Finding Unknown Angles
To find a missing angle of a right-angled triangle, pick the ratio that links the two sides you know, then use the inverse key on a calculator, for example \(\theta=\tan^{-1}\left(\dfrac{5}{8}\right)\). In NSW Year 10 Advanced maths angles are rounded to the nearest degree or minute, and the triangle is often hidden inside a bigger shape (NESA outcome MA5‑TRG‑C‑01).
You will change decimal degrees to degrees and minutes, solve equations such as \(3\cos\theta+1=2.5\) and \(4\sin\theta=7\cos\theta\), and find the angle a line makes with the \(x\)-axis from its gradient. Practical problems include planks, stairs, ramps, solar panels and ladders, and you will find angles inside rectangles, squares and trapeziums, or on a sports field, by adding or subtracting two angles. It is part of Year 10 Advanced Trigonometry.
Every question with a fully worked solution.
- Finding Unknown Angles - Video - Trigonometry: finding angles in right-angled triangles Watch
Theory
To find an unknown angle in a right-angled triangle, write the ratio of the two known sides, then use the inverse key: \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\). This Year 10 Advanced maths page covers answers in degrees and minutes, the angle a line makes with the \(x\)-axis, equations such as \(a\sin\theta = b\cos\theta\), and angles that you must build by halving, adding or subtracting.
If you know two sides of a right-angled triangle, you know a trigonometric ratio of each acute angle. The inverse ratio turns that ratio back into the angle: if \(\tan\theta = 0.75\) then \(\theta = \tan^{-1}(0.75)\). The \(-1\) is not a power. It means "the angle whose tan is".
Angles are measured in degrees and minutes, with \(1^\circ = 60'\). To change a decimal answer, multiply the decimal part by \(60\): \(24.35^\circ = 24^\circ\) and \(0.35 \times 60 = 21'\), so \(24^\circ 21'\).
The angle of inclination of a line is the angle it makes with the positive direction of the \(x\)-axis. Its gradient is rise over run, so \(\tan\theta = m\).
In any right-angled triangle the smallest angle is opposite the shortest side, and the two acute angles add to \(90^\circ\).
Choose the ratio from the two sides you know:
How to find an unknown angle
- Find a right-angled triangle that contains the angle. Draw a height, diagonal or perpendicular if you need one.
- Label the two known sides from the angle: opposite, adjacent or hypotenuse.
- Write the ratio, e.g. \(\cos\theta = \dfrac{A}{H}\), and use the matching inverse key in degree mode.
- Round once, at the end, to the nearest degree or minute as asked.
When the angle is not in a simple triangle
- Halve by symmetry: in an isosceles triangle, rhombus or rectangle, a line of symmetry halves the angle. Find the half, then double it.
- Isosceles trapezium: the overhang at each end is half the difference of the parallel sides. That and the height give the base angle.
- Add or subtract: an angle between two lines of sight, or between two lines from the origin, is the difference of two angles you can find (Example 3). Angles on a straight line add to \(180^\circ\).
- Algebraic sides: if a ratio is given and the sides contain \(x\), set up the ratio as an equation, e.g. \(\dfrac{x}{x + 2} = 0.6\), solve for \(x\), then find any other angle.
- Missing side first: use Pythagoras' theorem, or an area, to find a side, then find the angle. A staircase angle comes from one step: rise over going.
- Checks: compare with a given range ("is the ladder safe?") and check the two acute angles add to \(90^\circ\).
The string is the hypotenuse and the height is opposite \(\theta\). Opposite and hypotenuse means sine.
| \(\sin\theta\) | \(=\) | \(\dfrac{28}{45}\) |
| \(\theta\) | \(=\) | \(\sin^{-1}\left(\dfrac{28}{45}\right)\) |
| \(\) | \(=\) | \(38.478\ldots^\circ\) |
Change the decimal part to minutes:
| \(0.478\ldots \times 60\) | \(=\) | \(28.7\ldots'\) |
| \(\therefore\ \theta\) | \(\approx\) | \(\mathbf{38^\circ 29'}\) |
On a calculator, the degrees-minutes key converts the display for you.
