Lesson 6 ~25 min Unit 3 · Geometry +85 XP

Triangle Problem Solving

Combine the angle sum of a triangle, the exterior angle theorem, isosceles & equilateral properties, and algebraic equations to find unknown angles in multi-step diagrams. The skill is choosing the right rule for each step.

Today's hook: A bridge uses a network of triangles. Can you find every missing angle from just one given measurement?

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Think First

warm-up

An isosceles triangle has one base angle of $70^{\circ}$. A second triangle shares the apex of the first, with one of its own angles equal to the apex of the isosceles. The second triangle has another angle of $40^{\circ}$. Without doing the algebra yet: write down what rule you would use FIRST, and on which triangle. Then list every rule you'd need to find every angle.

Record your answer in your workbook.

The Big Idea

+5 XP

Triangle problems combine several rules at once: angle sum ($180^{\circ}$), exterior angle theorem ($e = a + b$), isosceles base angles equal, equilateral $= 60^{\circ}$, plus angles on parallel lines (alternate / co-interior). The skill is choosing which rule to apply first, then chaining the rest.

Use a four-step approach. (1) What do you KNOW? List every given angle, every "equal sides" mark, every parallel-line tick. (2) What do you WANT? Mark the angle you must find. (3) Which TRIANGLE does the unknown belong to? That tells you whether to use angle sum, exterior angle, or isosceles. (4) WRITE down the equation with the reason — never skip the reason.

Strategy: KNOW → WANT → WHICH TRIANGLE → RULE + REASON

Label everything

Add letters or numbers to unlabelled angles before solving — you can't refer to what you can't name.

Start where you have most info

Find the easiest unknown first. Each new angle unlocks the next.

Always state the reason

Each line of working should end with a reason in brackets — "(∠ sum of $\triangle$)", "(ext. ∠ of $\triangle$)", "(base ∠s isos.)".

What to write in your book

Use the four-step strategy: KNOW → WANT → WHICH TRIANGLE → RULE + REASON.
Common tools: angle sum ($180^{\circ}$), exterior angle theorem ($e = a + b$), base angles of isosceles equal, equilateral = $60^{\circ}$.
Every line of working ends with a reason in brackets — markers always look for it.

Quick check — which reason justifies "$a + b + c = 180^{\circ}$" in your working?

What You'll Master

objectives

Know

Angle sum of a triangle $= 180^{\circ}$
Exterior angle of a triangle $= $ sum of two remote interiors
Base angles of an isosceles triangle are equal
All angles of an equilateral triangle are $60^{\circ}$

Understand

Why a multi-step problem needs MORE than one rule
How to choose between angle sum and the exterior angle theorem
Why finding any one angle in a diagram often unlocks all the others

Can Do

Plan a solution: list givens, identify the unknown, choose the rule
Solve linear equations like $2x + 3x + 5x = 180$ for unknown angles
Justify each step with a geometric reason

Words You Need

vocabulary

Angle sumThe three interior angles of any triangle add to $180^{\circ}$.

Exterior angle theoremExterior angle of a triangle $=$ sum of the two non-adjacent (remote) interior angles.

Isosceles triangleA triangle with two equal sides; the angles opposite the equal sides (base angles) are equal.

Apex / vertex angleIn an isosceles triangle, the angle between the two equal sides (not a base angle).

Equilateral triangleAll three sides equal; all three angles $= 60^{\circ}$.

ReasonThe geometric rule that justifies a line of working — written in brackets at the end.

Spot the Trap

heads-up

Wrong: "I'll just write $x = 60^{\circ}$, done." No working, no reasons — even if the number is right, you lose most of the marks.

Right: Show the equation, solve, then write the reason: e.g. $x + 70 + 70 = 180$ (∠ sum of $\triangle$).

Wrong: Mixing two triangles into one equation. Each triangle has its own angle sum — don't put angles from different triangles into one $180^{\circ}$ statement.

Right: Solve one triangle, get an angle. Then use that angle as a known in the second triangle's equation.

Isosceles Problems

+5 XP

If you're told ONE base angle of an isosceles triangle, you immediately know the OTHER base angle (they're equal). Then use the angle sum to find the apex. Or, if you're told the apex angle, the two base angles share what's left equally.

