How to find the angle of a triangle trigonometry

In the module Further trigonometry (Year 10), we introduced and proved the sine rule, which is used to find sides and angles in non-right-angled triangles.

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In the triangle \(ABC\), labelled as shown, we have

\[ \dfrac{a}{\sin A} = \dfrac{b}{\sin B} = \dfrac{c}{\sin C}. \]

Clearly, we may also write this as

\[ \dfrac{\sin A}{a} = \dfrac{\sin B}{b} = \dfrac{\sin C}{c}. \]

In general, one of the three angles may be obtuse. The formula still holds true, although the geometric proof is slightly different.

Exercise 4

Find two expressions for \(h\) in the diagram below, and hence deduce the sine rule.

Detailed description of diagram
Repeat the method of part (a), using the following diagram, to show that the sine rule holds in obtuse-angled triangles.

The triangle \(ABC\) has \(AB = 9\text{ cm}\), \(\angle ABC = 76^\circ\) and \(\angle ACB = 58^\circ\).

Find, correct to two decimal places,

Solution

Applying the sine rule gives \[ \dfrac{AC}{\sin 76^\circ} = \dfrac{9}{\sin 58^\circ} \] and so \begin{align*} AC = \dfrac{9 \sin 76^\circ}{\sin 58^\circ}\\ \approx 10.30\text{ cm} \qquad \text{(to two decimal places).} \end{align*}
To find \(BC\), we first find the angle \(\angle CAB\) opposite it. \begin{align*} \angle CAB = 180^\circ - 58^\circ - 76^\circ\\ = 46^\circ. \end{align*} Thus, by the sine rule, \[ \dfrac{BC}{\sin 46^\circ} = \dfrac{9}{\sin58^\circ} \] and so \[ BC \approx 7.63\text{ cm} \qquad \text{(to two decimal places).} \]

The ambiguous case

In the module Congruence (Year 8), it was emphasised that, when applying the SAS congruence test, the angle in question has to be the angle included between the two sides. For example, the following diagram shows two non-congruent triangles \(ABC\) and \(ABC'\) having two pairs of matching sides and sharing a common (non-included) angle.

Detailed description of diagram

Suppose we are told that a triangle \(PQR\) has \(PQ = 9\), \(\angle PQR = 45^\circ\) and \(PR = 7\). Then the angle opposite \(PQ\) is not uniquely determined. There are two non-congruent triangles that satisfy the given data.

Detailed description of diagram

Applying the sine rule to the triangle, we have

\[ \dfrac{\sin \theta}{9} = \dfrac{\sin 45^\circ}{7} \] and so \begin{align*} \sin\theta = \dfrac{9\sin 45^\circ}{7}\\ \approx 0.9091. \end{align*}

Thus \(\theta \approx 65^\circ\), assuming that \(\theta\) is acute. But the supplementary angle is \(\theta' \approx 115^\circ\). The triangle \(PQR'\) also satisfies the given data. This situation is sometimes referred to as the ambiguous case.

Since the angle sum of a triangle is \(180^\circ\), in some circumstances only one of the two angles calculated is geometrically valid.

The cosine rule

We know from the SAS congruence test that a triangle is completely determined if we are given two sides and the included angle. However, if we know two sides and the included angle in a triangle, the sine rule does not help us determine the remaining side or the remaining angles.

The second important formula for general triangles is the cosine rule.

Suppose \(ABC\) is a triangle and that the angles \(A\) and \(C\) are acute. Drop a perpendicular from \(B\) to the line interval \(AC\) and mark the lengths as shown in the following diagram.

Detailed description of diagram

In the triangle \(ABD\), applying Pythagoras' theorem gives

\[ c^2 = h^2 + (b - x)^2. \]

Also, in the triangle \(BCD\), another application of Pythagoras' theorem gives

\[ h^2 = a^2 - x^2. \]

Substituting this expression for \(h^2\) into the first equation and expanding,

\begin{align*} c^2 = a^2 - x^2 + (b - x)^2\\ = a^2 - x^2 + b^2 - 2bx + x^2\\ = a^2 + b^2 - 2bx. \end{align*}

Finally, from triangle \(BCD\), we have \(x = a\,\cos C\) and so

\[ c^2 = a^2 + b^2 - 2ab\,\cos C. \]

This last formula is known as the cosine rule.

