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E6.1 Pythagoras’ theorem

Pythagoras' theorem links the three side lengths of a right-angled triangle. The hypotenuse is opposite the right angle and is always the longest side.

c^2=a^2+b^2

Identify the hypotenuse $c$ before substituting. To find $c$ , add the squares of the two shorter sides and take the positive square root. To find a shorter side, subtract the other shorter-side square from $c^2$ , then take the positive square root. Keep an exact surd when required and round only the final length.

For shorter sides $9.6$ cm and $18$ cm, $c=\sqrt{9.6^2+18^2}=20.4$ cm. If the hypotenuse is $125$ m and one shorter side is $100$ m, the other is $\sqrt{125^2-100^2}=75$ m. The first calculation adds because the hypotenuse is unknown; the second subtracts because a shorter side is unknown.

Use the theorem only when a right angle is known or established. Never subtract when finding the hypotenuse, and do not forget the square root. Conversely, if the square of the longest side equals the sum of the other two squares, the triangle is right-angled. Lengths take positive roots and use linear units.

E6.2 Right-angled triangles

Sine, cosine and tangent compare side lengths relative to one acute angle in a right-angled triangle. The hypotenuse is opposite the right angle; opposite and adjacent depend on the chosen angle.

Ratio	Relationship	Relevant sides
sine	$\sin\theta=\dfrac{O}{H}$	opposite and hypotenuse
cosine	$\cos\theta=\dfrac{A}{H}$	adjacent and hypotenuse
tangent	$\tan\theta=\dfrac{O}{A}$	opposite and adjacent

Mark the right angle, circle the reference angle and label O, A and H. Select the ratio containing the known and unknown sides. Rearrange it to find a side; use the matching inverse function to find an angle. Keep the calculator in degree mode and retain unrounded values until the final answer.

If $H=21.8$ cm and $\theta=56^\circ$ , then $O=21.8\sin56^\circ=18.1$ cm. If instead $O=11$ cm and $A=27$ cm, $\theta=\tan^{-1}(11/27)=22.2^\circ$ to one decimal place.

Adjacent is the non-hypotenuse side beside the chosen angle, so O and A can swap when the reference angle changes. Use inverse trig only for an unknown angle. These three ratios apply here to acute angles in right-angled triangles.

A two-dimensional problem may contain more than one right triangle. Treat each triangle as one step: a result from the first becomes a known value in the next.

Information in the current right triangle	Method
two side lengths	Pythagoras' theorem
one acute angle and one side	sine, cosine or tangent
two sides and an angle required	inverse sine, cosine or tangent

Sketch and label the diagram, mark the right angles, and identify which unknown must be found first. Work through connected right triangles in dependency order. Keep full calculator values between stages; rounding an intermediate side can move the final result.

Suppose a first right triangle has rise $6$ m and horizontal run $8$ m, so its diagonal is $\sqrt{6^2+8^2}=10$ m. If that diagonal is the adjacent side of a second right triangle with angle $35^\circ$ , the new opposite side is $10\tan35^\circ=7.00$ m. Each ratio uses sides from one triangle only.

Do not combine lengths from different triangles in one formula. Pythagoras requires a right angle, and the basic sine, cosine and tangent ratios require a right triangle. Non-right triangles need the later sine or cosine rule.

The shortest distance from a point to a line is the length of the perpendicular segment from the point to that line. Its foot meets the line at $90^\circ$ .

Draw or identify the perpendicular from the point to the line and mark its foot. This creates a right triangle. Choose sine, cosine, tangent or Pythagoras using only that triangle, then report the perpendicular length—not a sloping side or a distance to an endpoint.

\text{Area of triangle}=\frac12\times\text{base}\times\text{perpendicular height}

If a segment of length $80$ m meets the target line at an angle of $72^\circ$ , the perpendicular distance is opposite that angle: $d=80\sin72^\circ=76.1$ m. The same distance could be found from area when the triangle's area and base are known.

Shortest means perpendicular to the infinite line. The perpendicular foot may lie on an extension beyond a drawn segment. A sloping connector can never be shorter than the perpendicular distance.

An angle of elevation is measured upward from an observer's horizontal line to the line of sight. An angle of depression is measured downward from that horizontal line.

Draw a horizontal through the observer and a vertical height difference at the object. These form a right triangle with the line of sight. Use the height difference—not automatically the object's full height—and the horizontal separation in the appropriate trig ratio. Parallel horizontal lines make an angle of depression equal to the corresponding angle of elevation.

Two vertical poles are $33$ m apart. Their relevant points are $12.6$ m and $1.5$ m above level ground, so the vertical difference is $12.6-1.5=11.1$ m. The angle of elevation is $\tan^{-1}(11.1/33)=18.6^\circ$ to one decimal place.

