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Year 7 Maths Geometry

Exterior Angle of a Triangle

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Theory

When one side of a triangle is extended, the angle outside the triangle is an exterior angle. It equals the sum of the two interior opposite angles (the two angles inside the triangle that are not next to it). This Year 7 maths page shows why, and how to use the rule forwards and backwards with reasons.

An exterior angle is outside the triangle, between one side and the next side extended. The interior opposite angles are the two angles inside the triangle that are not at that vertex.

The rule. If \(d^\circ\) is an exterior angle and \(a^\circ\), \(b^\circ\) are its interior opposite angles, then \(d = a + b\). Reason: exterior angle of a triangle equals the sum of the two interior opposite angles.

Why it works. \(a + b + c = 180\) (angle sum of a triangle) and \(c + d = 180\) (angles on a straight line). Both sums are \(180\), so \(a + b + c = c + d\), and taking \(c\) from both sides gives \(d = a + b\).

The angle next to it. The exterior angle and the interior angle at the same vertex lie on a straight line, so they add to \(180^\circ\).

One at each vertex. Each vertex can have its own exterior angle. With one exterior angle at each vertex, the three exterior angles add to \(360^\circ\).

Exterior angle d equals a plus b Triangle ABC with BC extended to D. Interior angles a at A and b at B are blue, c at C is grey, and the exterior angle d at C is red. a° b° c° d° A B C D
\(BC\) is extended to \(D\). The exterior angle \(d^\circ\) (red) equals \(a^\circ + b^\circ\) (blue). The angle \(c^\circ\) next to it is not used in the rule: \(c + d = 180\).
An exterior angle at each vertex Triangle ABC with interior angles 40 at A, 65 at B and 75 at C. Each side is extended in turn, making exterior angles p at C, q at A and r at B. 40° 65° 75° p° q° r°
Extend each side in turn. \(p = 40 + 65 = 105\), \(q = 65 + 75 = 140\) and \(r = 40 + 75 = 115\). Together \(105 + 140 + 115 = 360\).

The rule and the straight-line fact:

\[ d = a + b \quad (\text{exterior angle of a triangle}) \]
d = a + b ( exterior angle of a triangle )
\[ c + d = 180 \quad (\text{angles on a straight line}) \]
c + d =180( angles on a straight line )
\[ \text{one exterior angle at each vertex: total } 360^\circ \]
one exterior angle at each vertex total 360
Use the two far angles. The interior angle next to the exterior angle is not an interior opposite angle. It adds to the exterior angle to make 180°.
Add, or take away? To find an exterior angle, add: \(40 + 75 = 115\). To find a missing interior opposite angle, subtract: \(115 - 40 = 75\).

How to use the exterior angle rule

  1. Find the exterior angle: it is between a side and the extended next side.
  2. Find its two interior opposite angles: the two angles not at that vertex.
  3. Write exterior \(=\) sum of the two, with the reason.
  4. Solve: add to find the exterior angle, subtract to find an interior one.

How to check

  1. The exterior angle and the interior angle next to it add to \(180^\circ\).
  2. The three interior angles add to \(180^\circ\).
Example 1 — Find the exterior angle
\(BC\) is extended to \(D\). Find the value of \(x\).
Exterior angle x with interior opposite angles 72 and 45 Triangle ABC with BC extended to D. Angle A is 72 degrees, angle B is 45 degrees and the exterior angle ACD is x degrees. 72° 45° x° A B C D
Solution
\(x\)\(=\)\(72 + 45\)
\(\)\(=\)\(117\)
x =72+45=117

\(x = 117\) (exterior angle of a triangle equals the sum of the two interior opposite angles). Check: \(\angle ACB = 180 - 117 = 63\) and \(72 + 45 + 63 = 180\).

Example 2 — Naming the angles
In \(\triangle ABC\), \(\angle BAC = 52^\circ\) and \(\angle CBA = 63^\circ\). \(BC\) is extended past \(C\) to \(D\). Find the size of \(\angle DCA\).
Solution
\(\angle DCA\)\(=\)\(\angle BAC + \angle CBA\)
\(\)\(=\)\(52^\circ + 63^\circ\)
\(\)\(=\)\(115^\circ\)
DCA = BAC + CBA =52+63=115

\(\angle DCA = 115^\circ\) (exterior angle of a triangle).

