What is 5.1: Triangle rules, area and radians?

For a unit-circle angle θ, cosθ and sinθ are the point's horizontal and vertical coordinates, while tanθ=sinθ/cosθ where cosθ0; this extends the ratios beyond acute angles.

How many marks are 5.1 questions worth?

5.1 practice drills on this page are worth 2–6 marks each; marks in a real exam vary by question.

Which past exam papers has 5.1 appeared in?

We have not yet indexed a verified past-paper appearance for 5.1 — check the Edexcel A-level Maths 9MA0 past-papers hub directly.

5.1

Understand and use the definitions of sine, cosine and tangent for all arguments; the sine and cosine rules; the area of a triangle in the form ½ab sin C; work with radian measure, including use for arc length and area of sector.

Draft — not yet indexed

Triangle rules, area and radians

Worked answers and methods for 5.1 on Edexcel A-level Maths 9MA0.

Explanation

For a unit-circle angle $\theta$ , $\cos\theta$ and $\sin\theta$ are the point's horizontal and vertical coordinates, while $\tan\theta=\sin\theta/\cos\theta$ where $\cos\theta\neq0$ ; this extends the ratios beyond acute angles.
For a triangle, use $a/\sin A=b/\sin B=c/\sin C$ when an opposite side-angle pair is known, $a^2=b^2+c^2-2bc\cos A$ for three sides or an included angle, and $\tfrac12ab\sin C$ for area.
In the ambiguous sine-rule case, test the supplementary angle because $\sin A=\sin(\pi-A)$ , then reject any triangle inconsistent with the given sides and angles.
Radian measure makes arc and sector formulae direct: an angle $\theta$ radians in a circle of radius $r$ gives arc length $s=r\theta$ and sector area $A=\tfrac12r^2\theta$ .
Keep the calculator in the required angle mode and label sides opposite their matching angles; using degrees in $s=r\theta$ or pairing the wrong side and angle is a common error.

A labelled triangle: each lower-case side is opposite its matching upper-case angle.

Worked example

Two sides of a triangular sail are $8\,\text{m}$ and $11\,\text{m}$ , with included angle $0.9$ radians. Calculate the third side and the area of the sail, giving each answer to $3$ significant figures.

1.The cosine rule gives $c^2=8^2+11^2-2(8)(11)\cos(0.9)=75.597\ldots$ , so $c=8.694\ldots\,\text{m}=8.69\,\text{m}$ .
2.The area is $\tfrac12(8)(11)\sin(0.9)=34.466\ldots\,\text{m}^2$ , hence $34.5\,\text{m}^2$ .

Answer: Third side $=8.69\,\text{m}$ Area $=34.5\,\text{m}^2$

Common mistakes

Don't use degrees in $s=r\theta$ or $A=\tfrac12r^2\theta$ , even though these formulae require radians.
Don't pair a side with a non-opposite angle in the sine rule.

Exam tip

Label each side opposite its matching angle and check the calculator angle mode before substituting.

Worked practice

Tier 1 · Easy

A circular arc has radius

7.5\,\text{cm}

and subtends

1.2

radians at the centre. Find its length.

(2)

(Total for Question 1 is 2 marks)

Mark scheme

Mark scheme for question 1
Question	Scheme	Marks
1	$9.0\,\text{cm}$	2
Notes
Use $s=r\theta$ with the angle already in radians: $s=7.5(1.2)=9.0\,\text{cm}$ .

(2 marks)

Tier 2 · Standard

A sector of a circle has radius

7\,\text{cm}

and arc length

11.2\,\text{cm}

. Find the angle of the sector in radians and its area.

(3)

(Total for Question 2 is 3 marks)

Mark scheme

Mark scheme for question 2
Question	Scheme	Marks
2	Angle $=1.6$ radians Area $=39.2\,\text{cm}^2$	3
Notes
Using $s=r\theta$ , the angle is $\theta=11.2/7=1.6$ radians. The sector area is $\tfrac12r^2\theta=\tfrac12(7^2)(1.6)=39.2\,\text{cm}^2$ .

