Finding an Angle
Every question with a fully worked solution.
- Finding an Angle - Video - Trigonometry: finding angles in right-angled triangles Watch
Theory
When two sides of a right-angled triangle are known, the angle can be found with an inverse trigonometric ratio. This Year 9 maths page shows how to choose sin, cos or tan from the two known sides, use \(\sin^{-1}\), \(\cos^{-1}\) or \(\tan^{-1}\), and give the answer to the nearest degree or minute.
To find an angle in a right-angled triangle, you need two sides. Label them from the unknown angle as opposite, adjacent or hypotenuse.
The two sides make one of the ratios: sin, cos or tan. Write that ratio as a fraction.
The inverse keys \(\sin^{-1}\), \(\cos^{-1}\) and \(\tan^{-1}\) do the reverse of sin, cos and tan: they turn a ratio into an angle.
Angles can be rounded to the nearest degree or the nearest minute. One degree is \(60\) minutes, so the decimal part of the angle is multiplied by \(60\).
Undo the ratio with its inverse:
Decimal degrees to minutes:
How to find an unknown angle
- Label the two known sides from the unknown angle.
- Choose the ratio those two sides make: SOH, CAH or TOA.
- Write the ratio as a fraction of the two lengths.
- Apply the inverse key, in degree mode.
- Round to a degree or a minute, and check the angle is between \(0^\circ\) and \(90^\circ\).
Opposite and hypotenuse, so the ratio is sin. The angle is wanted, so apply \(\sin^{-1}\):
| \(\sin \theta\) | \(=\) | \(\dfrac{6}{11}\) |
| \(\theta\) | \(=\) | \(\sin^{-1}\left(\dfrac{6}{11}\right)\) |
| \(\) | \(=\) | \(33.05\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{33^\circ}\) |
The right angle is at \(B\), so \(AC\) is the hypotenuse. From \(A\), \(AB\) runs to the right angle, so it is adjacent. Use cos:
| \(\cos \angle BAC\) | \(=\) | \(\dfrac{15}{19}\) |
| \(\angle BAC\) | \(=\) | \(\cos^{-1}\left(\dfrac{15}{19}\right)\) |
| \(\) | \(=\) | \(37.86\ldots^\circ\) |
| \(\) | \(\approx\) | \(\mathbf{38^\circ}\) |
(i) to the nearest degree and
(ii) to the nearest minute.
No hypotenuse, so use tan:
| \(\tan \theta\) | \(=\) | \(\dfrac{5.2}{8.7}\) |
| \(\theta\) | \(=\) | \(\tan^{-1}\left(\dfrac{5.2}{8.7}\right)\) |
| \(\) | \(=\) | \(30.8668\ldots^\circ\) |
(i) \(\theta \approx \mathbf{31^\circ}\). (ii) Keep the \(30^\circ\) and turn the decimal part into minutes:
| \(0.8668 \times 60\) | \(=\) | \(52.00\ldots\) |
| \(\theta\) | \(\approx\) | \(\mathbf{30^\circ 52'}\) |
The degrees-minutes key on the calculator does the same conversion.
(i) the angle \(\theta\) the ladder makes with the ground and
(ii) the angle \(\alpha\) it makes with the wall.
(i) From \(\theta\) at the foot, the height \(6.8\) is opposite and the ladder \(7.5\) is the hypotenuse:
| \(\sin \theta\) | \(=\) | \(\dfrac{6.8}{7.5}\) |
| \(\theta\) | \(=\) | \(65.0486\ldots^\circ\) |
| \(0.0486\ldots \times 60\) | \(=\) | \(2.92\ldots\) |
| \(\theta\) | \(\approx\) | \(\mathbf{65^\circ 3'}\) |
(ii) The two acute angles add to \(90^\circ\), so no calculator is needed:
| \(\alpha\) | \(=\) | \(90^\circ - 65^\circ 3'\) |
| \(\) | \(=\) | \(\mathbf{24^\circ 57'}\) |
Check: the larger angle, \(\theta\), faces the longer side, \(6.8\) m.
Common pitfalls
Frequently asked questions
How do you find an angle in a right-angled triangle?
Label two known sides from the angle, choose sin, cos or tan, write the ratio as a fraction, then use the inverse key on the calculator.
What does sin to the minus 1 mean?
It is the inverse sine. It takes a sine ratio and gives back the angle. It does not mean \(1\) divided by sine.
How do you round an angle to the nearest minute?
Multiply the decimal part of the angle by \(60\) to get minutes, then round. \(30.87^\circ\) is \(30^\circ 52'\).
How do you find the other angle in a right-angled triangle?
Subtract the angle you found from \(90^\circ\), because the two acute angles add to \(90^\circ\).