Triangles and Sectors in Radians

Sine and cosine rule with radian angles (radian mode, angle sum π, exact cos(2π/3) = −½ etc.)

In the exam for Edexcel International AS Maths · 9 real past-paper questions below

Sine and cosine rules in radians

You can already use the sine rule and the cosine rule in degrees. AS questions use the same two rules, but give the angles in radians. Three things change.

Steps

1
1. Set your calculator to radian mode before you start.
2. Angles in a triangle add up to $π$ , not 180°. Angles on a straight line also add up to $π$ .
3. Learn the exact values below. Questions that say "exact" expect them.

Formula

Angle	$\frac{π}{6}$	$\frac{π}{4}$	$\frac{π}{3}$	$\frac{2 π}{3}$	$\frac{5 π}{6}$
sin	$\frac{1}{2}$	$\frac{2}{2}$	$\frac{3}{2}$	$\frac{3}{2}$	$\frac{1}{2}$
cos	$\frac{3}{2}$	$\frac{2}{2}$	$\frac{1}{2}$	$- \frac{1}{2}$	$- \frac{3}{2}$

Memory trick

$π$ is the new 180. An obtuse angle like $\frac{2 π}{3}$ is $π - \frac{π}{3}$ : same sine as $\frac{π}{3}$ , and the cosine just flips sign.

Worked example

1
Two sides of a triangle are 4 cm and 6 cm, with an angle of $\frac{2 π}{3}$ between them. Find the exact length of the third side.
2
$a^{2} = 4^{2} + 6^{2} - 2 (4) (6) cos \frac{2 π}{3}$ — Cosine rule. The angle sits between the 4 and the 6.
3
$a^{2} = 52 - 48 (- \frac{1}{2})$ — $16 + 36 = 52$ , and $cos \frac{2 π}{3} = - \frac{1}{2}$ exactly.
4
$a^{2} = 52 + 24 = 76$ — Minus times minus is plus.
5
$a = 76 = 219 cm$ — Simplify the surd: $76 = 4 \times 19$ .

Worked example

1
A triangle has angles $A = 0.9$ and $B = 1.1$ radians. The side $c$ , facing angle $C$ , is 10 cm. Find side $a$ .
2
$C = π - 0.9 - 1.1 = 1.1416 \dots$ — Angles in a triangle add up to $π$ . Keep the full value in your calculator.
3
$\frac{a}{s i n 0.9} = \frac{10}{s i n 1.1416 \dots}$ — Sine rule, with the unknown side on top.
4
$a = \frac{10 s i n 0.9}{s i n 1.1416 \dots} = 8.61 cm$ — Radian mode. The answer is 8.6146…, so 8.61 to 3 s.f.

Common mistake

Leaving the calculator in degree mode. Then $cos \frac{2 π}{3}$ comes out as 0.9993 instead of −0.5, and every answer after it is wrong.

Key idea

The word "radian" is usually traced to 1873, when it is said to have appeared in exam questions set by James Thomson at Queen's College, Belfast. So it may have started life in an exam paper, just like the ones you will sit.

Key idea

Same rules, new units: radian mode, angle sum $π$ , and exact values for the special angles.

When the angle is obtuse

A sine gives two possible angles. If $sin x = 0.6$ , then $x = 0.6435$ works, and so does $π - 0.6435 = 2.498$ . Your calculator only shows the first one, the acute angle.

Rule

If the question tells you the angle is obtuse, or that it is greater than $\frac{π}{2}$ , use $π$ minus the calculator answer.

θ_{obtuse} = π - θ_{calculator}

Worked example

1
In triangle $P QR$ , angle $P = 0.5$ radians, $QR = 7$ cm and $P Q = 11$ cm. Angle $R$ is obtuse. Find angle $R$ .
2
$\frac{s i n R}{11} = \frac{s i n 0.5}{7}$ — Sine rule with sines on top: 11 faces $R$ , 7 faces $P$ .
3
$sin R = \frac{11 s i n 0.5}{7} = 0.7534 \dots$ — This is $sin R$ , not $R$ yet. Take the inverse sine next.
4
$sin^{- 1} (0.7534 \dots) = 0.8532$ — The acute answer from the calculator.
5
$R = π - 0.8532 = 2.288 rad$ — We were told $R$ is obtuse, so take $π$ minus it.

The same choice turns up when a question gives you the area of a triangle and asks for an angle.