Example 3 — Work backwards
\(BC\) is extended to \(D\). Find \(a\) and \(b\).
Exterior angle 130 at the base Triangle ABC with BC extended to D. The exterior angle ACD is 130 degrees, angle B is 58 degrees, angle A is a degrees and angle ACB is b degrees. a° 58° b° 130° A B C D
Solution
\(a + 58\)\(=\)\(130\)
\(a\)\(=\)\(72\)
\(b + 130\)\(=\)\(180\)
\(b\)\(=\)\(50\)
a +58=130, a =72 b +130=180, b =50

\(a = 72\) (exterior angle of a triangle) and \(b = 50\) (angles on a straight line). Check: \(72 + 58 + 50 = 180\).

Example 4 — Find an interior angle
In \(\triangle PQR\), \(\angle QPR = 38^\circ\). \(QR\) is extended to \(S\) and \(\angle PRS = 108^\circ\). Find the size of \(\angle PQR\).
Solution
\(\angle PRS\)\(=\)\(\angle QPR + \angle PQR\)
\(108^\circ\)\(=\)\(38^\circ + \angle PQR\)
\(\angle PQR\)\(=\)\(108^\circ - 38^\circ\)
\(\)\(=\)\(70^\circ\)
PRS = QPR + PQR ,108=38+ PQR , PQR =70

\(\angle PQR = 70^\circ\)

Example 5 — An isosceles triangle
\(AB = AC\) and \(BC\) is extended to \(D\). Find \(a\) and \(b\).
Isosceles triangle with an exterior angle Triangle ABC with AB and AC marked equal, BC extended to D. Angle A is 44 degrees, the base angles are a degrees and the exterior angle ACD is b degrees. 44° a° a° b° A B C D
Solution
\(a + a + 44\)\(=\)\(180\)
\(2a\)\(=\)\(136\)
\(a\)\(=\)\(68\)
\(b\)\(=\)\(44 + 68\)
\(\)\(=\)\(112\)
2 a +44=180, a =68 b =44+68=112

\(a = 68\) (base angles equal, angle sum \(180^\circ\)) and \(b = 112\) (exterior angle of a triangle). Check: \(68 + 112 = 180\).

Example 6 — Write an equation
Find \(x\), then the two interior opposite angles.
Exterior angle 136 with interior opposite angles x and x plus 24 Triangle with one side extended. The interior opposite angles are x degrees and (x + 24) degrees; the exterior angle is 136 degrees. (x + 24)° x° 136°
Solution
\(x + x + 24\)\(=\)\(136\)
\(2x\)\(=\)\(112\)
\(x\)\(=\)\(56\)
x + x +24=136,2 x =112, x =56

\(x = 56\). The angles are \(56^\circ\) and \(80^\circ\). Check: \(56 + 80 = 136\).

Example 7 — Two triangles on a line
\(B\), \(C\), \(D\) and \(E\) lie on a straight line, in that order. \(A\) is joined to \(B\), \(C\) and \(D\). \(\angle ABC = 35^\circ\), \(\angle BAC = 25^\circ\) and \(\angle CAD = 45^\circ\). Find
(i) \(\angle ACD\)
(ii) \(\angle ADE\)
Solution
\(\angle ACD\)\(=\)\(35^\circ + 25^\circ = 60^\circ\)
\(\angle ADE\)\(=\)\(60^\circ + 45^\circ = 105^\circ\)
ACD =35+25=60 in triangle ABC ADE =60+45=105 in triangle ACD

(i) \(60^\circ\) (exterior angle of \(\triangle ABC\))
(ii) \(105^\circ\) (exterior angle of \(\triangle ACD\)). Use the rule in one triangle at a time.

Common pitfalls

Adding the wrong angles. Do not use the interior angle next to the exterior angle in the sum.
Subtracting from 180 instead of adding. The exterior angle is the sum of the two far angles. Subtracting from 180 only finds the angle next to it.
A reflex angle is not the exterior angle. The exterior angle is between one side and the extension of the next side, not all the way around the outside.
No reason. Write the reason for each step: exterior angle of a triangle, angles on a straight line, or angle sum of a triangle.

Frequently asked questions

What is an exterior angle of a triangle?

The angle between one side of the triangle and the next side extended, outside the triangle.

What is the exterior angle rule?

An exterior angle of a triangle equals the sum of the two interior opposite angles.

Which angles are the interior opposite angles?

The two angles inside the triangle that are not at the vertex of the exterior angle.

How do you find a missing interior angle from an exterior angle?

Subtract the known interior opposite angle from the exterior angle. If the exterior angle is \(108^\circ\) and one angle is \(38^\circ\), the other is \(70^\circ\).

What do the exterior angles of a triangle add to?

With one exterior angle at each vertex, they add to \(360^\circ\).

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