(3 marks)

Tier 3 · Hard

A minor segment is cut from a circle of radius

6\,\text{cm}

by a chord whose endpoints subtend

1.4

radians at the centre. Determine the perimeter and area of the segment, giving both to

3

significant figures.

(5)

(Total for Question 3 is 5 marks)

Mark scheme

Mark scheme for question 3
Question	Scheme	Marks
3	Perimeter $=16.1\,\text{cm}$ Area $=7.46\,\text{cm}^2$	5
Notes
The arc length is $6(1.4)=8.4\,\text{cm}$ . Splitting the isosceles triangle in half gives chord length $2(6)\sin(0.7)=7.730\ldots\,\text{cm}$ , so the perimeter is $16.130\ldots\,\text{cm}$ . The sector area is $\tfrac12(6^2)(1.4)=25.2\,\text{cm}^2$ and the triangle area is $\tfrac12(6^2)\sin(1.4)=17.738\ldots\,\text{cm}^2$ . Their difference is $7.461\ldots\,\text{cm}^2$ , giving the stated answers.

(5 marks)

Tier 1 · Easy

Two sides of a triangle have lengths

10\,\text{cm}

and

13\,\text{cm}

. The included angle is

30^\circ

. Find the area of the triangle.

(2)

(Total for Question 4 is 2 marks)

Mark scheme

Mark scheme for question 4
Question	Scheme	Marks
4	$32.5\,\text{cm}^2$	2
Notes
Use $A=\tfrac12ab\sin C$ : $A=\tfrac12(10)(13)\sin30^\circ=65(1/2)=32.5\,\text{cm}^2$ .

(2 marks)

Tier 2 · Standard

In triangle

ABC

, side

a

has length

6\,\text{cm}

A=30^\circ

and

B=45^\circ

. Find the exact perimeter of the triangle.

(4)

(Total for Question 5 is 4 marks)

Mark scheme

Mark scheme for question 5
Question	Scheme	Marks
5	$6+9\sqrt2+3\sqrt6\,\text{cm}$	4
Notes
First $C=180^\circ-30^\circ-45^\circ=105^\circ$ . By the sine rule, $b=6\sin45^\circ/\sin30^\circ=6\sqrt2$ and $c=6\sin105^\circ/\sin30^\circ=3(\sqrt6+\sqrt2)$ . Therefore the perimeter is $6+6\sqrt2+3(\sqrt6+\sqrt2)=6+9\sqrt2+3\sqrt6\,\text{cm}$ .

(4 marks)

Tier 3 · Hard

Two concentric circles and two radial lines enclose an annular sector of angle

1.2

radians. The outer radius is

3\,\text{cm}

greater than the inner radius, and the area of the annular sector is

54\,\text{cm}^2

. Find both radii and the perimeter of the annular sector.

(5)

(Total for Question 6 is 5 marks)

Mark scheme

Mark scheme for question 6
Question	Scheme	Marks
6	Inner radius $=13.5\,\text{cm}$ and outer radius $=16.5\,\text{cm}$ Perimeter $=42\,\text{cm}$	5
Notes
Let the outer and inner radii be $R$ and $r$ . Then $R-r=3$ . The area difference is $\tfrac12(1.2)(R^2-r^2)=0.6(R-r)(R+r)=54$ , so $1.8(R+r)=54$ and $R+r=30$ . Solving gives $R=16.5$ and $r=13.5$ . The perimeter consists of both arcs and two radial lengths: $1.2R+1.2r+2(R-r)=1.2(30)+2(3)=42\,\text{cm}$ .

(5 marks)

Tier 2 · Standard

In triangle

ABC

, sides

a

and

b

are opposite angles

A

and

B

respectively. Given that

a=7\,\text{cm}

b=10\,\text{cm}

and

A=30^\circ

, find the two possible values of angle

B

and the corresponding areas of triangle

ABC

. Give the angles to

1

decimal place and the areas to

3

significant figures. Use unrounded values in your working.