Worked example

1
Two sides of a triangle are 8 cm and 10 cm. The angle $θ$ between them is obtuse, and the area is 30 cm². Find $θ$ , then the third side.
2
$\frac{1}{2} (8) (10) sin θ = 30$ — Area $= \frac{1}{2} ab sin θ$ , with $θ$ between the two sides.
3
$sin θ = \frac{30}{40} = 0.75$ — Divide by 40.
4
$θ = π - 0.8481 = 2.294 rad$ — $sin^{- 1} 0.75 = 0.8481$ is acute. The angle is obtuse.
5
$a^{2} = 8^{2} + 1 0^{2} - 2 (8) (10) cos 2.294$ — Cosine rule for the side facing $θ$ .
6
$a^{2} = 164 + 105.8 = 269.8$ — $cos 2.294$ is negative, so the last term is added.
7
$a = 16.4 cm$ — Square root, to 3 s.f.

Common mistake

Using the acute angle 0.8481. The third side then comes out as 7.63 cm. The method can still earn marks, but the accuracy marks are lost.

Rule

In a "show that" question, such as "show that the angle is 2.29 to 3 s.f.", write one more digit first (2.2935…), then round. The answer is given, so the mark is for your working.

Key idea

Calculator answer first. If the angle must be obtuse, take it away from $π$ .

Major sectors

Two radii cut a circle into two sectors. The small one is the minor sector. The big one, the rest of the circle, is the major sector.

Formula

major angle = 2 π - θ

$2 π$ — a full turn in radians.
$θ$ — the minor angle, the one usually marked on the diagram.

Then use the usual formulas, $s = r θ$ and $A = \frac{1}{2} r^{2} θ$ , with the major angle.

Worked example

1
A circle has centre $O$ and radius 6 m. Points $A$ and $B$ on the circle make a minor angle $A O B = \frac{π}{3}$ . Find the exact arc length, area and perimeter of the major sector.
2
$2 π - \frac{π}{3} = \frac{5 π}{3}$ — The major angle. Mark schemes often credit this angle on its own.
3
$s = 6 \times \frac{5 π}{3} = 10 π m$ — Major arc: $s = r θ$ .
4
$A = \frac{1}{2} (6^{2}) (\frac{5 π}{3}) = 30 π m^{2}$ — Major sector area: $\frac{1}{2} r^{2} θ$ .
5
$Perimeter = 10 π + 12 m$ — Major arc plus two radii.

Now close the gap with the straight line $A B$ . The shape is the major sector plus triangle $A O B$ .

Worked example

1
Find the exact area and perimeter of that new shape.
2
$Triangle A O B = \frac{1}{2} (6) (6) sin \frac{π}{3} = 93$ — Two radii with $\frac{π}{3}$ between them.
3
$Area = 30 π + 93 m^{2}$ — Add the pieces.
4
$A B = 6$ — $O A = O B$ and the angle is $\frac{π}{3}$ , so the triangle is equilateral.
5
$Perimeter = 10 π + 6 m$ — Major arc plus $A B$ . The radii are now inside the shape.

Common mistake

Putting the marked angle $\frac{π}{3}$ into $r θ$ or $\frac{1}{2} r^{2} θ$ . That gives the minor sector, the part that is missing.

Key idea

Major sector: use $2 π - θ$ . Check: the major and minor areas must add up to $π r^{2}$ .

Plan views: pieces and edges

Many AS questions show the plan view of a stage, a pond or a landing pad: a sector joined to one or two triangles. Plan the pieces before any numbers.

Steps

1
1. Find every missing angle. Angles on a straight line add up to $π$ .
2. Area: add the area of each piece.
3. Perimeter: trace round the outside. Count only the outer edges. A shared line inside the shape is not part of the perimeter.

