Without using a calculator, find the exact value of 2\cos2(30°) + \tan(45°).
Worked Solution & Answer
2(√(3)2)2 + 1 = 32 + 1 = 52 B1
Answer: 52
Exact trig. values (harder) — Notes & Worked Examples Show / Hide
Explanation:
On non-calculator Higher papers, you must know the exact values of sin, cos, and tan for 0°, 30°, 45°, 60°, and 90° and combine them with surd simplification, rationalising denominators, and the Sine or Cosine Rule.
Key exact values: sin 30° = cos 60° = 12, cos 30° = sin 60° = √32, sin 45° = cos 45° = √22, tan 30° = √33, tan 45° = 1, tan 60° = √3.
Example 1: Work out the exact value of tan 60° + sin 30°cos 30° in the form a + √3b.
Solution:
Substitute exact values: tan 60° = √3, sin 30° = 12, cos 30° = √32
Numerator: √3 + 12 = 2√3 + 12
Divide by √32: 2√3 + 1√3 × √3√3 = 6 + √33
Answer: 2 + √33
Example 2: In ΔABC, AB = 4 cm, AC = 6√3 cm, and ∠BAC = 30°. Find the exact length of BC.
Solution:
BC2 = 42 + (6√3)2 − 2(4)(6√3) cos 30°
BC2 = 16 + 108 − 48√3(√32) = 124 − 72 = 52
BC = √52 = 2√13 cm
Answer: 2√13 cm
Without using a calculator, find the exact value of \sin(60°)\cos(30°) + \tan2(60°).
Worked Solution & Answer
1 + (√(3))2 = 4 B1
Answer: 4
ABCDEFGH is a cuboid.
AF = 6.8 cm
FC = 13.6 cm
Work out the size of the angle between FC and the plane ABCD.
FC = 13.6 cm
Work out the size of the angle between FC and the plane ABCD.
Worked Solution & Answer
In right-angled triangle FAC: sin(∠FCA) = AF / FC = 6.8 / 13.6 = 0.5 M1
⇒ ∠FCA = 30° A1
⇒ ∠FCA = 30° A1
Answer: 30
Here are a triangle and a rectangle.
a and b are positive numbers.
a and b are positive numbers.
triangle rectangle cannot tell
Which shape has the larger perimeter?
You must work out expressions for both perimeters.
You must work out expressions for both perimeters.
Worked Solution & Answer
Perimeter of triangle = (5a + b) + (4a + 7b) + (2a + 3b) = 11a + 11b M1
Perimeter of rectangle = 2[(4a + 2b) + (a + 4b)] = 2(5a + 6b) = 10a + 12b M1
Difference = (11a + 11b) - (10a + 12b) = a - b
Since we do not know whether a > b, a < b, or a = b, it is impossible to determine which is larger. A1
Perimeter of rectangle = 2[(4a + 2b) + (a + 4b)] = 2(5a + 6b) = 10a + 12b M1
Difference = (11a + 11b) - (10a + 12b) = a - b
Since we do not know whether a > b, a < b, or a = b, it is impossible to determine which is larger. A1
Answer: cannot tell
ABC is a triangle.
Work out the length of AB.
cm
Worked Solution & Answer
AB / sin 45° = 3√2 / sin 30° M2
⇒ AB = 3√2 × (√2 / 2) ÷ (1 / 2) = 3 ÷ 0.5 = 6 cm A1
⇒ AB = 3√2 × (√2 / 2) ÷ (1 / 2) = 3 ÷ 0.5 = 6 cm A1
Answer: 6 cm
Here is a cuboid.
DH = 8 cm
DB = 13 cm
DH = 8 cm
DB = 13 cm
(a) Work out the size of angle DBH. [2 marks]
Answer
degrees
(b) Using your answer to part (a), work out the size of angle ECG. [1 mark]
Answer
degrees
Worked Solution & Answer
(a)
Triangle BDH is right-angled at D, because DH is vertical and BD is in the horizontal base. M1
\tan(∠ DBH) = DHDB = 813.
∠ DBH = \arctan(813) ≈ 31.6°. A1
\tan(∠ DBH) = DHDB = 813.