Given one base angle $= a$: the other base angle $= a$ too. Apex $= 180 - 2a$.
Given the apex $= v$: each base angle $= (180 - v) \div 2$.

Apex $= 180 - 2 \times (\text{base angle})$

Equal sides → equal angles

Tick marks tell you which sides are equal; the angles OPPOSITE those sides are equal.

"Base angles" reason

Write "(base ∠s isos.)" or "(base angles of isosceles triangle equal)".

Watch for the obtuse apex

If the apex is obtuse (e.g. $120^{\circ}$), each base angle is small: $(180 - 120) \div 2 = 30^{\circ}$.

What to write in your book

One base angle known → the other base angle is the same; apex $= 180 - 2a$.
Apex known → each base angle $= (180 - v) \div 2$.
Reason: "(base ∠s isos.)" or "(base angles of isosceles triangle equal)".

True or false?

An isosceles triangle with an apex angle of $120^{\circ}$ has two base angles of $30^{\circ}$ each.

Algebraic Angles

+5 XP

When angles are given as expressions in $x$ — for example $2x$, $3x$, $5x$ — add them and set the sum to $180^{\circ}$. Solve for $x$, then substitute back to find each angle. Always finish by checking the angles add to $180^{\circ}$.

Example: angles are $2x^{\circ}, 3x^{\circ}$ and $5x^{\circ}$. Step 1: $2x + 3x + 5x = 180$ (∠ sum of $\triangle$). Step 2: $10x = 180$, so $x = 18$. Step 3: the three angles are $36^{\circ}, 54^{\circ}, 90^{\circ}$. Check: $36 + 54 + 90 = 180^{\circ}$ ✓

Sum of expressions $= 180^{\circ}$, then solve for $x$.

Substitute back

$x$ alone is rarely the final answer — you usually need the actual angle sizes.

Collect like terms

Add the $x$ coefficients first, then the constants, before solving.

Check the sum

Last line: add your three angles — they must equal $180^{\circ}$.

What to write in your book

For algebraic angles, set the sum equal to $180^{\circ}$ then solve for the variable.
Collect like terms first (e.g. $2x + 3x + 5x = 10x$) before isolating $x$.
Always back-substitute — the question usually wants the actual angle sizes, not just $x$.

Fill in the blank.

A triangle has angles $2x^{\circ}, 3x^{\circ}$ and $5x^{\circ}$. Solving $10x = 180$ gives $x = $ .

Watch Me Solve It · 3 examples

Watch Me Solve It · Isosceles apex

+15 XP per step

PROBLEM

An isosceles triangle has one base angle of $70^{\circ}$. Find the apex angle.

1
Identify the other base angle
Other base angle $= 70^{\circ}$ (base ∠s isos.)
2
Use angle sum
Apex $= 180 - 70 - 70$ (∠ sum of $\triangle$)
3
Compute
Apex $= 40^{\circ}$
Check: $70 + 70 + 40 = 180^{\circ}$ ✓

AnswerApex angle $= 40^{\circ}$.

Watch Me Solve It · Algebraic angle sum

+15 XP per step

PROBLEM

A triangle has angles $2x^{\circ}$, $3x^{\circ}$ and $5x^{\circ}$. Find $x$ and state all three angles.

1
Set up the equation
$2x + 3x + 5x = 180$ (∠ sum of $\triangle$)
2
Collect & solve
$10x = 180 \Rightarrow x = 18$
3
State the angles
Angles $= 36^{\circ}, 54^{\circ}, 90^{\circ}$
Check: $36 + 54 + 90 = 180^{\circ}$ ✓

Answer$x = 18$; angles are $36^{\circ}, 54^{\circ}, 90^{\circ}$.

Watch Me Solve It · Two triangles, shared side

+15 XP per step

PROBLEM

Two triangles $ABD$ and $BCD$ share side $BD$, with $A, B, C$ collinear (on a straight line). In triangle $ABD$, $\angle DAB = 50^{\circ}$ and $\angle ABD = 80^{\circ}$. In triangle $BCD$, $\angle BCD = 40^{\circ}$. Find $\angle ADB$, then $\angle DBC$, then $\angle BDC$.