Worked example

1
Sector $O A B$ has centre $O$ , radius 6 m and angle $A O B = 0.9$ radians. Triangle $O B C$ is joined along $O B$ . $C$ lies on $A O$ extended, so $A O C$ is a straight line, and $O C = 4$ m. Find the area and perimeter of the shape.
2
$∠ B O C = π - 0.9 = 2.2416$ — $A O C$ is a straight line, so the two angles at $O$ add up to $π$ .
3
$Sector = \frac{1}{2} (6^{2}) (0.9) = 16.2$ — $\frac{1}{2} r^{2} θ$ .
4
$Triangle = \frac{1}{2} (6) (4) sin 2.2416 = 9.40$ — $\frac{1}{2} ab sin C$ : $O B = 6$ and $O C = 4$ , with the angle between them.
5
$Area = 16.2 + 9.40 = 25.6 m^{2}$ — Add the pieces.
6
$B C^{2} = 6^{2} + 4^{2} - 2 (6) (4) cos 2.2416 = 81.84$ — Cosine rule. The cosine is negative, so this is more than $36 + 16$ .
7
$B C = 9.046$ — Square root.
8
$Arc A B = 6 \times 0.9 = 5.4$ — $s = r θ$ .
9
$Perimeter = 5.4 + 9.046 + 4 + 6 = 24.4 m$ — Arc $A B$ , then $B C$ , then $C O$ and $O A$ . $O B$ is inside, so leave it out.

Common mistake

Adding $O B$ to the perimeter. It is the edge the two pieces share, so it is inside the shape.

Rule

Each piece usually earns its own method mark, even if you slip in another piece. Label each one ("sector =", "triangle =") so the examiner can find it.

Key idea

Area: add the pieces. Perimeter: walk round the outside only.

Two sectors, one centre

Some shapes use two sectors with the same centre and the same angle: a small one inside a big one. The region between them is like a slice of a ring.

Formula

ring area = \frac{1}{2} (R^{2} - r^{2}) θ

$R$ — the outer radius.
$r$ — the inner radius.
$θ$ — the shared angle, in radians.

Worked example

1
Two sectors share centre $O$ and angle 0.75 radians. The outer arc is 12 m long. The inner radius $O A$ and the gap $A C$ between the arcs are in the ratio $O A : A C = 2 : 3$ . Find the area between the arcs.
2
$12 = R \times 0.75 \Rightarrow R = 16$ — Outer arc: $s = R θ$ .
3
$O A = \frac{2}{5} \times 16 = 6.4$ — The ratio 2 : 3 splits $R$ into 5 parts. $O A$ is 2 of them.
4
$\frac{1}{2} (1 6^{2} - 6. 4^{2}) (0.75) = 80.6 m^{2}$ — Big sector minus small sector.

When the gap is unknown, call it $x$ . Then the outer radius is $r + x$ , and the area gives a quadratic.

Worked example

1
Two sectors share centre $O$ and angle 0.5 radians. The inner radius is 2 m. The region between the arcs has area 24 m². The gap between the arcs is $x$ m. Find $x$ , then the perimeter of the region.
2
$\frac{1}{2} (2 + x)^{2} (0.5) - \frac{1}{2} (2^{2}) (0.5) = 24$ — Outer radius $2 + x$ . Big sector minus small sector.
3
$\frac{1}{4} (2 + x)^{2} - 1 = 24$ — Tidy each term: $\frac{1}{2} \times 0.5 = \frac{1}{4}$ .
4
$(2 + x)^{2} = 100$ — Add 1, then multiply by 4.
5
$x^{2} + 4 x - 96 = 0$ — Expand: $x^{2} + 4 x + 4 = 100$ . A "show that" wants this line written out.
6
$(x + 12) (x - 8) = 0 \Rightarrow x = 8$ — A length cannot be negative, so reject $- 12$ .
7
$0.5 \times 2 + 0.5 \times 10 + 2 \times 8 = 22 m$ — Inner arc, outer arc and the two straight edges of length $x$ .

Common mistake

Squaring the gap instead of the whole radius: $\frac{1}{2} x^{2} θ$ is wrong. The outer radius is $r + x$ , so square $(r + x)$ .

Key idea

Same centre, same angle: big sector minus small sector. Unknown gap? Outer radius $r + x$ , then solve the quadratic.

Which method? Read the question

The question's wording tells you which tool to pick.

You see	You do
Angles given in radians	Radian mode; angles in a triangle add to $π$
"Exact" with $\frac{π}{3}$ , $\frac{2 π}{3}$ and so on	Exact sin and cos values, surds
"Given the angle is obtuse"	$π$ minus the calculator angle
Area of a triangle given	$\frac{1}{2} ab sin θ$ for $sin θ$ first
"Major sector" or the big part of a circle	Angle $2 π - θ$
Sector joined to triangles	Add areas; perimeter = outer edges only
Two arcs with the same centre	Big sector minus small sector; gap $x$ gives a quadratic

Key idea

Before any working, mark the angle you need and write it in radians. Most lost marks in these questions start with the wrong angle.

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