Blaaabla

. O is the centre of the circle which has a radius of 5.4 cm.
. O is the centre of the circle which has a radius of 5.4 cm.


The area of the shaded sector OAB is 21.6 cm2. Find the length of the minor arc AB.
RESPUESTA AB = rθ
= (M1)(A1)
= 21.6 × (A1)
= 8 cm (A1)

OR × (5.4)2θ = 21.6
⇒ θ = (= 1.481 radians) (M1)
AB = rθ (A1)
= 5.4 × (M1)
= 8 cm (A1) (C4)
[4]


2. Thecircle shown has centre O and radius 6. is the vector , is the vector and is the vector .


(a) Verify that A, B and C lie on the circle.
(3)
(b) Find the vector .
(2)

(c) Using an appropriate scalar product, or otherwise, find the cosine of angle .
(3)
(d) Find the area of triangle ABC, giving your answer in the form a, where a ∈ .
(4)
(Total 12 marks)
RESPUESTA
(a) = 6 ⇒ A ison the circle (A1)
 = 6 ⇒ B is on the circle. (A1)

= 
= 6 ⇒ C is on the circle. (A1) 3

(b) 
= (M1)
= (A1) 2

(c) (M1)
= 
= (A1)
= (A1)

OR (M1)(A1)
= as before (A1)
OR using the triangle formed by and its horizontal and
vertical components:
(A1)
 (M1)(A1) 3
Note: The answer is 0.289 to 3 sf

(d) A number of possible methods here

= (A1)
= (A1)
⎢BC ⎢ = 
⎢ΔABC ⎢=(A1)
= (A1)

OR ΔABC has base AB = 12 (A1)
and height = (A1)
⇒ area = (A1)
= (A1)

OR Given 
 (A1)(A1)(A1)
= (A1) 4
[12]


3. Solve the equation 3 sin2 x = cos2 x, for 0° ≤ x ≤ 180°.
(Total 4 marks)
RESPUESTA
tan2 x = (M1)
⇒ tan x = ± (M1)
⇒ x = 30° or x = 150° (A1)(A1) (C2)(C2)
[4]
4. If A is an obtuse angle in a triangle and sin A = , calculate the exact value of sin 2A.RESPUESTA
sin A = ⇒ cos A = ± (A1)
But A is obtuse ⇒ cos A = – (A1)
sin 2A = 2 sin A cos A (M1)
= 2 × 
= – (A1) (C4)
[4]
5. Given that sin θ = , cos θ = – and 0° ≤θ ≤360°,
(a) find the value of θ;
(b) write down the exact value of tan θ.
(Total 4 marks)

respuesta (a)

Acute angle 30° (M1)
Note: Award the (M1) for 30° and/or quadrant diagram/graph seen.
2nd quadrant since sine positiveand cosine negative
⇒ θ = 150° (A1) (C2)

(b) tan 150° = –tan 30° or tan 150° = (M1)
tan 150° = – (A1) (C2)
[4]


6. In the following diagram, O is the centre of the circle and (AT) is the tangent to the circle at T.
Diagram not to scale
If OA = 12 cm, and the circle has a radius of 6 cm, fid the area of the shaded region.
(Total 4 marks)


RESPUESTA = 90° (A1)
AT = 
=
=60° = (A1)
Area = area of triangle – area of sector
 = × 6 × – × 6 × 6 × (M1)
 = 12.3 cm2 (or – 6π) (A1) (C4)
OR
= 60° (A1)
Area of Δ = × 6 × 12 × sin 60 (A1)
Area of sector = × 6 × 6 × (A1)
Shaded area = – 6π = 12.3 cm2 (3 sf) (A1) (C4)
[4]


. O is the centre of the circle which has a radius of 5.4 cm.


The area of the shaded sector OAB is 21.6 cm2. Find thelength of the minor arc AB.
RESPUESTA AB = rθ
= (M1)(A1)
= 21.6 × (A1)
= 8 cm (A1)

OR × (5.4)2θ = 21.6
⇒ θ = (= 1.481 radians) (M1)
AB = rθ (A1)
= 5.4 × (M1)
= 8 cm (A1) (C4)
[4]


2. The circle shown has centre O and radius 6. is the vector , is the vector and is the vector .


(a) Verify that A, B and C lie on the circle.
(3)
(b) Find the vector .
(2)

(c) Usingan appropriate scalar product, or otherwise, find the cosine of angle .
(3)
(d) Find the area of triangle ABC, giving your answer in the form a, where a ∈ .
(4)
(Total 12 marks)
RESPUESTA
(a) = 6 ⇒ A is on the circle (A1)
 = 6 ⇒ B is on the circle. (A1)

= 
= 6 ⇒ C is on the circle. (A1) 3

(b) 
= (M1)
= (A1) 2

(c) (M1)
= 
= (A1)
= (A1)

OR (M1)(A1)
= as before (A1)
ORusing the triangle formed by and its horizontal and
vertical components:
(A1)
 (M1)(A1) 3
Note: The answer is 0.289 to 3 sf

(d) A number of possible methods here

= (A1)
= (A1)
⎢BC ⎢ = 
⎢ΔABC ⎢= (A1)
= (A1)

OR ΔABC has base AB = 12 (A1)
and height = (A1)
⇒ area = (A1)
= (A1)

OR Given 
 (A1)(A1)(A1)
= (A1) 4
[12]


3. Solve the equation 3 sin2 x = cos2 x, for...