Trig Problems
Trig Problems mixes everything in the topic, so the first job is finding the right-angled triangle hidden in each situation. This NSW Year 9 Advanced maths subtopic applies the trig ratios to practical and geometric problems (NESA outcome MA5‑TRG‑C‑01).
You will find the depth of water under an anchored boat, the angle of a slide and of a ramp with a gradient of \(1\) in \(14\), and the height of a roof truss. You will find exact values such as \(\tan 60^\circ = \sqrt{3}\), the diagonals of a rhombus, a chord of a circle and the area of a regular pentagon, and show that a triangle's area is \(\frac{1}{2}ab\sin C\). It is part of Year 9 Advanced Trigonometry.
Theory
Trigonometry problems ask you to find the right-angled triangle inside a real situation or another shape, then solve it. This Year 9 Advanced maths page covers sides and angles in context, gradients, exact values, and hidden right angles in rhombuses, regular polygons, circles, trapeziums and triangle areas.
Every problem on this page comes down to a right-angled triangle. Label its sides from the angle you are using, choose sine, cosine or tangent, and solve for a side or, with an inverse ratio such as \(\sin^{-1}\), for an angle.
In context, decide which length is the hypotenuse: a ladder, rope, cable, slide, slant height or sloping face is usually the hypotenuse; heights and depths are vertical; distances along the ground are horizontal.
Watch where the angle is measured from. An anchor rope at \(31^\circ\) to the vertical has the depth adjacent to the angle, not opposite it.
A gradient of \(1\) in \(n\) means a rise of \(1\) for every \(n\) horizontally, so \(\tan\theta = \dfrac{1}{n}\).
Many shapes hide right angles: the height of an isosceles triangle, the diagonals of a rhombus, the line from a circle's centre to the middle of a chord, and the line from the centre of a regular polygon to the middle of a side.
Exact values for \(30^\circ\), \(45^\circ\) and \(60^\circ\) come from two special triangles: half an equilateral triangle, and an isosceles right-angled triangle with equal sides \(1\) and hypotenuse \(\sqrt{2}\).
| Shape | Where the right angle is | What gets halved |
|---|---|---|
| Isosceles triangle | height from the apex to the base | apex angle and base |
| Rhombus | where the diagonals cross | both diagonals and the angles |
| Circle and chord | centre to the midpoint of the chord | chord and the angle at the centre |
| Regular \(n\)-gon | centre to the midpoint of a side | side and the angle \(\dfrac{360^\circ}{n}\) |
| Cone | axis meets the base radius | vertical angle and diameter |
| Trapezium | a perpendicular dropped between the parallel sides | nothing; it splits off a rectangle |
How to tackle a trig problem
- Draw a diagram and mark every given length and angle.
- Find the right angle: vertical meets horizontal, or use the shape facts in the table on the Formulas tab.
- Halve if needed: in isosceles triangles, rhombuses, chords, cones and regular polygons, use half the angle and half the length.
- Label and choose a ratio; use \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\) for an angle.
- Finish the question: double a half-length, add a rectangle part, multiply for an area or volume. Keep full values until the end.
- Sense-check: no side can be longer than the hypotenuse, and an angle in a right-angled triangle is less than \(90^\circ\).
Particular situations
- Angle to the vertical: the vertical side is adjacent to the angle.
- Gradient: \(\tan\theta\) \(= \dfrac{\text{rise}}{\text{run}}\). A distance along the slope is the hypotenuse (Example 3).
- Rectangle and diagonal: find a missing side from the area or perimeter first, then use tangent.
- Two positions: for two ramps to the same height, find each run and subtract.
- Across a river: the width is opposite the angle at a point on the bank; a second point further along gives a second triangle with the same width.
- Exact values: draw half an equilateral triangle (sides \(1\), \(\sqrt{3}\), \(2\)) or an isosceles right-angled triangle (sides \(1\), \(1\), \(\sqrt{2}\)) and read off the ratio (Example 1).
All angles of an equilateral triangle are \(60^\circ\). A line from the top to the middle of the base is perpendicular to it, halves the base (\(1\)) and halves the top angle (\(30^\circ\)).