∠ DBH = \arctan(813) ≈ 31.6°. A1
Answer (a): Answer 31.6 degrees
(b)
By symmetry of the cuboid, right triangle ECG has base diagonal EC = DB = 13 cm and vertical height CG = DH = 8 cm, so ∠ ECG = ∠ DBH = 31.6°. B1
Answer (b): Answer 31.6 degrees
(a) Write down the exact value of tan 45° (1) (b) Right-angled triangle with adjacent 4 cm, angle 60°, hypotenuse x cm. cos 60° = 0.5. Work out x. (2)
(a) Write down the exact value of \tan 45° [1 mark]
(b) Work out the value of x. [2 marks]
Worked Solution & Answer
(a)
tan 45° = 1 B1
Answer (a): 1
(b)
cos 60° = 4 / x M1
⇒ 0.5 = 4 / x
⇒ x = 8 A1
⇒ 0.5 = 4 / x
⇒ x = 8 A1
Answer (b): 8
A, B, C, D and E are points on a circle.
BFD and AFC are straight lines.
DC = DF
BFD and AFC are straight lines.
DC = DF
Work out the size of angle x.
You must show your working which may be on the diagram.
You must show your working which may be on the diagram.
Answer
degrees
Worked Solution & Answer
Angle ∠ BDC and angle ∠ BAC are subtended by the same arc BC M1
Angles in the same segment are equal, so ∠ BDC = ∠ BAC = 24°
Since BFD is a straight line, ∠ FDC = 24° M1
In △ CDF, DC = DF, so the triangle is isosceles with base angles equal:
∠ DCF = ∠ DFC = 2x M1
The angles in △ CDF sum to 180°:
24° + 2x + 2x = 180° ⇒ 4x = 156° ⇒ x = 39° A1
Angles in the same segment are equal, so ∠ BDC = ∠ BAC = 24°
Since BFD is a straight line, ∠ FDC = 24° M1
In △ CDF, DC = DF, so the triangle is isosceles with base angles equal:
∠ DCF = ∠ DFC = 2x M1
The angles in △ CDF sum to 180°:
24° + 2x + 2x = 180° ⇒ 4x = 156° ⇒ x = 39° A1
Answer: Answer 39 degrees
Here is triangle ABC.
Find the size of angle ABC.
You must show all your working.
You must show all your working.
Worked Solution & Answer
cos B = (10² + 5² – (5√7)²) / (2 × 10 × 5) = (100 + 25 – 175) / 100 = –50 / 100 = –0.5 M3
⇒ ∠ABC = 120°. A1
⇒ ∠ABC = 120°. A1
Answer: 120
Here is the plan of the floor of an L-shaped room.
All lengths are in metres.
All lengths are in metres.
(a) The area of the floor is 75 m2
Show that x2 + x - 90 = 0 [3 marks]
Show that x2 + x - 90 = 0 [3 marks]
(b) By factorising x2 + x - 90 work out the value of x.
You must show your working [2 marks]
You must show your working [2 marks]
x =
Worked Solution & Answer
(a)
Split the L-shape horizontally into two rectangles: M1
- Bottom rectangle: width x, height x - 2
Area1 = x(x - 2) = x2 - 2x
- Top rectangle: width x - 5, height 3
Area2 = 3(x - 5) = 3x - 15 M1
Total area:
Total Area = x2 - 2x + 3x - 15 = x2 + x - 15
Equate to 75 m2:
x2 + x - 15 = 75 ⇒ x2 + x - 90 = 0 C1
- Bottom rectangle: width x, height x - 2
Area1 = x(x - 2) = x2 - 2x
- Top rectangle: width x - 5, height 3
Area2 = 3(x - 5) = 3x - 15 M1
Total area:
Total Area = x2 - 2x + 3x - 15 = x2 + x - 15
Equate to 75 m2:
x2 + x - 15 = 75 ⇒ x2 + x - 90 = 0 C1
Answer (a): Area = x(x - 2) + 3(x - 5) = x² + x - 15 = 75 => x² + x - 90 = 0
(b)
Find two numbers that multiply to -90 and add to +1: +10 and -9. M1
(x + 10)(x - 9) = 0
x = -10 \quador x = 9
Since length must be positive (x - 5 > 0), x = 9. A1
(x + 10)(x - 9) = 0
x = -10 \quador x = 9
Since length must be positive (x - 5 > 0), x = 9. A1
Answer (b): x = 9