(5)

(Total for Question 7 is 5 marks)

Mark scheme

Mark scheme for question 7
Question	Scheme	Marks
7	$B=45.6^\circ$ , area $=33.9\,\text{cm}^2$ $B=134.4^\circ$ , area $=9.40\,\text{cm}^2$	5
Notes
By the sine rule, $\sin B=10\sin30^\circ/7=5/7$ . Hence $B=45.584\ldots^\circ$ or $180^\circ-45.584\ldots^\circ=134.415\ldots^\circ$ . The corresponding values of $C$ are $104.415\ldots^\circ$ and $15.584\ldots^\circ$ . Using $\tfrac12ab\sin C$ , the areas are $\tfrac12(7)(10)\sin(104.415\ldots^\circ)=33.898\ldots\,\text{cm}^2$ and $\tfrac12(7)(10)\sin(15.584\ldots^\circ)=9.403\ldots\,\text{cm}^2$ .

(5 marks)

Tier 3 · Hard

A sector of a circle has perimeter

20\,\text{cm}

and area

24\,\text{cm}^2

. Find all possible values of its radius and its angle in radians.

(5)

(Total for Question 8 is 5 marks)

Mark scheme

Mark scheme for question 8
Question	Scheme	Marks
8	Radius $4\,\text{cm}$ and angle $3$ radians Radius $6\,\text{cm}$ and angle $4/3$ radians	5
Notes
Let the radius be $r$ cm and the angle be $\theta$ radians. The perimeter gives $r\theta+2r=20$ , so $\theta=20/r-2$ . Substitution into $\tfrac12r^2\theta=24$ gives $10r-r^2=24$ , hence $(r-4)(r-6)=0$ . If $r=4$ , then $\theta=3$ ; if $r=6$ , then $\theta=4/3$ . Both radii and angles are positive, so both sectors are valid.

(5 marks)

Tier 3 · Hard

Two sides of a triangle have lengths

8\,\text{cm}

and

13\,\text{cm}

, and its area is

26\,\text{cm}^2

. Find the two possible included angles. For each angle, find the exact length of the third side.

(5)

(Total for Question 9 is 5 marks)

Mark scheme

Mark scheme for question 9
Question	Scheme	Marks
9	Included angle $30^\circ$ , third side $=\sqrt{233-104\sqrt3}\,\text{cm}$ Included angle $150^\circ$ , third side $=\sqrt{233+104\sqrt3}\,\text{cm}$	5
Notes
If the included angle is $C$ , then $26=\tfrac12(8)(13)\sin C$ , so $\sin C=1/2$ . The two possible angles in a triangle are $C=30^\circ$ and $C=150^\circ$ . By the cosine rule, the third side $c$ satisfies $c^2=8^2+13^2-2(8)(13)\cos C$ . This gives $c^2=233-104\sqrt3$ when $C=30^\circ$ and $c^2=233+104\sqrt3$ when $C=150^\circ$ . Both values are positive and each angle, together with the two given sides, constructs a valid triangle.

(5 marks)

Q10

Tier 3 · Hard

10.

A chord of a circle has length

16\,\text{cm}

. The perpendicular distance from the centre of the circle to the chord is

6\,\text{cm}

. Calculate the perimeter and area of the major segment cut off by the chord, giving each answer to

3

significant figures. Use unrounded values in your working.

(6)

(Total for Question 10 is 6 marks)

Mark scheme

Mark scheme for question 10
Question	Scheme	Marks
10	Perimeter $=60.3\,\text{cm}$ Area $=269\,\text{cm}^2$	6
Notes
The perpendicular from the centre bisects the chord, so the radius is $r=\sqrt{8^2+6^2}=10\,\text{cm}$ . If the minor central angle is $\theta$ , then $\theta=2\arctan(8/6)=1.854590\ldots$ radians. The major angle is $2\pi-\theta=4.428594\ldots$ radians, so the major-segment perimeter is $10(2\pi-\theta)+16=60.285948\ldots\,\text{cm}$ . The minor triangle has area $\tfrac12(10^2)\sin\theta=48\,\text{cm}^2$ . Hence the major-segment area is $\tfrac12(10^2)(2\pi-\theta)+48=269.429743\ldots\,\text{cm}^2$ , giving the stated answers.

(6 marks)

Verified exam appearances

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