If a bearing is supplied, measure it clockwise from north using three figures and use angle facts to locate the horizontal direction before forming the right triangle. Keep the calculator in degree mode and round a decimal angle to one decimal place unless instructed otherwise.

Elevation and depression are measured from a horizontal, not from a vertical pole or the line of sight itself. Subtract observer height when the observer is above ground, and keep intermediate lengths unrounded.

E6.3 Exact trigonometric values

Exact trigonometric values use fractions and square roots instead of rounded calculator decimals. They let an expression be simplified without losing accuracy.

$x$	$0^\circ$	$30^\circ$	$45^\circ$	$60^\circ$	$90^\circ$
$\sin x$	$0$	$\frac12$	$\frac{\sqrt2}{2}$	$\frac{\sqrt3}{2}$	$1$
$\cos x$	$1$	$\frac{\sqrt3}{2}$	$\frac{\sqrt2}{2}$	$\frac12$	$0$
$\tan x$	$0$	$\frac{\sqrt3}{3}$	$1$	$\sqrt3$	not required

For $0^\circ,30^\circ,45^\circ,60^\circ,90^\circ$ , the sine numerators follow $\sqrt0,\sqrt1,\sqrt2,\sqrt3,\sqrt4$ , all over $2$ . Cosine uses the same list in reverse. For the required tangent angles, use $\tan x=\sin x/\cos x$ and simplify exactly.

Simplify $\dfrac{4.5\cos30^\circ}{\sqrt3}-2$ . Substitute $\cos30^\circ=\dfrac{\sqrt3}{2}$ : $\dfrac{4.5(\sqrt3/2)}{\sqrt3}-2=2.25-2=0.25=\dfrac14$ . The $\sqrt3$ factors cancel before any decimal approximation is needed.

Do not replace exact fractions or surds with rounded decimals unless the question asks for an approximation. Since $\cos90^\circ=0$ , $\tan90^\circ=\sin90^\circ/\cos90^\circ$ would divide by zero; this is why $90^\circ$ is not in the required tangent list.

E6.4 Trigonometric functions

From $0^\circ$ to $360^\circ$ , sine and cosine complete one smooth cycle between $-1$ and $1$ , while tangent repeats every $180^\circ$ in separate increasing branches.

Graph	Zeros	Maxima / minima	Other defining features
$y=\sin x$	$0^\circ,180^\circ,360^\circ$	max $(90^\circ,1)$ ; min $(270^\circ,-1)$	period $360^\circ$
$y=\cos x$	$90^\circ,270^\circ$	max $(0^\circ,1)$ and $(360^\circ,1)$ ; min $(180^\circ,-1)$	period $360^\circ$
$y=\tan x$	$0^\circ,180^\circ,360^\circ$	no maximum or minimum	vertical asymptotes $x=90^\circ,270^\circ$ ; period $180^\circ$

Mark the axes and anchor points first. Join sine and cosine anchors with smooth curves, without sharp corners. For tangent, draw dashed vertical asymptotes and four separate branches; each branch rises from negative toward positive values and approaches but never touches or crosses an asymptote.

A graph solves $f(x)=k$ where the curve meets the horizontal line $y=k$ . It also shows sign: sine is positive from $0^\circ$ to $180^\circ$ ; cosine is positive before $90^\circ$ and after $270^\circ$ ; tangent is positive in the first and third quadrants.

Do not join tangent across an asymptote or mark $x=90^\circ$ and $270^\circ$ as points on its graph. Sine and cosine stay within $-1\leq y\leq1$ ; tangent is unbounded. All angles here are degrees, not radians.

A trigonometric equation can have more than one solution between $0^\circ$ and $360^\circ$ . First isolate the trig function, then use its sign and symmetry to find every angle in the interval.

Rearrange to $\sin x=k$ , $\cos x=k$ or $\tan x=k$ . 2. Check the calculator is in degree mode. 3. Find the reference angle $\beta$ from $|k|$ . 4. Use the function's sign to select the correct quadrants. 5. List every distinct solution in $0^\circ\leq x\leq360^\circ$ and substitute or compare with the graph to check.

Function	$k>0$	$k<0$
sine	$\beta,\ 180^\circ-\beta$	$180^\circ+\beta,\ 360^\circ-\beta$
cosine	$\beta,\ 360^\circ-\beta$	$180^\circ-\beta,\ 180^\circ+\beta$
tangent	$\beta,\ 180^\circ+\beta$	$180^\circ-\beta,\ 360^\circ-\beta$

Solve $3\sin x+1=0$ . Rearranging gives $\sin x=-\frac13$ . The reference angle is $\beta=\sin^{-1}(1/3)=19.47\ldots^\circ$ . Sine is negative in the third and fourth quadrants, so $x=180^\circ+\beta=199.5^\circ$ or $x=360^\circ-\beta=340.5^\circ$ to one decimal place.

