Geometry and Trigonometry | Leaving Cert

Geometry and Trigonometry

Geometry and trigonometry form a significant part of the Leaving Certificate syllabus, particularly Paper 2. This topic covers coordinate geometry, trigonometric functions, identities, and geometric Theorems and proofs.

Coordinate Geometry

Distance Formula (OL/HL)

The distance between two points $A(x_1, y_1)$ and $B(x_2, y_2)$ :

D = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

This is the Pythagorean theorem applied to the horizontal and vertical displacements.

Midpoint Formula (OL/HL)

The midpoint $M$ of $AB$ :

M = \left(\frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2}\right)

Slope (Gradient) (OL/HL)

The slope of the line through $A(x_1, y_1)$ and $B(x_2, y_2)$ :

M = \frac{y_2 - y_1}{x_2 - x_1}

A vertical line has undefined slope. A horizontal line has slope $0$ .

Equation of a Line (OL/HL)

Point-slope form:

Y - y_1 = m(x - x_1)

General form:

Ax + by + c = 0

Slope-intercept form:

Y = mx + c

Where $m$ is the slope and $c$ is the $y$ -intercept.

Perpendicular and Parallel Lines (OL/HL)

Parallel lines have equal slopes: $m_1 = m_2$ .
Perpendicular lines: $m_1 \cdot m_2 = -1$ .

Proof of the perpendicular condition. If two lines with slopes $m_1$ and $m_2$ are Perpendicular, then the angle between them is $90^\circ$ . Using the tangent addition formula: $\tan(\alpha + \beta) = \frac{m_1 + m_2}{1 - m_1 m_2}$ . Setting $\alpha + \beta = 90^\circ$ : $\tan 90^\circ$ Is undefined, so $1 - m_1 m_2 = 0$ Giving $m_1 m_2 = -1$ .

Example (OL): Find the equation of the line through $(1, 3)$ perpendicular to $y = 2x + 1$ .

The slope of the given line is $m_1 = 2$ So $m_2 = -\frac{1}{2}$ .

Y - 3 = -\frac{1}{2}(x - 1) \implies y = -\frac{1}{2}x + \frac{7}{2}

Example (HL): Find the equation of the perpendicular bisector of the segment joining $A(2, 5)$ And $B(6, 1)$ .

Midpoint: $M = \left(\frac{2+6}{2}, \frac{5+1}{2}\right) = (4, 3)$ .

Slope of $AB$ : $m = \frac{1 - 5}{6 - 2} = -1$ .

Slope of perpendicular bisector: $m_{\perp} = 1$ .

Equation: $y - 3 = 1(x - 4) \implies y = x - 1$ .

Distance from a Point to a Line (HL)

The perpendicular distance from $(x_0, y_0)$ to $ax + by + c = 0$ :

D = \frac{|ax_0 + by_0 + c|}{\sqrt{a^2 + b^2}}

Example (HL): Find the distance from $(3, 2)$ to $2x + y - 5 = 0$ .

D = \frac{|6 + 2 - 5|}{\sqrt{4 + 1}} = \frac{3}{\sqrt{5}} = \frac{3\sqrt{5}}{5}

Proof sketch. Let $P(x_0, y_0)$ be the point and $ax + by + c = 0$ the line. The closest point $Q$ on the line to $P$ lies along the perpendicular. The line through $P$ perpendicular to $ax + by + c = 0$ has equation $b(x - x_0) - a(y - y_0) = 0$ . Solving the two equations Simultaneously gives $Q$ And the distance $PQ$ simplifies to the formula above.

Area of a Triangle (HL)

The area of a triangle with vertices $(x_1, y_1)$$(x_2, y_2)$$(x_3, y_3)$ :

A = \frac{1}{2}|x_1(y_2 - y_3) + x_2(y_3 - y_1) + x_3(y_1 - y_2)|

This is derived from the shoelace formula (also known as Gauss”s area formula).

Example (HL): Find the area of the triangle with vertices $(1, 2)$$(4, 6)$$(3, -1)$ .