Find the height with Pythagoras' theorem:
| \(h^2\) | \(=\) | \(2^2 - 1^2 = 3\) |
| \(h\) | \(=\) | \(\sqrt{3}\) |
Now read the ratios from the right-angled triangle.
| \(\sin 60^\circ\) | \(=\) | \(\dfrac{\text{opp}}{\text{hyp}} = \boldsymbol{\dfrac{\sqrt{3}}{2}}\) |
| \(\tan 30^\circ\) | \(=\) | \(\dfrac{\text{opp}}{\text{adj}} = \boldsymbol{\dfrac{1}{\sqrt{3}}}\) |
Check on a calculator: \(\dfrac{\sqrt{3}}{2}\) \(\approx 0.866\) \(= \sin 60^\circ\). The opposite side changes with the angle you stand at: from \(30^\circ\) it is \(1\), from \(60^\circ\) it is \(\sqrt{3}\).
(i) the distance \(a\) from the centre to the middle of a side
(ii) the area of the octagon.
Joining the centre to every vertex makes \(8\) identical isosceles triangles. Each has an angle of \(\dfrac{360^\circ}{8} = 45^\circ\) at the centre. The height halves it to \(22.5^\circ\) and halves the side to \(3\) cm.
(i) \(3\) cm is opposite \(22.5^\circ\) and \(a\) is adjacent.
| \(\tan 22.5^\circ\) | \(=\) | \(\dfrac{3}{a}\) |
| \(a\) | \(=\) | \(\dfrac{3}{\tan 22.5^\circ} = 7.2426\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{7.2}\text{ cm}\) |
(ii) Area of one triangle \(= \dfrac{1}{2} \times 6 \times a\). There are \(8\) of them.
| \(A\) | \(=\) | \(8 \times \dfrac{1}{2} \times 6 \times 7.2426\ldots\) |
| \(\) | \(=\) | \(173.8233\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{173.8}\text{ cm}^2\) |
Use the unrounded \(a\) in part (ii); \(7.2\) would give \(172.8\).
(i) Find the angle it makes with the horizontal, to the nearest minute.
(ii) The driveway is \(26\) m long. How high does it rise, to one decimal place?
(i) \(1\) in \(8\) means a rise of \(1\) for a horizontal run of \(8\): opposite over adjacent.
| \(\tan\theta\) | \(=\) | \(\dfrac{1}{8}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\left(\dfrac{1}{8}\right) = 7.1250\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{7^\circ 8'}\) |
(ii) The \(26\) m is along the slope, so it is the hypotenuse. Use the unrounded angle.
| \(\text{rise}\) | \(=\) | \(26\sin\theta\) |
| \(\) | \(=\) | \(3.2249\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{3.2}\text{ m}\) |
The \(26\) m is not the run: the run is the horizontal distance, which is a little shorter.
Area \(= \dfrac{1}{2} \times \text{base} \times \text{height}\), but the height is not given. Draw it: it makes a right-angled triangle with the \(7.5\) m side as hypotenuse.
| \(h\) | \(=\) | \(7.5\sin 52^\circ = 5.9100\ldots\) |
| \(A\) | \(=\) | \(\dfrac{1}{2} \times 11 \times 5.9100\ldots\) |
| \(\) | \(=\) | \(32.5054\ldots\) |
| \(\) | \(\approx\) | \(\mathbf{32.5}\text{ m}^2\) |
In general, for sides \(a\) and \(b\) with angle \(C\) between them, \(h = a\sin C\), so \(A = \dfrac{1}{2}ab\sin C\).
Common pitfalls
Frequently asked questions
How do you know which trig ratio to use in a word problem?
Draw the right-angled triangle, mark the angle, then name the two sides involved. Opposite and hypotenuse: sine. Adjacent and hypotenuse: cosine. Opposite and adjacent: tangent.
How do you find an angle with trigonometry?
Write the ratio of the two known sides, then use the inverse function, such as \(\theta\) \(= \tan^{-1}\left(\dfrac{O}{A}\right)\). Make sure the calculator is in degree mode.
How do you turn a gradient into an angle?
A gradient of \(1\) in \(n\) is a rise of \(1\) over a run of \(n\), so \(\theta\) \(= \tan^{-1}\left(\dfrac{1}{n}\right)\).
What are exact values in trigonometry?
They are the ratios for \(30^\circ\), \(45^\circ\) and \(60^\circ\) written with whole numbers and surds instead of decimals. They come from half an equilateral triangle and from an isosceles right-angled triangle.
How do you use trigonometry when there is no right angle?
Look for one: draw a height, a diagonal, or a line from a centre to the middle of a side or chord. Symmetric shapes split into identical right-angled triangles.