Notice that, if \(C = 90^\circ\), then since \(\cos C = 0\) we obtain Pythagoras' theorem, and so we can regard the cosine rule as Pythagoras' theorem with a correction term.

The cosine rule is also true when the angle \(C\) is obtuse. But note that, in this case, the final term in the formula will produce a positive number, because the cosine of an obtuse angle is negative. Some care must be taken in this instance.

By relabelling the sides and angles of the triangle, we can also write the cosine rule as \(a^2 = b^2 + c^2 - 2bc\,\cos A\) and \(b^2 = a^2 + c^2 - 2ac\,\cos B\).

Find the value of \(x\) to one decimal place.

Solution

Applying the cosine rule gives

\begin{align*} x^2 = 7^2 + 8^2 - 2 \times 7 \times 8 \times \cos 110^\circ\\ = 113 + 112\cos70^\circ\\ \approx 151.306, \end{align*}

so \(x \approx 12.3\) (to one decimal place).

Finding angles

If the three sides of a triangle are known, then the three angles are uniquely determined. (This is the SSS congruence test.) Again, the sine rule is of no help in finding the three angles, since it requires the knowledge of (at least) one angle, but we can use the cosine rule instead.

We can substitute the three side lengths \(a\), \(b\), \(c\) into the formula \(c^2 = a^2 + b^2 - 2ab\,\cos C\), where \(C\) is the angle opposite the side \(c\), and then rearrange to find \(\cos C\) and hence \(C\).

Alternatively, we can rearrange the formula to obtain

\[ \cos C = \dfrac{a^2+b^2-c^2}{2ab} \]

and then substitute. Students may choose to rearrange the cosine rule or to learn this further formula. Using this form of the cosine rule often reduces arithmetical errors.

Recall that, in any triangle \(ABC\) labelled as shown, if \(a < b\), then \(\text{angle } A < \text{angle } B\).

A triangle has side lengths 6 cm, 8 cm and 11 cm. Find the smallest angle in the triangle.

Solution

The smallest angle in the triangle is opposite the smallest side.

Applying the cosine rule:

\begin{align*} 6^2 = 8^2 + 11^2 - 2 \times 8 \times 11 \times \cos \theta \\ \cos\theta = \dfrac{8^2+11^2-6^2}{2\times 8 \times 11}\\ =\dfrac{149}{176}. \end{align*}

So \(\theta \approx 32.2^\circ\) (correct to one decimal place).

The area of a triangle

We saw in the module Introductory trigonometry (Years 9--10) that, if we take any triangle with two given sides \(a\) and \(b\) about a given (acute) angle \(\theta\), then the area of the triangle is

\[ \text{Area} = \dfrac{1}{2} ab\,\sin\theta. \]

This formula also holds when \(\theta\) is obtuse.

Exercise 5

A triangle has two sides of length 5 cm and 4 cm containing an angle \(\theta\). Its area is 5 cm\(^2\). Find the two possible (exact) values of \(\theta\) and draw the two triangles that satisfy the given information.

Exercise 6

Write down two different expressions for the area of a triangle \(ABC\) and derive the sine rule from them.

Next page - Content - Trigonometric identities

In this explainer, we will learn how to find a missing angle in a right triangle using the appropriate inverse trigonometric function given two side lengths.

Recall that, when working with right triangles, we can use either the Pythagorean theorem or the trigonometric ratios to find unknown sides or angles. When we have two known sides and need to find a third side, we use the Pythagorean theorem. However, if we have a right triangle with one known side and one known angle, we can find an unknown side using the trigonometric ratios sine, cosine, and tangent. Let’s recap these below.

For a right triangle with a nonright angle 𝜃, hypotenuse H, the side opposite to 𝜃, O, and the side adjacent to 𝜃, A, as seen in the diagram below, the following trigonometric ratios hold true: sinOHcosAHtanOA𝜃=,𝜃=,𝜃=.

For example, if we are given the measures of an angle, 𝜃, and the side adjacent to it, A, and wanted to find the opposite side, O, we could rearrange the trigonometric ratio tanOA𝜃= to get OAtan=𝜃.

