Finding an Angle from a Ratio

Two sides known, the angle wanted

Solving a right triangle so far has started from an angle and found a side. This time it runs the other way: two sides are known and an angle is not.

Here the two shorter sides are 3 and 4, and $\theta$θ is the angle across from the side of 3.

A right triangle with sides of 3 and 4 at the right angle. The angle $\theta$θ across from the side of 3 is unknown.

The ratio is known

Name the sides from $\theta$θ: 3 is the opposite side and 4 is the adjacent side. Opposite and adjacent make the tangent, so $\tan \theta =\frac{3}{4}=0.75$tan θ = 3/4 = 0.75.

That is a fact about $\theta$θ: its tangent is 0.75. The question is now which angle has a tangent of 0.75.

Running the ratio backwards

The tangent takes an angle and gives a ratio. The inverse tangent, written ${\tan }^{-1}$tan⁻¹, goes the other way: it takes a ratio and gives the angle with that tangent. So ${\tan }^{-1}0.75$tan⁻¹ 0.75 means the angle whose tangent is 0.75, and $\theta ={\tan }^{-1}0.75$θ = tan⁻¹ 0.75.

On a calculator the inverse tangent is usually the tan key pressed after a key marked SHIFT or 2nd. With the calculator in degrees, ${\tan }^{-1}0.75=36.8698{\dots }^{\circ }$tan⁻¹ 0.75 = 36.8698…°, so $\theta =36.9^{\circ }$θ = 36.9° to one decimal place.

Check by putting the angle back: tan 36.87° = 0.7500, which is $\frac{3}{4}$3/4.

The inverse sine, ${\sin }^{-1}$sin⁻¹, and the inverse cosine, ${\cos }^{-1}$cos⁻¹, work in the same way: ${\sin }^{-1}x$sin⁻¹ x is the angle whose sine is x, and ${\cos }^{-1}x$cos⁻¹ x is the angle whose cosine is x. The −1 is not a power here. ${\sin }^{-1}x$sin⁻¹ x is an angle, and it is not $1/\sin x$1 / sin x.

Choosing ${\sin }^{-1}$sin⁻¹, ${\cos }^{-1}$cos⁻¹ or ${\tan }^{-1}$tan⁻¹

The choice is the same as for finding a side: the two sides you know decide the ratio. Opposite and hypotenuse give the sine, adjacent and hypotenuse give the cosine, and opposite and adjacent give the tangent.

A right triangle has a hypotenuse of 10, and the side opposite $\theta$θ is 5. Then $\sin \theta =\frac{5}{10}=\frac{1}{2}$sin θ = 5/10 = 1/2, so $\theta ={\sin }^{-1}\left(\frac{1}{2}\right)=30^{\circ }$θ = sin⁻¹(1/2) = 30°, one of the exact values.

Another has a hypotenuse of 9, and the side adjacent to $\theta$θ is 7. Then $\cos \theta =\frac{7}{9}=0.7778$cos θ = 7/9 = 0.7778, so $\theta ={\cos }^{-1}\left(\frac{7}{9}\right)=38.9424{\dots }^{\circ }$θ = cos⁻¹(7/9) = 38.9424…°, which is 38.9° to one decimal place. Check: cos 38.94° = 0.7778.

When dividing the sides, keep the full value, or better still type the fraction straight into the inverse function: ${\cos }^{-1}\left(\frac{7}{9}\right)$cos⁻¹(7/9) rather than ${\cos }^{-1}$cos⁻¹ of a rounded decimal.

Angles worth knowing

Some ratios give an angle without a calculator. If the two shorter sides are equal, $\tan \theta =1$tan θ = 1, and the triangle is half a square, so $\theta ={\tan }^{-1}1=45^{\circ }$θ = tan⁻¹ 1 = 45°.

In the same way, $\sin \theta =\frac{1}{2}$sin θ = 1/2 gives 30° and $\cos \theta =\frac{1}{2}$cos θ = 1/2 gives 60°, both from half an equilateral triangle.

Equal shorter sides make $\tan \theta =\frac{1}{1}=1$tan θ = 1/1 = 1, and ${\tan }^{-1}1=45^{\circ }$tan⁻¹ 1 = 45°.

The corner opens at 60°, where tan 60° = 1.732. Swing it round until the tangent reads 1, and read off the angle that has that tangent: 45°. Pull the corner outward at any angle, and the ratios do not change.

The other angle, and a ratio that cannot be

Once one acute angle is found, the other is 90° minus it. In the 3-4 triangle it is 90 − 36.87 = 53.13°, which is 53.1° to one decimal place. As a check, the tangent of that angle is $\frac{4}{3}$4/3, the sides the other way up: tan 53.13° = 1.3333.

A sine or cosine of more than 1 has no angle. The hypotenuse is the longest side, so opposite/hypotenuse and adjacent/hypotenuse are always less than 1. A calculator asked for ${\sin }^{-1}1.2$sin⁻¹ 1.2 gives an error, and in a problem that usually means the sides were put the wrong way up.

The usual mistakes

Pressing tan instead of ${\tan }^{-1}$tan⁻¹. tan 0.75 treats 0.75 as an angle of 0.75° and gives 0.0131, a ratio, not an angle.

Using the wrong inverse. Opposite over adjacent is a tangent. ${\sin }^{-1}\left(\frac{3}{4}\right)=48.6^{\circ }$sin⁻¹(3/4) = 48.6° is the angle whose sine is $\frac{3}{4}$3/4, which is not this triangle’s angle.

Turning the ratio upside down. ${\tan }^{-1}\left(\frac{4}{3}\right)=53.1^{\circ }$tan⁻¹(4/3) = 53.1° is the other acute angle, across from the side of 4.

A calculator set to radians. ${\tan }^{-1}0.75$tan⁻¹ 0.75 then shows 0.6435, the same angle measured in radians. Set it to degrees.