Area of a triangle

½ab sin C, forwards and backwards

In 9 of 20 recent exams
1% of the marks
Paper 1 or 2
2 to 7 marks

How it is set in the exam

Paper 1 (no calculator) mostly (8 of the 11 parts set since 2021), 2–7 marks, exact: a special angle, or a sine found from a given cosine, often after a cosine-rule step. On Paper 2 (calculator) it is a quick area once a side or an angle has been found, or the formula used backwards together with the cosine rule. Common (9 of 20 exam versions).

Spot it in the exam

Exam phrases to spot (5)

“find the area of triangle $ABC$ ”
“find the exact area”
“the area of triangle $PQR$ is $18$ cm²; find $PQ$ ”
“show that the area can be written as …”
“express the area in terms of $x$ ”

In disguise (3)

only the cosine (or the tangent) of the angle between the two sides is given: the sine comes first
one of the two sides must first come from the cosine rule, with sides in terms of $x$
an area, one angle and the side facing it are given, and both sides next to that angle are unknown: the area gives their product and the cosine rule gives a second equation

Variants (6)

forwards on , after a side found by the sine or cosine rule
exact on : a special angle ( $135°$ and $150°$ included), or a sine found earlier in the question
an area in terms of a letter, then set equal to a value: a quadratic
the area back to one side, as a surd
area plus cosine rule: a relation between the two unknown sides, then a system on the GDC
a right-angled triangle with sides in terms of a letter: $\frac{1}{2} \times$ base $\times$ height

Looks like it, but isn't (7)

a region bounded by an arc: Chords, segments and regions
two areas, one for each of two triangles that fit SSA data: The ambiguous case: two angles from one sine, for different reasons)
the sine itself (or $\sin 2\theta$ ) is what is asked: Exact values without the angle: from one ratio to another
two parts of equal area, used only to fix a point before a length or an angle is found: Find a side or an angle: choose the rule
the area is a model to maximise or minimise: Optimise with a fixed total (quadratic model) when the model is a quadratic, otherwise Optimisation: use the constraint, then optimise; this type's formula is the set-up
a triangle with two radii of a circle as sides, or a shape split into such triangles, including a triangle or polygon inscribed in a circle whose radius is known and none of whose sides is: Chords, segments and regions
a triangle cut off by a tangent or normal and the axes: The tangent or normal meets something

Worked example

An area in terms of x, then back to x

Paper 1 (no calculator)

Question

In triangle $ABC$ , $AB = 2x$ cm, $AC = (x + 5)$ cm and angle $BAC = 135°$ .
(a) Show that the area of triangle $ABC$ is $\frac{\sqrt{2}}{2}\left(x^2 + 5x\right)$ cm². [3]
(b) The area of the triangle is $12\sqrt{2}$ cm². Find the value of $x$ . [4]

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Which method?

The only real choice is how to get the sine of the angle between the two sides.

What you know about the angle $C$ between the two sides	How to get $\sin C$
$C$ itself, or two other angles (then the angle sum)	directly; exact values on ( $\sin 135° = \sin 45°$ )
$\cos C$ only	$\sin C = \sqrt{1 - \cos^2 C}$ , or a right triangle; positive even when $\cos C < 0$
$\tan C$ only	a right triangle with shorter sides $\lvert \tan C \rvert$ and $1$ , and Pythagoras for the hypotenuse; the sine is positive even when $\tan C < 0$ ( $C$ obtuse)
the three sides	the cosine rule for $\cos C$ first, then as above (or $C$ itself on the GDC)
$C$ , the side facing it and the area, but not the other two sides	no choice needed: $\frac{1}{2}ab\sin C = \text{area}$ gives $ab$ , and the cosine rule gives $a^2 + b^2$

Steps

Find two sides and the angle between them.
If the given angle is not between the given sides, find the one that is (angle sum, sine rule).
Get $\sin C$
from the table. It is always positive.
Forwards:
substitute into $\frac{1}{2}ab\sin C$ , exactly on .
Backwards:
set $\frac{1}{2}ab\sin C$ equal to the area and solve for the unknown: a side, the product $ab$ , or $\sin C$ (then both $C$ and $180° - C$ , each checked: angle from a ratio).
Two unknown sides:
pair the area equation with the cosine rule and solve the system (solve an equation numerically, or zeros and intersections with both equations in the same subject: Solving equations graphically).

Common mistakes

An angle that is not between the two sides
in $\frac{1}{2}ab\sin C$ , $C$ is the angle where $a$ and $b$ meet.
The cosine in place of the sine
with only $\cos C$ given, find $\sin C$ first. The method mark needs the sine.
A negative sine
for an obtuse angle: every angle of a triangle has a positive sine.

More mistakes (5)

Only one angle from $\sin C$
when the area is given: $180° - C$ fits as well, unless the question rules it out (Your turn 4).
Surds left unsimplified or unrationalised
when a form is asked for: $3\sqrt{5}$ , not $\frac{15}{\sqrt{5}}$ .
A rejected root kept
there is no follow-through from an $x$ that makes a side zero or negative.
Rounding in a show-that
keep $\frac{36}{\sin 52°}$ exact until the given result appears.
The wrong mode
with the angles in degrees, radian mode turns $\sin 52°$ into $\sin 52 = 0.987$ , and Example 2.4(a) gives $xy = 36.5$ instead of $45.7$ . It keeps the method marks and loses the answers (check the settings).

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