A = \frac{1}{2}|1(6 - (-1)) + 4((-1) - 2) + 3(2 - 6)| = \frac{1}{2}|7 - 12 - 12| = \frac{1}{2}|-17| = 8.5

Intersection of Two Lines (OL/HL)

Two lines $a_1 x + b_1 y + c_1 = 0$ and $a_2 x + b_2 y + c_2 = 0$ :

If $\frac{a_1}{a_2} \neq \frac{b_1}{b_2}$ : lines intersect at a unique point.
If $\frac{a_1}{a_2} = \frac{b_1}{b_2} \neq \frac{c_1}{c_2}$ : lines are parallel (no intersection).
If $\frac{a_1}{a_2} = \frac{b_1}{b_2} = \frac{c_1}{c_2}$ : lines are coincident (infinitely many intersections).

The Circle

Equation of a Circle (OL/HL)

Centre-radius form: $(x - h)^2 + (y - k)^2 = r^2$ with centre $(h, k)$ and radius $r$ .

General form: $x^2 + y^2 + 2gx + 2fy + c = 0$ with centre $(-g, -f)$ and radius $\sqrt{g^2 + f^2 - c}$ .

The circle exists only if $g^2 + f^2 - c > 0$ .

Example (OL): Find the centre and radius of $x^2 + y^2 - 4x + 6y - 3 = 0$ .

Completing the square:

(x - 2)^2 - 4 + (y + 3)^2 - 9 - 3 = 0 \implies (x - 2)^2 + (y + 3)^2 = 16

Centre $(2, -3)$ Radius $4$ .

Tangent to a Circle (HL)

The tangent at a point $(x_1, y_1)$ on the circle $x^2 + y^2 = r^2$ has equation:

X_1 x + y_1 y = r^2

Proof. The radius to $(x_1, y_1)$ has slope $y_1/x_1$ . The tangent is perpendicular, so its Slope is $-x_1/y_1$ . Using point-slope form: $y - y_1 = -\frac{x_1}{y_1}(x - x_1)$ Which simplifies To $x_1 x + y_1 y = x_1^2 + y_1^2 = r^2$ .

Example (HL): Find the equation of the tangent to $x^2 + y^2 = 25$ at the point $(3, 4)$ .

3x + 4y = 25

Example (HL): Show that the line $3x - 4y + 25 = 0$ is a tangent to $x^2 + y^2 = 25$ .

Substitute $y = \frac{3x + 25}{4}$ into $x^2 + y^2 = 25$ :

X^2 + \left(\frac{3x + 25}{4}\right)^2 = 25

16x^2 + 9x^2 + 150x + 625 = 400

25x^2 + 150x + 225 = 0

X^2 + 6x + 9 = 0 \implies (x + 3)^2 = 0

The discriminant is $\Delta = 0$ Confirming a tangent. The point of tangency is $x = -3$ $y = \frac{-9 + 25}{4} = 4$ .

Intersection of Line and Circle

Substitute the line into the circle equation. The discriminant of the resulting quadratic tells you:

$\Delta > 0$ : two intersection points
$\Delta = 0$ : tangent (one intersection point)
$\Delta < 0$ : no intersection

Circle Through Three Points (HL)

Example (HL): Find the equation of the circle through $(0, 0)$$(4, 0)$ And $(0, 4)$ .

Let the circle be $x^2 + y^2 + 2gx + 2fy + c = 0$ .

Substituting $(0,0)$ : $c = 0$ .

Substituting $(4,0)$ : $16 + 8g = 0 \implies g = -2$ .

Substituting $(0,4)$ : $16 + 8f = 0 \implies f = -2$ .

The circle is $x^2 + y^2 - 4x - 4y = 0$ With centre $(2, 2)$ and radius $\sqrt{4+4} = 2\sqrt{2}$ .

Trigonometry

Trigonometric Ratios (OL/HL)

For a right-angled triangle with angle $\theta$ :

Ratio	Definition
$\sin\theta$	$\frac{\mathrm{opposite}{\mathrm{hypotenuse}$
$\cos\theta$	$\frac{\mathrm{adjacent}{\mathrm{hypotenuse}$
$\tan\theta$	$\frac{\mathrm{opposite}{\mathrm{adjacent}$

The Unit Circle (OL/HL)

On the unit circle, a point at angle $\theta$ has coordinates $(\cos\theta, \sin\theta)$ .

