Finding Unknown Angles in Triangles

Apply triangle properties to find unknown angles in complex figures

Exterior Angles of a Triangle

When a side of a triangle is extended, it forms an exterior angle. The exterior angle has a special property:

Exterior Angle Property:

Exterior angle = Sum of the two interior opposite angles

Example 1: BC extended beyond C

Side BC is extended beyond point C to point D, forming exterior angle ACD.

BC extended to D: Exterior angle ACD = 120°

Why does this work?

Interior angle ACB + Exterior angle ACD = 180° (angles on a straight line)

Interior angle ACB = 180° − 50° − 70° = 60°

So exterior angle ACD = 180° − 60° = 120°

Notice: $120° = 50° + 70°$ (sum of interior opposite angles!)

Example 2: AB extended beyond B

Side AB is extended beyond point B to point E, forming exterior angle CBE.

AB extended to E: Exterior angle CBE = 130°

Exterior angle CBE = $\angle A + \angle C$

CBE = $45° + 85° = 130°$

Example 3: CA extended beyond A

Side CA is extended beyond point A to point F, forming exterior angle BAF.

CA extended to F: Exterior angle BAF = 130°

Exterior angle BAF = $\angle B + \angle C$

BAF = $55° + 75° = 130°$

Using the Exterior Angle Property

We can use the exterior angle property to find unknown interior angles:

Worked Example: Find angle x

In the figure, BC is extended to D. Find angle x.

Find angle x using the exterior angle property

Solution:

Exterior angle = Sum of interior opposite angles

$x = 110° - 35°$

$x = 75°$

Adjacent Triangles

Sometimes two triangles share a common side. We can use properties of both triangles to find unknown angles.

Example 1: Equilateral Triangle Adjacent to Another Triangle

Triangle PQR is equilateral. Triangle QRS shares side QR, with S on the other side of QR from P. Angle PQS = 130° and angle QRS = 50°. Find angle QSR.

Equilateral triangle PQR adjacent to triangle QRS

Solution:

Step 1: In equilateral triangle PQR, all angles = 60°, so ∠PQR = 60°

Step 2: ∠PQS is made up of ∠PQR and ∠RQS

$\angle RQS = 130° - 60° = 70°$

Step 3: In triangle QRS, use angle sum = 180°

$\angle QSR = 180° - 70° - 50° = 60°$

Example 2: Two Isosceles Triangles

Triangle ABD is isosceles with AB = AD. Triangle BCD is also isosceles with BC = CD. They share side BD. Angle ABD = 65° and angle DBC = 40°. Find (a) angle BAD, (b) angle BCD.

Two isosceles triangles sharing side BD

Solution:

(a) Triangle ABD: AB = AD, so the base angles are equal

$\angle ABD = \angle ADB = 65°$

$\angle BAD = 180° - 65° - 65° = 50°$

(b) Triangle BCD: BC = CD, so the base angles are equal

$\angle BDC = \angle DBC = 40°$

$\angle BCD = 180° - 40° - 40° = 100°$

Example 3: Right-angled Triangle Next to Another Triangle

Triangle XYZ has a right angle at Y and angle YXZ = 30°. Triangle YZW shares side YZ, with angle ZYW = 45° and angle YWZ = 50°. Find angle XZW.

Right-angled triangle XYZ next to triangle YZW

In triangle XYZ: ∠XZY = 180° − 90° − 30° = 60°. In triangle YZW: ∠YZW = 180° − 45° − 50° = 85°. So ∠XZW = 60° + 85° = 145°.

Practice Problems

Practice 1: Find angle y using exterior angle property

BC extended to D. Find angle y.

ShowHide solution

Exterior angle = Sum of interior opposite angles

$y = 95° - 40°$

$y = 55°$

Practice 2: PQ extended beyond Q

PQ extended to S. Find exterior angle z.

ShowHide solution

Exterior angle z = Sum of interior opposite angles

$z = \angle P + \angle R$

$z = 65° + 50°$

$z = 115°$

Practice 3: Isosceles triangle with exterior angle

Triangle DEF is isosceles with DE = DF. EF is extended to G. If the exterior angle DFG = 130°, find all angles of triangle DEF.

Isosceles triangle with DE = DF

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Step 1: Interior angle DFE + Exterior angle DFG = 180°

$\angle DFE = 180° - 130° = 50°$

Step 2: Since DE = DF, base angles are equal

$\angle DEF = \angle DFE = 50°$

Step 3: Find angle D using angle sum

$\angle D = 180° - 50° - 50° = 80°$

Answer: $\angle D = 80°$ , $\angle E = \angle F = 50°$

Practice 4: Two adjacent triangles

Triangle ABC is equilateral. Triangle ACD shares side AC. Angle CAD = 40° and angle ACD = 80°. Find angle ADC.

Equilateral triangle ABC with triangle ACD

ShowHide solution

In triangle ACD:

$\angle ADC = 180° - 40° - 80°$

$\angle ADC = 60°$

Practice 5: Isosceles triangles sharing a side

Triangle PQR is isosceles with PQ = PR and angle QPR = 40°. Triangle PRS is isosceles with PR = PS and angle RPS = 30°. Find (a) angle PQR, (b) angle PSR.

Two isosceles triangles sharing PR

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(a) Triangle PQR: PQ = PR, angle QPR = 40°

Base angles: $\angle PQR = \angle PRQ = (180° - 40°) \div 2 = 70°$

(b) Triangle PRS: PR = PS, angle RPS = 30°

Base angles are equal: $\angle PRS = \angle PSR$

$\angle PRS + \angle PSR = 180° - 30° = 150°$

$\angle PSR = 150° \div 2 = 75°$

$\angle PQR = 70°$ , $\angle PSR = 75°$

Practice 6: Challenge Problem

In the figure, ABC is an equilateral triangle. BC is extended to D. Find angle ACD (the exterior angle at C).

Equilateral triangle ABC with BC extended to D

ShowHide solution

Method 1: Using angles on a straight line

Interior angle ACB = 60° (equilateral triangle)

$\angle ACB + \angle ACD = 180°$ (angles on a straight line)

$\angle ACD = 180° - 60° = 120°$

Method 2: Using exterior angle property

Exterior angle = Sum of interior opposite angles

$\angle ACD = \angle A + \angle B = 60° + 60° = 120°$

$\angle ACD = 120°$

Key Takeaways

Exterior Angle Property: Exterior angle = Sum of two interior opposite angles
Angles on a straight line: Interior angle + Exterior angle = 180°
For adjacent triangles, apply angle sum (180°) to each triangle separately
In isosceles triangles, base angles are always equal
Look for relationships between angles when triangles share sides