(i) \(7\sin\theta = 2\cos\theta\)
(ii) \(5\tan\theta = 8\sin\theta\)
(i) Divide both sides by \(7\cos\theta\). For an acute angle \(\cos\theta \neq 0\), so this is allowed. Then use \(\dfrac{\sin\theta}{\cos\theta} = \tan\theta\).
| \(\dfrac{\sin\theta}{\cos\theta}\) | \(=\) | \(\dfrac{2}{7}\) |
| \(\tan\theta\) | \(=\) | \(\dfrac{2}{7}\) |
| \(\theta\) | \(=\) | \(15.945\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{15^\circ 57'}\) |
(ii) Replace \(\tan\theta\) by \(\dfrac{\sin\theta}{\cos\theta}\). An acute angle has \(\sin\theta \gt 0\), so you may divide both sides by \(\sin\theta\).
| \(\dfrac{5\sin\theta}{\cos\theta}\) | \(=\) | \(8\sin\theta\) |
| \(\dfrac{5}{\cos\theta}\) | \(=\) | \(8\) |
| \(\cos\theta\) | \(=\) | \(\dfrac{5}{8}\) |
| \(\theta\) | \(=\) | \(51.317\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{51^\circ 19'}\) |
No triangle has \(\alpha\) in it with two known sides. Find the angle up to the top edge and the angle up to the bottom edge, then subtract.
The top edge is \(4 + 6 = 10\) m above eye level.
| \(\tan(\alpha + \beta)\) | \(=\) | \(\dfrac{10}{15}\) |
| \(\alpha + \beta\) | \(=\) | \(33.690\ldots^\circ\) |
| \(\tan\beta\) | \(=\) | \(\dfrac{4}{15}\) |
| \(\beta\) | \(=\) | \(14.931\ldots^\circ\) |
| \(\alpha\) | \(=\) | \(33.690\ldots^\circ - 14.931\ldots^\circ\) |
| \(\) | \(=\) | \(18.758\ldots^\circ\) |
| \(\therefore\ \alpha\) | \(\approx\) | \(\mathbf{18^\circ 46'}\) |
Subtract the full calculator values and round once, at the end.
The diagonals of a rhombus cross at right angles, halve each other and halve the angles. Triangle \(AOB\) is right-angled at \(O\) with \(BO = 8\) cm and hypotenuse \(17\) cm.
| \(AO\) | \(=\) | \(\sqrt{17^2 - 8^2} = 15\text{ cm}\) |
| \(\tan x\) | \(=\) | \(\dfrac{8}{15}\) |
| \(x\) | \(=\) | \(28.072\ldots^\circ\) |
| \(\angle DAB = 2x\) | \(=\) | \(56.144\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{56^\circ 9'}\) |
Co-interior angles in a rhombus add to \(180^\circ\), so the other angle is
| \(\angle ABC\) | \(=\) | \(180^\circ - 56.144\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{123^\circ 51'}\) |
The angles are \(56^\circ 9'\), \(123^\circ 51'\), \(56^\circ 9'\) and \(123^\circ 51'\). Round the doubled angle, not the half.
Common pitfalls
Frequently asked questions
How do you find an angle using trigonometry?
Label the two known sides from the angle, write the matching ratio (sine, cosine or tangent), then use the inverse key on the calculator, such as inverse tan of opposite over adjacent.
How do you convert decimal degrees to degrees and minutes?
Keep the whole number of degrees and multiply the decimal part by \(60\) to get minutes. For example \(24.35^\circ\) is \(24^\circ\) and \(0.35 \times 60 = 21'\), so \(24^\circ 21'\).
What does sin to the power of minus 1 mean?
\(\sin^{-1}\) is the inverse sine, read as "the angle whose sine is". It undoes sine and gives an angle. It does not mean \(\dfrac{1}{\sin\theta}\).
How do you find the angle a line makes with the x-axis?
Find the gradient \(m\) of the line, then \(\theta = \tan^{-1}(m)\), because the gradient is rise over run, which is opposite over adjacent.
How do you solve a sin x = b cos x?
Divide both sides by \(a\cos x\), which is not zero for an acute angle. Since \(\dfrac{\sin x}{\cos x} = \tan x\), you get \(\tan x = \dfrac{b}{a}\), then use \(\tan^{-1}\).
Which angle is the smallest in a right-angled triangle?
The smallest angle is opposite the shortest side. If only the hypotenuse and one side are given, find the third side with Pythagoras' theorem first to see which side is shortest.