Key values:

$\theta$	$0$	$\frac{\pi}{6}$	$\frac{\pi}{4}$	$\frac{\pi}{3}$	$\frac{\pi}{2}$
$\sin\theta$	$0$	$\frac{1}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{\sqrt{3}}{2}$	$1$
$\cos\theta$	$1$	$\frac{\sqrt{3}}{2}$	$\frac{\sqrt{2}}{2}$	$\frac{1}{2}$	$0$
$\tan\theta$	$0$	$\frac{1}{\sqrt{3}}$	$1$	$\sqrt{3}$	undefined

Trigonometric Identities (OL/HL)

Pythagorean identities:

\sin^2\theta + \cos^2\theta = 1

1 + \tan^2\theta = \sec^2\theta

1 + \cot^2\theta = \csc^2\theta

Proof of the second identity. Divide $\sin^2\theta + \cos^2\theta = 1$ by $\cos^2\theta$ : $\tan^2\theta + 1 = \sec^2\theta$ .

Example (HL): Given $\sin\theta = \frac{3}{5}$ and $\theta$ is in the second quadrant, find $\cos\theta$ and $\tan\theta$ .

\cos^2\theta = 1 - \sin^2\theta = 1 - \frac{9}{25} = \frac{16}{25}

Since $\theta$ is in the second quadrant, $\cos\theta < 0$ So $\cos\theta = -\frac{4}{5}$ .

\tan\theta = \frac{\sin\theta}{\cos\theta} = \frac{3/5}{-4/5} = -\frac{3}{4}

Compound Angle Formulae (HL)

\sin(A \pm B) = \sin A \cos B \pm \cos A \sin B

\cos(A \pm B) = \cos A \cos B \mp \sin A \sin B

\tan(A \pm B) = \frac{\tan A \pm \tan B}{1 \mp \tan A \tan B}

Example (HL): Find the exact value of $\sin 75^\circ$ .

\sin 75° = \sin(45° + 30°) = \sin 45°\cos 30° + \cos 45°\sin 30°

= \frac{\sqrt{2}}{2} \cdot \frac{\sqrt{3}}{2} + \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{6} + \sqrt{2}}{4}

Example (HL): Find the exact value of $\tan 15^\circ$ .

\tan 15° = \tan(45° - 30°) = \frac{1 - \frac{1}{\sqrt{3}}}{1 + \frac{1}{\sqrt{3}}} = \frac{\sqrt{3} - 1}{\sqrt{3} + 1} = \frac{(\sqrt{3} - 1)^2}{3 - 1} = \frac{4 - 2\sqrt{3}}{2} = 2 - \sqrt{3}

Double Angle Formulae (HL)

\sin 2A = 2\sin A \cos A

\cos 2A = \cos^2 A - \sin^2 A = 2\cos^2 A - 1 = 1 - 2\sin^2 A

\tan 2A = \frac{2\tan A}{1 - \tan^2 A}

Proof of $\cos 2A = 2\cos^2 A - 1$ . Using the compound angle formula:

\cos(A + A) = \cos A \cos A - \sin A \sin A = \cos^2 A - \sin^2 A

Since $\sin^2 A = 1 - \cos^2 A$ :

\cos 2A = \cos^2 A - (1 - \cos^2 A) = 2\cos^2 A - 1

Triple Angle Formulae (HL)

Proof that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ :

\sin 3\theta = \sin(2\theta + \theta) = \sin 2\theta \cos\theta + \cos 2\theta \sin\theta

= 2\sin\theta\cos^2\theta + (1 - 2\sin^2\theta)\sin\theta

= 2\sin\theta(1 - \sin^2\theta) + \sin\theta - 2\sin^3\theta

= 2\sin\theta - 2\sin^3\theta + \sin\theta - 2\sin^3\theta = 3\sin\theta - 4\sin^3\theta

Similarly, $\cos 3\theta = 4\cos^3\theta - 3\cos\theta$ :

\cos 3\theta = \cos(2\theta + \theta) = \cos 2\theta \cos\theta - \sin 2\theta \sin\theta

= (2\cos^2\theta - 1)\cos\theta - 2\sin^2\theta\cos\theta

= 2\cos^3\theta - \cos\theta - 2(1 - \cos^2\theta)\cos\theta

= 2\cos^3\theta - \cos\theta - 2\cos\theta + 2\cos^3\theta = 4\cos^3\theta - 3\cos\theta

Factor Formulae (HL)

\sin A + \sin B = 2\sin\frac{A+B}{2}\cos\frac{A-B}{2}

\sin A - \sin B = 2\cos\frac{A+B}{2}\sin\frac{A-B}{2}

\cos A + \cos B = 2\cos\frac{A+B}{2}\cos\frac{A-B}{2}

\cos A - \cos B = -2\sin\frac{A+B}{2}\sin\frac{A-B}{2}

Example (HL): Evaluate $\sin 75° - \sin 15^\circ$ .

