Find an acute angle
Use the two given lengths to form a ratio for the requested angle. Apply the matching inverse in degree mode
- Choose from the two given sides Name the sides from the requested angle, then use the ratio containing both lengths \(\displaystyle \begin{aligned}\sin\theta&=\frac{\text{opposite}}{\text{hypotenuse}}\\\cos\theta&=\frac{\text{adjacent}}{\text{hypotenuse}}\\\tan\theta&=\frac{\text{opposite}}{\text{adjacent}}\end{aligned}\)
- Use the matching inverse Enter the fraction directly with the matching inverse key in degree mode. Keep the calculator value until the final rounding
- Check the angle The result must be acute. A tangent ratio greater than one is valid and gives the larger acute angle. The two acute angles together make a right angle \(\displaystyle \alpha+\beta=90^{\circ}\)
Worked example
The opposite length and hypotenuse of a right triangle are given for the acute angle. Find the angle to one decimal place in degree mode
- \(\displaystyle \sin\theta=\frac{18.7}{31.4}\)Sine uses the opposite side and hypotenuse
- \(\displaystyle \theta=\sin^{-1}\left(\frac{18.7}{31.4}\right)\)Use the inverse key in degree mode. Enter the original fraction directly
- \(\displaystyle \theta=36.551236\ldots^{\circ}\)Keep the calculator result before rounding
- \(\displaystyle \theta\approx36.6^{\circ}\)Round once to one decimal place. The result is acute
Common mistakes
Difficulty levels
Questions about this topic
Should I round the side quotient first?+
No. Enter the original fraction directly or keep the full calculator value. Round only the requested angle
Can a tangent ratio exceed one?+
Yes. It means the opposite leg is longer than the adjacent leg. The angle is still acute
How can I find the other acute angle?+
Subtract the unrounded first angle from a right angle and round the requested result. You can also swap the side roles and calculate the other angle directly
Can I use a calculator?+
A calculator is allowed unless a question is marked no calculator
Which year level and chapter is this?+
Year 10 Mathematics, chapter 9: Pythagoras and trigonometry