For sine or cosine, no real solution exists when $|k|>1$ . Tangent can equal any real value. Endpoint solutions need care: $0^\circ$ and $360^\circ$ are distinct allowed inputs but may give the same trig value, so include each only when it satisfies the equation and interval.

E6.5 Non-right-angled triangles

The sine and cosine rules extend trigonometry to triangles without a right angle. Method choice depends on which sides and angles are known—not on the triangle's orientation.

Known information	Rule and useful form
a side and its opposite angle, plus another side or angle	sine rule: $\dfrac{a}{\sin A}=\dfrac{b}{\sin B}=\dfrac{c}{\sin C}$
two sides and their included angle; find the third side	cosine rule: $c^2=a^2+b^2-2ab\cos C$
all three sides; find an angle	cosine rule: $\cos C=\dfrac{a^2+b^2-c^2}{2ab}$

Label each side opposite its matching capital angle. For the sine rule, use two complete opposite pairs and rearrange. For the cosine rule, make the required side or angle the $c,C$ pair; $C$ must be the angle between sides $a$ and $b$ . Keep unrounded values until the final answer.

With sides $6.4$ cm and $10.9$ cm enclosing $38^\circ$ , the third side is $c=\sqrt{6.4^2+10.9^2-2(6.4)(10.9)\cos38^\circ}=7.06$ cm. If instead $A=50^\circ$ , $B=100^\circ$ and $b=12$ cm, then $a=12\sin50^\circ/\sin100^\circ=9.33$ cm.

The longest side must face the largest angle, which is a useful check. When inverse sine gives an angle, its supplement has the same sine; use the stated geometry and angle sum to decide whether an acute or obtuse value is valid. Cosine rule resolves an SSS angle directly.

Two sides and their included angle determine a triangle's area because one side contributes the perpendicular height $b\sin C$ to the other side used as the base.

K=\frac12ab\sin C

Choose two known sides $a$ and $b$ and use the angle $C$ between them. Substitute consistently and attach square units. For a reverse problem, rearrange to $\sin C=2K/(ab)$ , find the acute calculator angle, then check its supplement $180^\circ-C$ because both angles have the same sine.

For sides $8$ cm and $9$ cm with included angle $50^\circ$ , $K=\frac12(8)(9)\sin50^\circ=27.6$ cm $^2$ . If sides $10$ cm and $14$ cm enclose an unknown angle and $K=45$ cm $^2$ , then $\sin C=90/140$ . This gives $C=40.0^\circ$ or $140.0^\circ$ : an acute/obtuse ambiguous pair.

The angle must be included between the two sides used in the formula. Since $\sin C=\sin(180^\circ-C)$ , area alone may not determine a unique angle. Reject any candidate that conflicts with other given lengths, angles or the triangle angle sum.

E6.6 Pythagoras’ theorem and trigonometry

A three-dimensional problem is solved by locating one or more two-dimensional right triangles inside the solid. A line's angle with a plane is the angle between the line and its perpendicular projection onto that plane.

Mark the target line and the relevant plane. 2. Drop the line's endpoint perpendicularly to the plane; join the other endpoint to this foot to form the projection. 3. Find the projection length within the plane, often by Pythagoras. 4. Use the projection, perpendicular height and target line as one right triangle. 5. Apply Pythagoras or the appropriate trig ratio and keep intermediate values unrounded.

d=\sqrt{l^2+w^2+h^2}

A cuboid is $20$ cm long, $5.5$ cm wide and has volume $495$ cm $^3$ , so its height is $495/(20\times5.5)=4.5$ cm. The projection of the space diagonal onto the base is $\sqrt{20^2+5.5^2}$ cm. Therefore $\tan\theta=4.5/\sqrt{20^2+5.5^2}$ , giving the line-base angle $\theta=12.2^\circ$ .

Do not use an arbitrary visible edge as the projection: the projection must lie in the named plane and connect to the perpendicular foot. The requested line-plane angle is the smaller angle in this right triangle, not the complementary angle with the vertical. Use consistent linear units and round only the final result.

6. Trigonometry

Exam analysis

Most tested topics

E6.1 Pythagoras’ theorem

Use Pythagoras' theorem

E6.2 Right-angled triangles

Choose sine, cosine or tangent

Sequence two-dimensional right-triangle problems

Find the shortest distance from a point to a line

Calculate angles of elevation and depression

E6.3 Exact trigonometric values

Know the exact trigonometric values

E6.4 Trigonometric functions

Recognise and sketch trigonometric graphs

Solve trigonometric equations from 0° to 360°

E6.5 Non-right-angled triangles

Choose the sine rule or cosine rule

Use the sine area formula and handle ambiguity

E6.6 Pythagoras’ theorem and trigonometry

Solve three-dimensional Pythagoras and trigonometry problems