\sin 75° - \sin 15° = 2\cos\frac{90°}{2}\sin\frac{60°}{2} = 2\cos 45°\sin 30° = 2 \cdot \frac{\sqrt{2}}{2} \cdot \frac{1}{2} = \frac{\sqrt{2}}{2}

Solving Trigonometric Equations (OL/HL)

Example (OL): Solve $\sin\theta = \frac{1}{2}$ for $0 \leq \theta \leq 2\pi$ .

\theta = \frac{\pi}{6}, \frac{5\pi}{6}

Example (HL): Solve $2\cos^2\theta + 3\cos\theta - 2 = 0$ for $0 \leq \theta \leq 2\pi$ .

Let $u = \cos\theta$ : $2u^2 + 3u - 2 = 0 \implies (2u - 1)(u + 2) = 0$ .

$u = \frac{1}{2}$ or $u = -2$ (rejected since $|\cos\theta| \leq 1$ ).

$\cos\theta = \frac{1}{2} \implies \theta = \frac{\pi}{3}, \frac{5\pi}{3}$ .

Example (HL): Solve $\sin 2\theta = \sin \theta$ for $0 \le \theta \lt 2\pi$ .

2\sin\theta\cos\theta = \sin\theta

\sin\theta(2\cos\theta - 1) = 0

$\sin\theta = 0$ : $\theta = 0, \pi$ .

$2\cos\theta - 1 = 0$ : $\cos\theta = \frac{1}{2}$ So $\theta = \frac{\pi}{3}, \frac{5\pi}{3}$ .

Solutions: $0, \frac{\pi}{3}, \pi, \frac{5\pi}{3}$ .

Functions can be zero. If they can, you lose solutions. Instead, factorise.

Example (HL): Solve $2\sin^2 x + 3\cos x - 3 = 0$ for $0 \le x \le 2\pi$ .

Replace $\sin^2 x = 1 - \cos^2 x$ :

2(1 - \cos^2 x) + 3\cos x - 3 = 0

-2\cos^2 x + 3\cos x - 1 = 0

2\cos^2 x - 3\cos x + 1 = 0

(2\cos x - 1)(\cos x - 1) = 0

$\cos x = \frac{1}{2}$ : $x = \frac{\pi}{3}, \frac{5\pi}{3}$ .

$\cos x = 1$ : $x = 0$ .

The Sine Rule (OL/HL)

\frac{a}{\sin A} = \frac{b}{\sin B} = \frac{c}{\sin C}

Use when you know: two sides and a non-included angle, or two angles and one side.

Ambiguous case (HL): When given two sides and a non-included angle, there may be two solutions Or none. If $a > b$ and $A$ is acute, there is exactly one solution. If $a < b$ and $A$ is acute, There may be two solutions (the “ambiguous case”).

Example (HL) — Ambiguous case: In $\triangle ABC$$a = 8$$b = 10$$A = 40^\circ$ . Find all Possible values of $B$ .

By the sine rule: $\sin B = \frac{b \sin A}{a} = \frac{10 \sin 40°}{8} = \frac{10 \times 0.6428}{8} = 0.8035$ .

$B = \arcsin(0.8035) \approx 53.5^\circ$ or $B \approx 180° - 53.5° = 126.5^\circ$ .

Check: $A + B = 40° + 126.5° = 166.5° < 180^\circ$ So both solutions are valid.

The Cosine Rule (OL/HL)

A^2 = b^2 + c^2 - 2bc\cos A

Example (OL): In triangle $\triangle ABC$$a = 7$$b = 5$$c = 8$ . Find angle $A$ .

49 = 25 + 64 - 80\cos A \implies \cos A = \frac{40}{80} = \frac{1}{2} \implies A = 60°

Area of a Triangle Using Trigonometry (OL/HL)

A = \frac{1}{2}ab\sin C

Proof. Drop altitude $h$ from $B$ to side $b$ . Then $h = a\sin C$ So $A = \frac{1}{2} \times b \times h = \frac{1}{2}ab\sin C$ .

Example (HL): In $\triangle ABC$$a = 8$$b = 6$ And $C = 50^\circ$ . Find the area.