In addition to finding an unknown side, we can use the trigonometric ratios to find an unknown angle when two sides are known. To do this we use the trigonometric inverse of sine, cosine, and tangent in order to rearrange to make the angle the subject of the equation.

For an angle 𝜃 the trigonometric inverse for sine, cosine, and tangent can be written as follows:

For sinOH𝜃=, the inverse can be written as 𝜃=sinOH.
For cosAH𝜃=, the inverse can be written as 𝜃=cosAH.
For tanOA𝜃=, the inverse can be written as 𝜃=tanOA.

As with solving for an unknown side, when solving for an unknown angle, we start by labeling the sides of the triangle in correspondence with the angle we are trying to find. Recall that, for an angle 𝜃 in a right triangle, we label the hypotenuse as H, the side opposite the angle 𝜃 as O, and the side adjacent to the angle 𝜃 as A. Once we have labeled the sides, we can then determine which trigonometric ratio to use and substitute into it accordingly. We then use the trigonometric inverse to solve for the angle.

We will consider how to find an unknown angle given the diagram in the first example.

In the given figure, find the measure of angle 𝜃, in degrees, to two decimal places.

Answer

In this question we are given a right triangle with two known sides and an unknown angle. Since we are required to find the missing angle, we need to use right angle trigonometry.

When solving problems involving right angle trigonometry, we need to label the figure in order to determine which trigonometric ratio to use. We label according to the angle we know or are trying to find, which, in this case, is 𝜃. We are given the length of the side opposite the right angle, which is the hypotenuse, and the length of the side opposite the angle 𝜃. We label these as H and O accordingly, as seen below.

We can see from the figure that Ocm=5 and Hcm=11. Therefore, we need to use the trigonometric ratio that contains O and H. This is sinOH𝜃=.

Substituting O=5 and H=11, we get sin𝜃=511.

Now that we have substituted into the sine ratio, we need to solve for the unknown angle 𝜃. To do this, we rearrange the sine ratio by using the trigonometric inverse sin. Doing so, we get sinsin𝜃=511𝜃=511.

And then, solving using a calculator, we get 𝜃=27.03569….∘

Therefore, the measure of angle 𝜃 in degrees correct to 2 decimal places is 27.04∘.

In the next example, we will again find an unknown angle when given a figure but are told which angles to find in angle notation.

For the given figure, find the measure of ∠𝐵𝐴𝐶, in degrees, to two decimal places.

Answer

In this question, we are asked to find the measure of ∠𝐵𝐴𝐶 and, from the figure, have been given the lengths of two sides in a right triangle. As such, we need to use right angle trigonometry to find the unknown angle.

When solving problems involving right angle trigonometry, we label the angle we are trying to find 𝜃 and then the corresponding known sides. We are given the side opposite to ∠𝐵𝐴𝐶 and the side adjacent to ∠𝐵𝐴𝐶. Therefore, we label ∠𝐵𝐴𝐶 as 𝜃, the opposite side as O, and the adjacent side as A. This is shown in the diagram below.

We can see from the figure that Ocm=7 and Acm=5. Therefore, we need to use the trigonometric ratio that contains A and O. This is tanOA𝜃=.

Substituting O=7 and A=5, we get tan𝜃=75.

Now that we have substituted into the tangent ratio, we need to solve for the unknown angle 𝜃. To do this, we rearrange the tangent ratio by using the trigonometric inverse tan. Doing so, we get tantan𝜃=75𝜃=75.

And, solving using a calculator, we get 𝜃=54.4623….∘

Therefore, the measure of ∠𝐵𝐴𝐶 is 54.46∘ to 2 decimal places.

In the following example, we are asked to find the measures of two unknown angles instead of just one.

For the given figure, find the measures of ∠𝐴𝐵𝐶 and ∠𝐴𝐶𝐵, in degrees, to two decimal places.

Answer

In this question, we are asked to find two unknown angles in a right triangle while given two sides in the figure. As such, we need to use right angle trigonometry to solve for the unknown angles.

