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\[\text{Geometry of Triangles}\]
Triangles are everywhere in mathematics and the real world. In this lesson we explore the angle properties of triangles — including the angle sum, exterior angles, and special triangles — and then apply the Theorem of Pythagoras to solve problems. Work through each step carefully.
\[\hat{A} + \hat{B} + \hat{C} = 180°\]
The angles of any triangle always add up to 180°. This is called the angle sum of a triangle. The reason: if you extend the base of a triangle and draw a parallel line through the apex, the three angles form a straight line (180°).
\[\hat{A} = 50°,\ \hat{B} = 70°,\ \hat{C} = ?\]
Find the missing angle: \(\hat{C} = 180° - 50° - 70° = 60°\). Always subtract the two known angles from 180°.
\[x + 2x + 3x = 180°\]
If the three angles of a triangle are \(x\), \(2x\) and \(3x\): combine like terms → \(6x = 180°\) → \(x = 30°\). The angles are 30°, 60° and 90°. This is a right-angled triangle!
\[\hat{A} + \hat{B} + \hat{C} = 180°\]
Key fact: the three interior angles of any triangle sum to 180°. To find a missing angle, subtract the other two from 180°. This works for all triangles — scalene, isosceles and equilateral.
\[\text{exterior angle} = \hat{A} + \hat{B}\]
An exterior angle of a triangle equals the sum of the two remote interior angles (the two angles NOT adjacent to it). This follows directly from the angle sum: if interior angles sum to 180° and the exterior + adjacent = 180°, then exterior = the sum of the other two.
\[\text{ext} = 130°,\ \hat{A} = 70°,\ \hat{B} = ?\]
The exterior angle at C is 130°. One remote interior angle is 70°. Find the other: \(\hat{B} = 130° - 70° = 60°\). Check: \(70° + 60° = 130°\)
\[y + (y - 20°) = 130°\]
Exam question: The exterior angle at A is 130°. The two remote interior angles are \(y\) and \((y - 20°)\). Find \(x\) and \(y\).
Step 1: \(x = 180° - 130° = 50°\) (angles on a straight line).
Step 2: \(y + (y - 20°) = 130°\) → \(2y = 150°\) → \(y = 75°\).
Check: \(75° + 55° + 50° = 180°\)
\[\text{exterior angle of equilateral} = 120°\]
Each angle of an equilateral triangle is 60°. So each exterior angle = 180° − 60° = 120°. The three exterior angles of any triangle always sum to 360°.
\[\text{ext} = \hat{A} + \hat{B}\]
The exterior angle of a triangle equals the sum of the two remote interior angles. Use this rule whenever you see an exterior angle in a question — set up an equation and solve.
\[AB = AC \Rightarrow \hat{B} = \hat{C}\]
An isosceles triangle has two equal sides. The angles opposite those equal sides (the base angles) are also equal. This is one of the most useful properties in geometry.
\[\hat{P} = 40°,\ \hat{Q} = \hat{R} = ?\]
In \(\triangle PQR\), \(PQ = PR\), so \(\hat{Q} = \hat{R}\). The apex angle \(\hat{P} = 40°\). Find the base angles: \(\hat{Q} = \hat{R} = \dfrac{180° - 40°}{2} = 70°\).
\[\hat{B} = 55°,\ AB = AC \Rightarrow \hat{A} = ?\]
In \(\triangle ABC\), \(AB = AC\), so base angles are equal: \(\hat{B} = \hat{C} = 55°\). Find \(\hat{A}\): \(\hat{A} = 180° - 55° - 55° = 70°\).
\[AB = AC \Rightarrow \hat{B} = \hat{C}\]
In an isosceles triangle, equal sides are opposite equal angles. To find a missing angle: identify which sides are equal, set the base angles equal, then use the angle sum. Always state the reason: "base angles of isosceles triangle".
\[AB = BC = CA \Rightarrow \hat{A} = \hat{B} = \hat{C} = 60°\]
An equilateral triangle has all three sides equal. Since all three angles must also be equal and sum to 180°, each angle = 60°. Equilateral triangles are also isosceles (a special case).