A = \frac{1}{2}(8)(6)\sin 50° = 24 \times 0.766 = 18.39 \mathrm{ square units

R-Addition Formula (HL)

An expression of the form $a\sin\theta + b\cos\theta$ can be written as $R\sin(\theta + \alpha)$ Where $R = \sqrt{a^2 + b^2}$ and $\alpha = \arctan\frac{b}{a}$ .

Example (HL): Express $3\sin\theta - 4\cos\theta$ in the form $R\sin(\theta + \alpha)$ .

R = \sqrt{9 + 16} = 5

3\sin\theta - 4\cos\theta = 5\sin(\theta + \alpha)

Where $\tan\alpha = \frac{-4}{3}$ So $\alpha = \arctan(-4/3) \approx -53.1^\circ$ .

Application — finding maximum value: The maximum of $R\sin(\theta + \alpha)$ is $R$ and the Minimum is $-R$ . So the maximum of $3\sin\theta - 4\cos\theta$ is $5$ and the minimum is $-5$ .

Example (HL): Find the maximum and minimum of $5\sin\theta + 12\cos\theta$ .

R = \sqrt{25 + 144} = 13

So $5\sin\theta + 12\cos\theta = 13\sin(\theta + \alpha)$ where $\tan\alpha = 12/5$ .

Maximum $= 13$ Minimum $= -13$ .

Vectors (HL)

Vector Operations

For vectors $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j}$ and $\mathbf{b} = b_1\mathbf{i} + b_2\mathbf{j}$ :

Scalar (dot) product:

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 = |\mathbf{a}||\mathbf{b}|\cos\theta

Magnitude:

|\mathbf{a}| = \sqrt{a_1^2 + a_2^2}

3D Vectors (HL)

For $\mathbf{a} = a_1\mathbf{i} + a_2\mathbf{j} + a_3\mathbf{k}$ :

\mathbf{a} \cdot \mathbf{b} = a_1 b_1 + a_2 b_2 + a_3 b_3

Cross product (HL):

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \end{vmatrix}

$|\mathbf{a} \times \mathbf{b}| = |\mathbf{a}||\mathbf{b}|\sin\theta$ gives the area of the Parallelogram spanned by $\mathbf{a}$ and $\mathbf{b}$ .

Example (HL): Given $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ Find $\mathbf{a} \times \mathbf{b}$ and the Angle between them.

\mathbf{a} \times \mathbf{b} = \begin{vmatrix} \mathbf{i} & \mathbf{j} & \mathbf{k} \\ 2 & -1 & 3 \\ 1 & 2 & -1 \end{vmatrix} = \mathbf{i}(1 - 6) - \mathbf{j}(-2 - 3) + \mathbf{k}(4 + 1) = -5\mathbf{i} + 5\mathbf{j} + 5\mathbf{k}

$|\mathbf{a}| = \sqrt{4 + 1 + 9} = \sqrt{14}$$|\mathbf{b}| = \sqrt{1 + 4 + 1} = \sqrt{6}$ .

$\mathbf{a} \cdot \mathbf{b} = 2 - 2 - 3 = -3$ .

$\cos\theta = \frac{-3}{\sqrt{14}\sqrt{6}} = \frac{-3}{2\sqrt{21}}$ .

Scalar Triple Product (HL)

The scalar triple product $\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c})$ gives the volume of the Parallelepiped spanned by $\mathbf{a}$$\mathbf{b}$ And $\mathbf{c}$ .

\mathbf{a} \cdot (\mathbf{b} \times \mathbf{c}) = \begin{vmatrix} a_1 & a_2 & a_3 \\ b_1 & b_2 & b_3 \\ c_1 & c_2 & c_3 \end{vmatrix}

If the scalar triple product is zero, the three vectors are coplanar.

Example (HL): Determine whether the vectors $\mathbf{a} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ $\mathbf{b} = 3\mathbf{i} - \mathbf{j} + 2\mathbf{k}$ $\mathbf{c} = 2\mathbf{i} + 3\mathbf{j} + \mathbf{k}$ are coplanar.

\begin{vmatrix} 1 & 2 & -1 \\ 3 & -1 & 2 \\ 2 & 3 & 1 \end{vmatrix} = 1(-1 - 6) - 2(3 - 4) + (-1)(9 - (-2)) = -7 + 2 - 11 = -16

Since the scalar triple product is $-16 \neq 0$ The vectors are not coplanar.

