.
• A1 Correct answer: $74$°

(b) $53$°

1. OA and OB are equal radii, so the base angles share the remaining angle equally.
   $$\frac{180-74}{2}$$
2. Therefore $53$°.

• M1 OA and OB are equal radii, so the base angles share the remaining angle equally.
• A1 Correct answer: $53$°

(a) $80$°

1. Opposite angles in a cyclic quadrilateral sum to 180 degrees.
   $$180-100$$
2. Therefore $80$°.

• M1 Opposite angles in a cyclic quadrilateral sum to 180 degrees.
• A1 Correct answer: $80$°

(b) $50$°

1. AB=AD makes triangle ABD isosceles.
   $$\frac{180-80}{2}$$
2. Therefore $50$°.

• M1 AB=AD makes triangle ABD isosceles.
• A1 Correct answer: $50$°