1
Angle sum of triangle ABD
$\angle ADB = 180 - 50 - 80 = 50^{\circ}$ (∠ sum of $\triangle$)
2
Straight line at B gives angle DBC
$\angle DBC = 180 - 80 = 100^{\circ}$ (∠s on str. line)
3
Angle sum of triangle BCD
$\angle BDC = 180 - 100 - 40 = 40^{\circ}$ (∠ sum of $\triangle$)
Each triangle gets its own equation. Use the straight line at $B$ to carry information across.

Answer$\angle ADB = 50^{\circ}$; $\angle DBC = 100^{\circ}$; $\angle BDC = 40^{\circ}$.

Common Pitfalls

heads-up

Mixing two triangles in one equation

Adding angles from triangle 1 with angles from triangle 2 and setting the sum to $180^{\circ}$.

Fix: Each triangle gets its own angle-sum equation. Solve one, then carry the result into the next.

Ignoring the tick marks

Missing that the diagram shows equal sides (isosceles) or that all three sides have a mark (equilateral).

Fix: First thing to check on every triangle diagram: are there tick marks?

Forgetting to find the actual angles

Stopping after finding $x$ when the question asks for the angle sizes.

Fix: Always substitute $x$ back into each expression — that's the angle.

Copy Into Your Books

The Four Steps

(1) What do I KNOW?
(2) What do I WANT?
(3) WHICH triangle?
(4) Rule + reason

Rules toolbox

(∠ sum of $\triangle$): $a + b + c = 180^{\circ}$
(ext. ∠ of $\triangle$): $e = a + b$
(base ∠s isos.): equal sides ⇒ equal angles
Equilateral: each angle $= 60^{\circ}$

Isosceles

Apex $= 180 - 2 \times (\text{base})$
Base $= (180 - \text{apex}) \div 2$

Algebra

Sum expressions $= 180^{\circ}$
Solve, then sub $x$ back
Final check: angles sum to $180^{\circ}$

How are you completing this lesson?

Brain Trainer · 4 problems

Brain Trainer · Mixed Triangle Problems

4 problems

Four mixed drills using angle sum, exterior angle, isosceles and algebra. Solve, then reveal.

1 Isosceles triangle has one base angle of $55^{\circ}$. Find the apex angle.

Other base $= 55^{\circ}$. Apex $= 180 - 55 - 55$.$= 70^{\circ}$
2 Angles of a triangle are $x^{\circ}, 2x^{\circ}, 3x^{\circ}$. Find $x$.

$x + 2x + 3x = 180 \Rightarrow 6x = 180$.$x = 30$
3 Each angle of an equilateral triangle equals?

All three equal, sum $= 180$.$60^{\circ}$
4 Isosceles triangle has apex $40^{\circ}$. Find each base angle.

$(180 - 40) \div 2 = 140 \div 2$.$= 70^{\circ}$ each

Complete in your workbook.

Quick Check · 5 questions

An isosceles triangle has one base angle of $65^{\circ}$. The apex angle is:

+10 XP

The angles of a triangle are $x^{\circ}, 2x^{\circ}, 3x^{\circ}$. Find $x$.

+10 XP

An isosceles triangle has an apex angle of $100^{\circ}$. Each base angle is:

+10 XP

In a triangle with interior angles $40^{\circ}, 70^{\circ}, 70^{\circ}$, the exterior angle at one of the $70^{\circ}$ vertices equals:

+10 XP

In an equilateral triangle inscribed inside a larger figure, each angle of the equilateral triangle measures:

+10 XP

Show Your Working · 3 questions

Show Your Working

9 marks total

Apply Easy 3 MARKS

Q6. Solve for the unknowns with reasons.
(a) Isosceles triangle: one base angle $= 75^{\circ}$. Find the apex.
(b) Triangle with angles $x^{\circ}, (x + 20)^{\circ}, (x + 40)^{\circ}$. Find $x$.
(c) Right-angled isosceles triangle: find each non-right angle.

Answer in your workbook.

Apply Medium 3 MARKS

Q7. A triangle has angles labelled $(3x + 10)^{\circ}, (2x + 20)^{\circ}$ and $(x + 30)^{\circ}$.
(a) Set up an equation.
(b) Solve for $x$.
(c) State the size of each angle and verify they sum to $180^{\circ}$.