Area of a Triangle Using Vectors (HL)

The area of triangle $\triangle ABC$ with position vectors $\mathbf{a}$$\mathbf{b}$$\mathbf{c}$ :

\mathrm{Area = \frac{1}{2}|\overrightarrow{AB} \times \overrightarrow{AC}|

Where $\overrightarrow{AB} = \mathbf{b} - \mathbf{a}$ and $\overrightarrow{AC} = \mathbf{c} - \mathbf{a}$ .

Geometric Proofs (HL)

Theorem: Angles in a Triangle

The sum of the interior angles of a triangle is $180^\circ$ .

Proof: Let $\triangle ABC$ have vertices $A$$B$$C$ . Draw a line through $A$ parallel to $BC$ . Then $\angle B = \angle BAX$ (alternate angles) and $\angle C = \angle CAY$ (alternate Angles). Since $BAX$ and $CAY$ together with $\angle A$ form a straight line:

\angle A + \angle BAX + \angle CAY = 180°

\angle A + \angle B + \angle C = 180°

Theorem: Pythagoras’ Theorem

In a right-angled triangle, $a^2 + b^2 = c^2$ .

Proof (using similar triangles): Let $\triangle ABC$ be right-angled at $C$ With altitude $CD$ To the hypotenuse $AB$ . Then $\triangle ABC \sim \triangle ACD \sim \triangle CBD$ . From $\triangle ABC \sim \triangle ACD$ : $\frac{AC}{AB} = \frac{AD}{AC}$ Giving $AC^2 = AB \cdot AD$ . From $\triangle ABC \sim \triangle CBD$ : $\frac{BC}{AB} = \frac{BD}{BC}$ Giving $BC^2 = AB \cdot BD$ . Adding:

AC^2 + BC^2 = AB \cdot AD + AB \cdot BD = AB(AD + BD) = AB^2

Theorem: Angle at the Centre

The angle at the centre of a circle is twice the angle at the circumference subtended by the same Arc.

Proof. Let $O$ be the centre and $A, B$ points on the circumference. Join $OA$ and $OB$ . If $C$ Is on the circumference on the same side of $AB$ as $O$ Then $\triangle OAC$ is isosceles with $OA = OC$ So $\angle OAC = \angle OCA$ . Similarly $\angle OBC = \angle OCB$ . The exterior angle of $\triangle OAC$ at $O$ equals $\angle AOC = 2\angle OAC$ . The full angle $AOB = 2\angle OAC + 2\angle OCB = 2\angle ACB$ .

Theorem: Angle in a Semicircle

The angle in a semicircle is a right angle.

Proof. If $AB$ is the diameter and $C$ is on the circumference, then the angle at the centre $AOB = 180^\circ$ . By the angle-at-centre theorem, the angle at the circumference $ACB = 90^\circ$ .

Theorem: Tangent-Radius Property

The tangent to a circle at a point is perpendicular to the radius at that point.

Proof (by contradiction). Suppose the tangent at $P$ is not perpendicular to the radius $OP$ . Then the perpendicular from $O$ to the tangent meets it at some point $Q \neq P$ . Since $OQ \lt OP$ (by the shortest distance property), $Q$ is closer to $O$ than $P$ . But $P$ lies on the circle and $Q$ is outside the perpendicular from the centre, so $Q$ must be outside the circle. If $Q$ is Outside the circle, the line through $P$ and $Q$ (the tangent) must cross the circle at $P$ and some Other point, contradicting that it is a tangent. Hence the tangent is perpendicular to the radius.

Intuition

Coordinate geometry is like a map with a mathematical grid — every point has an address (x, y), every line has an equation, and every shape can be described algebraically. Trigonometry connects angles to distances: the sine, cosine, and tangent ratios let you calculate the height of a building from its shadow or the distance across a river from the opposite bank. The unit circle is the master key that unlocks trigonometric functions for any angle, revealing their periodic, wave-like nature. Geometric proofs are like legal arguments: you start with what you know (axioms and theorems) and build step by step to what you want to prove.

Worked Examples

See the examples integrated throughout the sections above.