When solving problems involving right angle trigonometry, we label the angle we are trying to find 𝜃 and then the corresponding known sides. As we are trying to find two unknown angles, we will start by trying to find angle ∠𝐴𝐵𝐶. We are given the length of the side opposite the right angle, the hypotenuse, and the length of the side adjacent to the angle ∠𝐴𝐵𝐶. We therefore label ∠𝐴𝐵𝐶 as 𝜃, the hypotenuse as H, and the adjacent as A, as shown below.

We can see from the figure that Acm=4 and Hcm=9. Therefore, we need to use the trigonometric ratio that contains A and H. This is cosAH𝜃=.

Substituting A=4 and H=9, we get cos𝜃=49.

Now that we have substituted into the cosine ratio, we need to solve for the unknown angle 𝜃. To do this, we rearrange the cosine ratio by using the trigonometric inverse cos. Doing so, we get coscos𝜃=49𝜃=49.

And, solving using a calculator, we get 𝜃=63.6122….∘

So, the measure of ∠𝐴𝐵𝐶 is 63.61∘ correct to 2 decimal places.

To find ∠𝐴𝐶𝐵 we use the property that the sum of the interior angles in a triangle is 180∘, since we know the other two angles. This gives us 𝑚∠𝐴𝐶𝐵+63.61…+90=180𝑚∠𝐴𝐶𝐵=180−90−63.61…𝑚∠𝐴𝐶𝐵=26.392.∘∘∘∘∘∘∘correcttodecimalplaces

Therefore, ∠𝐴𝐵𝐶 is 63.61∘ and ∠𝐴𝐶𝐵 is 26.39∘ correct to 2 decimal places.

In the following example, we are given the measures of two sides and are required to find an unknown side and unknown angles without a figure being provided.

𝐴𝐵𝐶 is a right triangle at 𝐵, where 𝐵𝐶=10cm and 𝐴𝐶=18cm. Find the length 𝐴𝐵, giving the answer to the nearest centimetre, and the measure of angles 𝐴 and 𝐶, giving the answer to the nearest degree.

Answer

In this question, we are required to find an unknown side and two unknown angles in a right triangle. To do this, it is helpful to first draw a diagram to represent the information given. We will start by drawing a right triangle 𝐴𝐵𝐶 and labeling the known sides, 𝐵𝐶=10cm and 𝐴𝐶=18cm, as shown below.

We can solve this question using right triangle trigonometry. As we are given two sides, we will start by finding the unknown angles. Let’s start by finding the angle 𝐴.

To do this, it helps to label our diagram with angle 𝐴 as our unknown angle 𝜃, 𝐴𝐶 as the hypotenuse H, and 𝐵𝐶 as the opposite side O. The diagram below shows how 𝜃, H, and O have been labeled.

From the diagram we can see that Ocm=10 and Hcm=18. Therefore, we need to use the trigonometric ratio that contains O and H, which is sinOH𝜃=.

Substituting O=10 and H=18, we get sin𝜃=1018.

Now that we have substituted into the sine ratio, we need to solve for the unknown angle 𝜃. To do this we rearrange the sine ratio by using the trigonometric inverse sin. Doing so, we get sinsin𝜃=1018𝜃=1018.

And, solving using a calculator, we get 𝜃=33.749….∘

Therefore, 𝑚∠𝐴 is 34∘ to the nearest degree.

Next, to find angle 𝐶, we use the property that the sum of the interior angles in a triangle is 180∘. Since we know the other two angles, this gives us 𝑚∠𝐶+33.749…+90=180𝑚∠𝐶=180−90−33.749…𝑚∠𝐶=56.251….∘∘∘∘∘∘∘

Therefore, 𝑚∠𝐶 is 56∘ to the nearest degree.

Lastly, we need to find side 𝐴𝐵. Using right triangle trigonometry, we can use one of the known sides and one of the known angles to do this. As we have two known sides and two known angles, we can choose which trigonometric ratio to use. As we found 𝑚∠𝐴 first, we will use this angle as 𝜃. We will use the opposite side, O, leaving us with the side we are trying to find as the adjacent side A. We can see this below.

As we know, Ocm=10, 𝜃=33.749…∘, and we are trying to find A. We need to use the trigonometric ratio that contains O and A. This is tanOA𝜃=.