\[(2y)° = 60° \Rightarrow y = 30°\]
If one angle of an equilateral triangle is \((2y)°\), then \(2y = 60\), so \(y = 30\). All algebraic equilateral questions use the same idea: each angle equals 60°.
\[\hat{A} = \hat{B} = \hat{C} = 60°\]
All sides and all angles of an equilateral triangle are equal. Each angle is exactly 60°. Each exterior angle is 120°. Reason to state: "equilateral triangle, all angles equal 60°".
\[c^2 = a^2 + b^2\]
In a right-angled triangle, the square of the hypotenuse (the side opposite the 90° angle) equals the sum of the squares of the other two sides. The hypotenuse is always the longest side.
\[c^2 = 3^2 + 4^2 = 9 + 16 = 25\]
Find the hypotenuse when the legs are 3 and 4: \(c^2 = 9 + 16 = 25\), so \(c = \sqrt{25} = 5\). The 3-4-5 triangle is the most common Pythagorean triple — memorise it!
\[a^2 = c^2 - b^2\]
To find a shorter side (leg), rearrange: \(a^2 = c^2 - b^2\). Example: hypotenuse = 13, one leg = 5. Then \(a^2 = 169 - 25 = 144\), so \(a = 12\). Always subtract the known leg squared from the hypotenuse squared.
\[a^2 + b^2 = c^2 \Rightarrow \hat{C} = 90°\]
The converse: if the square of the longest side equals the sum of the squares of the other two sides, then the triangle is right-angled. Test: 3, 4, 5 → \(3^2 + 4^2 = 9 + 16 = 25 = 5^2\). Test: 2, 4, 6 → \(4 + 16 = 20 \neq 36\).
\[c^2 = a^2 + b^2\]
In a right-angled triangle: find the hypotenuse by adding the squares of the legs. Find a leg by subtracting. Common triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25. Always identify which side is the hypotenuse first.
\[d^2 = 6^2 + 8^2 = 100 \Rightarrow d = 10\]
A ladder 10 m long leans against a wall with its base 6 m away. How high does it reach? The ladder is the hypotenuse: \(h^2 = 10^2 - 6^2 = 100 - 36 = 64\), so \(h = 8\) m. Draw a sketch first — identify which measurement is the hypotenuse.
\[PQ = QR = 15,\ PS = 9,\ PS \perp QR\]
Exam question: \(PQ = QR = 15\) cm (isosceles). \(PS \perp QR\), \(PS = 9\) cm. Find \(PR\).
In \(\triangle PQS\): \(QS^2 = 15^2 - 9^2 = 225 - 81 = 144\), so \(QS = 12\) cm.
\(SR = QR - QS = 15 - 12 = 3\) cm.
In \(\triangle PSR\): \(PR^2 = 9^2 + 3^2 = 81 + 9 = 90\), so \(PR = \sqrt{90} = 3\sqrt{10}\) cm.
\[\text{Draw} \rightarrow \text{Label} \rightarrow \text{Apply}\]
For complex Pythagoras problems: (1) Draw and label the diagram. (2) Identify every right angle — there may be more than one triangle to work through. (3) Apply \(c^2 = a^2 + b^2\) step by step. In isosceles triangles, the perpendicular from the apex always bisects the base.
\[\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}\]
When the answer is a surd, simplify by finding the largest perfect square factor. \(\sqrt{90} = \sqrt{9 \times 10} = 3\sqrt{10}\). Always leave surds in simplest form unless told to give a decimal.
\[PR = 3\sqrt{10}\ \text{cm}\]
In multi-step problems, break the diagram into individual right-angled triangles and apply Pythagoras to each one. Key insight: a perpendicular from an apex to the base of an isosceles triangle creates two congruent right-angled triangles. Always simplify your final surd answer.
\[\hat{A}+\hat{B}+\hat{C}=180° \quad c^2=a^2+b^2\]
What you have learned:
1. Angle sum: all three angles of a triangle sum to 180°.
2. Exterior angle: equals the sum of the two remote interior angles.
3. Isosceles: two equal sides → two equal base angles.
4. Equilateral: all sides equal → each angle = 60°.
5. Pythagoras: \(c^2 = a^2 + b^2\) in any right-angled triangle.
6. Apply: identify right angles, work through each triangle step by step.