Answer in your workbook.

Reason Hard 3 MARKS

Q8. Two triangles $PQR$ and $QRS$ share side $QR$, with $P, R, S$ collinear (in that order on a straight line). In triangle $PQR$, $\angle QPR = 50^{\circ}$ and $\angle PQR = 60^{\circ}$. In triangle $QRS$, $\angle QSR = 40^{\circ}$. Find $\angle PRQ$, then $\angle QRS$, then $\angle RQS$. Justify every step.

Answer in your workbook.

Comprehensive Answers

Quick Check

1. C — $50^{\circ}$. Both base angles $= 65^{\circ}$, so apex $= 180 - 130$.

2. A — $x = 30$. $x + 2x + 3x = 6x = 180$.

3. B — $40^{\circ}$. $(180 - 100) \div 2$.

4. D — $110^{\circ}$. Remote interiors $40 + 70 = 110$.

5. C — $60^{\circ}$. Equilateral, $180 \div 3$.

Show Your Working Model Answers

Q6 (3 marks): (a) Other base $= 75^{\circ}$ (base ∠s isos.). Apex $= 180 - 75 - 75 = 30^{\circ}$ (∠ sum of $\triangle$) [1]. (b) $x + (x + 20) + (x + 40) = 180 \Rightarrow 3x + 60 = 180 \Rightarrow x = 40$ [1]. (c) Right angle $= 90^{\circ}$. Other two equal, so each $= (180 - 90) \div 2 = 45^{\circ}$ (base ∠s isos.) [1].

Q7 (3 marks): (a) $(3x + 10) + (2x + 20) + (x + 30) = 180$ (∠ sum of $\triangle$) [1]. (b) $6x + 60 = 180 \Rightarrow 6x = 120 \Rightarrow x = 20$ [1]. (c) Angles: $3(20) + 10 = 70^{\circ}, 2(20) + 20 = 60^{\circ}, 20 + 30 = 50^{\circ}$. Check: $70 + 60 + 50 = 180^{\circ}$ ✓ [1].

Q8 (3 marks): $\angle PRQ = 180 - 50 - 60 = 70^{\circ}$ (∠ sum of $\triangle PQR$) [1]. $\angle QRS = 180 - 70 = 110^{\circ}$ (∠s on str. line, $P, R, S$ collinear) [1]. $\angle RQS = 180 - 110 - 40 = 30^{\circ}$ (∠ sum of $\triangle QRS$) [1].

Stretch Challenge · +25 XP, +10 coins

Truss Bridge Angles

A truss bridge has two triangles side by side, meeting at a vertical strut. The left triangle is isosceles with the bottom side as base; one base angle is $55^{\circ}$. The right triangle shares the strut as one of its sides and has a top-corner angle of $80^{\circ}$. The top vertices of the two triangles lie on a straight horizontal line. Find every angle in both triangles, justifying every step. (Use both isosceles and straight-line reasoning.)

Reveal solution

Left isosceles triangle: both base angles $= 55^{\circ}$ (base ∠s isos.); apex $= 180 - 110 = 70^{\circ}$ (∠ sum of $\triangle$). The apex sits on the straight horizontal top, so the angle the right triangle makes with that horizontal line at the shared apex $= 180 - 70 = 110^{\circ}$ (∠s on str. line). Right triangle: top corner $= 80^{\circ}$ (given). Angle at the shared apex (inside right triangle, between strut and horizontal) $= 180 - 110 = 70^{\circ}$ (∠s on str. line down the strut). Bottom angle $= 180 - 80 - 70 = 30^{\circ}$ (∠ sum of $\triangle$). Check each triangle sums to $180^{\circ}$ ✓

Quick Review

Four-step plan

KNOW → WANT → WHICH triangle → rule + reason.

Angle sum

$a + b + c = 180^{\circ}$ in every triangle.

Exterior angle

$e = a + b$ (remote interiors).

Isosceles

Base angles equal; apex $= 180 - 2(\text{base})$.

Algebra

Sum expressions, equate to $180^{\circ}$, solve, sub back.

Two triangles

Each gets its own equation; carry results forward.

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