Choose a ratio using the sides relative to the stated angle.
Core & Extended · Syllabus C6.2, E6.2
Notes and key methods
For an acute angle θ in a right triangle, sin θ=opposite/hypotenuse, cos θ=adjacent/hypotenuse, and tan θ=opposite/adjacent. The adjacent and opposite sides depend on which acute angle is named.
Use degree mode. An inverse trigonometric function recovers an angle from its ratio. Keep full precision during a calculation and round only the requested final value. Every calculator question here states its own precision.
Check the conditions
Use the stated units and precision.
Show the method as well as the final answer.
Three worked examples
Example 1 · Right-triangle trigonometry in degrees
A right triangle has hypotenuse 15 cm and acute angle \(\displaystyle \theta =58\)°. “Opposite” and “adjacent” refer to θ; the adjacent side is not the hypotenuse.
Find the opposite length in cm, to 3 significant figures.
Find the adjacent length in cm, to 3 significant figures.
Find the triangle area in \(\displaystyle \mathrm{cm}^{2}\), to 3 significant figures.
In a different right triangle, \(\displaystyle \sin \phi =\)\(\displaystyle \frac{5}{12}\) for acute φ. Find φ in degrees, to 1 decimal place.
Show the reasoning
(a)
Multiply the hypotenuse by sin θ in degree mode, then round the final result.
\(\displaystyle 15 \sin{\left(\frac{29 \pi}{90} \right)}\) rounds to \(12.7\) cm.
Answer: \(12.7\).
(b)
Multiply the hypotenuse by cos θ in degree mode.
\(\displaystyle 15 \cos{\left(\frac{29 \pi}{90} \right)}\) rounds to \(7.95\) cm.
Answer: \(7.95\).
(c)
Use both unrounded perpendicular legs before the final rounding.
Half the product of the two unrounded legs rounds to \(50.6\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(50.6\).
(d)
Apply inverse sine in degree mode to the supplied exact ratio.
The acute inverse-sine angle rounds to \(24.6\)°.
Answer: \(24.6\).
Example 2 · Right-triangle trigonometry in degrees
A right triangle has hypotenuse 16 cm and acute angle \(\displaystyle \theta =59\)°. “Opposite” and “adjacent” refer to θ; the adjacent side is not the hypotenuse.
Find the opposite length in cm, to 3 significant figures.
Find the adjacent length in cm, to 3 significant figures.
Find the triangle area in \(\displaystyle \mathrm{cm}^{2}\), to 3 significant figures.
In a different right triangle, \(\displaystyle \sin \phi =\)\(\displaystyle \frac{6}{13}\) for acute φ. Find φ in degrees, to 1 decimal place.
Show the reasoning
(a)
Multiply the hypotenuse by sin θ in degree mode, then round the final result.
\(\displaystyle 16 \sin{\left(\frac{59 \pi}{180} \right)}\) rounds to \(13.7\) cm.
Answer: \(13.7\).
(b)
Multiply the hypotenuse by cos θ in degree mode.
\(\displaystyle 16 \cos{\left(\frac{59 \pi}{180} \right)}\) rounds to \(8.24\) cm.
Answer: \(8.24\).
(c)
Use both unrounded perpendicular legs before the final rounding.
Half the product of the two unrounded legs rounds to \(56.5\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(56.5\).
(d)
Apply inverse sine in degree mode to the supplied exact ratio.
The acute inverse-sine angle rounds to \(27.5\)°.
Answer: \(27.5\).
Example 3 · Right-triangle trigonometry in degrees
A right triangle has hypotenuse 17 cm and acute angle \(\displaystyle \theta =25\)°. “Opposite” and “adjacent” refer to θ; the adjacent side is not the hypotenuse.
Find the opposite length in cm, to 3 significant figures.
Find the adjacent length in cm, to 3 significant figures.
Find the triangle area in \(\displaystyle \mathrm{cm}^{2}\), to 3 significant figures.
In a different right triangle, \(\displaystyle \sin \phi =\)\(\displaystyle \frac{1}{2}\) for acute φ. Find φ in degrees, to 1 decimal place.
Show the reasoning
(a)
Multiply the hypotenuse by sin θ in degree mode, then round the final result.
\(\displaystyle 17 \sin{\left(\frac{5 \pi}{36} \right)}\) rounds to \(7.18\) cm.
Answer: \(7.18\).
(b)
Multiply the hypotenuse by cos θ in degree mode.
\(\displaystyle 17 \cos{\left(\frac{5 \pi}{36} \right)}\) rounds to \(15.4\) cm.
Answer: \(15.4\).
(c)
Use both unrounded perpendicular legs before the final rounding.
Half the product of the two unrounded legs rounds to \(55.3\) \(\displaystyle \mathrm{cm}^{2}\).
Answer: \(55.3\).
(d)
Apply inverse sine in degree mode to the supplied exact ratio.
The acute inverse-sine angle rounds to \(30.0\)°.
Answer: \(30.0\).
Common mistakes
Keep intermediate values exact until the final step.