Substituting O=10 and 𝜃=33.749…∘, we get tanA(33.749…)=10.∘

Rearranging for A, we get AtanAtan(33.749…)=10=10(33.749…).∘∘

And, solving using a calculator, we get Acm=14.9666….

Therefore, the length of 𝐴𝐵 is 15 cm correct to the nearest cm.

So, the length of 𝐴𝐵 to the nearest centimetre is 15 cm, the measure of angle 𝐴 to the nearest degree is 34∘, and the measure of angle 𝐶 to the nearest degree is 56∘, as required.

In the next example, we will consider how to find an unknown angle in the context of a real-life problem.

A 5 m ladder is leaning against a vertical wall such that its base is 2 m from the wall. Work out the angle between the ladder and the floor, giving your answer to two decimal places.

Answer

In this question we are given information about the length of a ladder and how far the base is from a wall. To solve this, it is helpful to first draw a diagram to illustrate the information given.

As the question asks for the angle between the ladder and the floor and we know the length of two sides, we use right angle trigonometry to find the unknown angle. We can label the unknown angle as 𝜃, the side opposite the right angle, the hypotenuse, as H, and the side adjacent to the angle as A.

From the diagram, we can see that Am=2 and Hm=5. Therefore, we need to use the trigonometric ratio that contains A and H. This is cosAH𝜃=.

Substituting A=2 and H=5, we get cos𝜃=25.

And, solving using a calculator, we get 𝜃=66.4218….∘

Therefore, the angle between the ladder and the floor is 66.42∘ to 2 decimal places.

In the last example, we will discuss how to find an unknown angle in a real-life problem.

A car is going down a ramp that is 10 metres high and 71 metres long. Find the angle between the ramp and the horizontal, giving the answer to the nearest second.

Answer

In this question we are required to find the angle between the ramp and the horizontal and are told that the ramp is 10 metres high and 71 metres long. We can draw a right triangle, with height 10 metres and hypotenuse 71 metres to illustrate this.

Next, we want to identify the unknown we are looking for, which is the angle between the ramp and the horizontal. We will label this as 𝜃 on our diagram, as follows.

As we have two known sides and one unknown angle in a right triangle, we can use right triangle trigonometry to find the unknown angle. Since the known sides are the height, which is the side opposite the angle, and the ramp, which is the hypotenuse, we can label the height and ramp as O and H respectively.

We can see from the figure that Om=10 and Hm=71. Therefore, we need to use the trigonometric ratio that contains O and H. This is sinOH𝜃=.

Substituting O=10 and H=71, we get sin𝜃=1071.

And, solving using a calculator, we get 𝜃=8.09674….∘

As the question requires the answer to the nearest second, we need to calculate the number of minutes and seconds given rather than the decimal of a degree.

Since we have 0.09674… degrees, then, by multiplying by 60, we get 5.805 minutes. Again, since we have 0.805 minutes, by multiplying by 60, we get 48.3 seconds, which is 48 to the nearest second. This gives us an answer of 8 degrees, 5 minutes, and 48 seconds to the nearest second, or 85′48′′∘.

In this explainer we have learned how to solve for an angle when using right triangle trigonometry. We have done this using the trigonometric inverses of sine, cosine, and tangent. Let’s recap the key points.

To solve for an unknown angle in a right triangle, where two sides are known, we use right triangle trigonometry. This involves using one of the following trigonometric ratios:
- sinOH𝜃=,
- cosAH𝜃=,
- tanOA𝜃=,
where we have a nonright angle 𝜃, a hypotenuse of length H, the side opposite to 𝜃 of length O, and the side adjacent to 𝜃 of length A, as seen below.
For the sine, cosine, and tangent of an angle 𝜃, we use the trigonometric inverse to solve for 𝜃 as follows:
- For sinOH𝜃=, the inverse can be written as 𝜃=sinOH.
- For cosAH𝜃=, the inverse can be written as 𝜃=cosAH.
- For tanOA𝜃=, the inverse can be written as 𝜃=tanOA.
When solving problems, it is helpful to draw a diagram if one is not given and then to label the sides and unknown angle in order to correctly identify which trigonometric ratio to use.