Common Pitfalls

Degrees vs radians — the Leaving Certificate uses radians unless stated otherwise. Always check.
CAST diagram — remember All, Sine, Tan, Cos for determining the sign of trig functions in each quadrant.
Compound angle formulae — the signs in $\cos(A + B)$ and $\cos(A - B)$ are swapped compared to $\sin$ .
Vector cross product is not commutative: $\mathbf{a} \times \mathbf{b} = -\mathbf{b} \times \mathbf{a}$ .
Distance from a point to a line — the absolute value in the numerator is essential.
Ambiguous case of the sine rule — always check whether a second solution exists.
Completing the square for circles — remember to add the constants to both sides.
Dividing by trig functions in equations — you may lose solutions. Factorise instead.
R-addition formula — be careful with the sign of $\alpha$ . If $a$ is negative, the reference angle calculation needs adjustment.
Circle general form — the centre is $(-g, -f)$ Not $(g, f)$ . The negative signs are a common source of error.

Practice Questions

Ordinary Level

Find the equation of the line through $(2, -1)$ and $(4, 5)$ .
Find the centre and radius of $x^2 + y^2 + 6x - 2y + 6 = 0$ .
Solve $2\sin\theta = 1$ for $0 \leq \theta \leq 360^\circ$ .
In $\triangle ABC$$a = 10$$b = 7$$C = 45^\circ$ . Find $c$ using the cosine rule.
Prove that $\sin^2\theta + \cos^2\theta = 1$ .
Find the area of $\triangle ABC$ where $a = 8$$b = 5$ And $C = 60^\circ$ .
Find the midpoint and length of the segment joining $(-2, 3)$ and $(4, -1)$ .

Higher Level

Prove that $\sin 3\theta = 3\sin\theta - 4\sin^3\theta$ using compound angle formulae.
Find the shortest distance from $(1, -2)$ to the line $3x - 4y + 5 = 0$ .
Solve $\cos 2\theta = \cos\theta$ for $0 \leq \theta \leq 2\pi$ .
Find the area of the triangle with vertices $(1, 2)$$(4, 6)$$(3, -1)$ .
Given $\mathbf{a} = 2\mathbf{i} - \mathbf{j} + 3\mathbf{k}$ and $\mathbf{b} = \mathbf{i} + 2\mathbf{j} - \mathbf{k}$ Find $\mathbf{a} \times \mathbf{b}$ and the angle between $\mathbf{a}$ and $\mathbf{b}$ .
Express $\cos 3\theta$ in terms of $\cos\theta$ .
Prove that $\sin(A+B)\sin(A-B) = \sin^2 A - \sin^2 B$ .
Find the equation of the tangent to the circle $x^2 + y^2 - 4x + 6y + 9 = 0$ at the point $(1, -1)$ .
Two ships leave a port. Ship A sails on a bearing of $030^\circ$ at 20 km/h. Ship B sails on a bearing of $110^\circ$ at 15 km/h. Find the distance between them after 3 hours.
Prove that the angle at the centre of a circle is twice the angle at the circumference.
Express $4\sin\theta + 3\cos\theta$ in the form $R\sin(\theta + \alpha)$ and hence find its maximum value.
Find the area of the triangle with vertices at the points with position vectors $\mathbf{i} + 2\mathbf{j} + 3\mathbf{k}$$2\mathbf{i} - \mathbf{j} + \mathbf{k}$ And $3\mathbf{i} + \mathbf{j} - 2\mathbf{k}$ .
Solve $2\cos^2 x + \sin x = 2$ for $0 \le x \le 2\pi$ .
Find the equation of the circle passing through $(1, 0)$$(0, 1)$ And $(2, 3)$ .
Determine whether the vectors $\mathbf{i} + \mathbf{j} + \mathbf{k}$ $2\mathbf{i} - \mathbf{j} + \mathbf{k}$ And $3\mathbf{i} + 4\mathbf{k}$ are coplanar.

flowchart TD
    A[3_Geometry Trig] --> B[Key Concepts]
    A --> C[Core Principles]
    A --> D[Practical Applications]
    B --> E[Fundamental definitions]
    C --> F[Design patterns]
    D --> G[Real-world usage]

Summary

This topic covers the mathematical techniques and concepts related to geometry and trigonometry, including key theorems, methods, and problem-solving approaches.

Key concepts include:

sine, cosine, and tangent functions
trigonometric identities
solving trigonometric equations
the sine and cosine rules
radian measure and arc length

Regular practice with a variety of question types is essential to build fluency and confidence in applying these mathematical techniques.

Geometry and Trigonometry | Leaving Cert