Lesson Explainer: Right Triangle Trigonometry: Solving for an Angle | Nagwa Lesson Explainer: Right Triangle Trigonometry: Solving for an Angle | Nagwa

Lesson Explainer: Right Triangle Trigonometry: Solving for an Angle Mathematics • Third Year of Preparatory School

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

When working with right triangle trigonometry, it is useful to recall the acronym “SOH CAH TOA.” This helps us remember the definitions of the trigonometric ratios sine, cosine, and tangent in terms of the sides relative to an angle that we call the opposite, adjacent, and hypotenuse. Let us list the ratios here.

Trigonometric Ratios

The hypotenuse is always the longest side of the right triangle, the opposite side is the side directly opposite the angle concerned, and the adjacent is the side next to the angle (which is not the hypotenuse). An example of this is shown here.

In order to find the measures of unknown angles in right triangles (using trigonometry), we need to be confident in our ability to correctly label the triangle in terms of the opposite, adjacent, and hypotenuse and correctly remember the trigonometric ratios. Once we are happy with these two things, we are in a position to start solving trigonometry problems involving finding the measure of an unknown angle. We can always find the missing measure of an angle in a right triangle from two of its side lengths using the following process.

How To: Finding the Missing Angle in a Right Triangle From Two of Its Side Lengths

  1. If we are not given a diagram of the right triangle, we should begin by sketching the information given.
  2. Label the sides of the right triangle based on their positions relative to the angle we want to determine.
  3. Use the acronym SOH CAH TOA to determine which trigonometric ratio includes the known side lengths.
  4. We use the shift button and then the relevant trigonometric ratio on the calculator followed by the ratio of the know lengths to determine the measure of the angle.
    For example, if we know that cos𝜃=12, then we press the following shiftcos12=

Let us start by looking at an example.

Example 1: Finding the Measure of an Unknown Angle in a Right Triangle

For the given figure, find the measure of angle 𝜃, to the nearest second.

Answer

Our first step in answering this question is to label the triangle relative to the angle 𝜃.

Notice here that we have circled A and H, as these are the two sides whose lengths we know. If we then recall the acronym “SOH CAH TOA,” we can see that “CAH” is the only one that contains the letters A and H, which means that we need to use the cosine ratio. Recall that cosAH𝜃=.

We now substitute the values of A and H to find that cos𝜃=38.

Using our calculators, we can find 𝜃 by calculating cos138, an example of the sequence of buttons that we might need to press is shiftcos38=. However, the exact buttons will vary from calculator to calculator.

If we then calculate this, we get 675832.tothenearestsecond

In some questions, we may be asked to calculate the measures of all the unknown angles in a right triangle. In this case, we have to use trigonometry to find one of the unknown angles and then we can use the fact that the measures of the angles in a triangle sum to 180. Let us look at an example where this is the case.

Example 2: Finding the Measures of all the Unknown Angles in a Right Triangle

For the given figure, find the measures of 𝐴𝐶𝐵 and 𝐵𝐴𝐶, to the nearest second.

Answer

Our first step is to choose one of the two unknown angles to calculate first. Here, we are going to start by finding 𝐴𝐶𝐵, which we will call 𝑥. We can then label the sides of the triangle in relation to angle 𝑥 as shown.

We have circled O and A, as these are the lengths that we know. If we then recall the acronym “SOH CAH TOA,” we can see that we need to use the tangent ratio, as “TOA” contains the letters O and A. Recall that tanOA𝑥=.

Substituting the lengths O and A, we get tan𝑥=45.

Using our calculators, we can find 𝑥 by calculating tan45, an example of the sequence of buttons that we might need to press is shifttan45=. However, the exact buttons will vary from calculator to calculator.

If we calculate this, we find that 𝑥=383935.

To find the measure of the second unknown angle in the triangle, we need to use the fact that the sum of the measures of the angles in a triangle is 180. If we call 𝐵𝐴𝐶 𝑦, we have that 𝑦+38.66+90=180.

This simplifies to 𝑦+128.66=180, and subtracting 128.66 from both sides, we find that 𝑦=512025.

In some trigonometry questions, we are not given a diagram and part of the skill of the question is drawing an appropriate diagram. In the following example, we will demonstrate this skill.

Example 3: Solving Triangles with Trigonometry

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

Answer

We begin by making a diagram. It is usually helpful to try and draw approximately to scale. It is not completely necessary, but it does help us check that our answers are reasonable when comparing them to the diagram. We, therefore, draw a triangle 𝐴𝐵𝐶 and label the edge lengths we know.

The first thing we have been asked to find is the length 𝐴𝐵. To do this, we can use the Pythagorean theorem, which states that 𝑐=𝑎+𝑏, where 𝑐 is the length of the hypotenuse. In the triangle we have been given, 𝐴𝐶 is the hypotenuse. Hence, we can write the Pythagorean theorem for the triangle as 𝐴𝐶=𝐴𝐵+𝐵𝐶.

Therefore, 𝐴𝐵=𝐴𝐶𝐵𝐶.

Substituting in 𝐵𝐶=10 and 𝐴𝐶=18, we have 𝐴𝐵=1810=324100=224.

Taking the square root, we have 𝐴𝐵=224=14.966=15cm to the nearest centimetre.

We now need to find the measures of the angles at 𝐴 and 𝐶. To do this, we can find one of the angles and then use the fact that the angles in a triangle sum to 180. We will find the measure of 𝐴, which we will denote 𝜃. To know which trigonometric ratio we should use, we first need to label the sides of the triangle. We know that 𝐴𝐶 is the hypotenuse. Since we are considering 𝐴, 𝐵𝐶 is the opposite and 𝐴𝐵 is the adjacent.

Since we know the lengths of all the sides, we could use any trigonometric ratio. However, it is best to use the two lengths we were given in the question. There are two good reasons for this. First, this means that if we made a mistake calculating the third side, it will not affect our answer to this part of the question. Second, we could easily make rounding errors if we use the length of the third side since its exact form is not a whole number. Therefore, we would like to calculate 𝐴 using the opposite and the hypotenuse. This means that we will use the sine ratio: sinOH𝜃=.

Substituting in the lengths of the opposite (𝐵𝐶=10) and the hypotenuse (𝐴𝐶=18), we have sin𝜃=1018=59.

We can find 𝜃 by using our calculators to evaluate sin59, an example of the sequence of buttons that we might need to press is shiftsin59=. However, the exact buttons will vary from calculator to calculator.

If we calculate this, we find that 𝜃=33.748=34 to the nearest degree. Therefore, 𝑚𝐴=34 to the nearest degree.

We can now use the fact that the angles in a triangle sum to 180 to find 𝑚𝐶. Since 𝑚𝐴+𝑚𝐵+𝑚𝐶=180, we have 𝑚𝐶=180𝑚𝐵𝑚𝐴.

Substituting in the values of 𝑚𝐵 and 𝑚𝐴, we have 𝑚𝐶=1809033.748=56.251=56 to the nearest degree.

Trigonometry questions may also be presented as story problems. When this is the case, if an associated diagram is not given, it is always worth drawing one. An example of this type of question would be the following.

Example 4: Using Trigonometry to Find the Measure of Angles in Trapezoids

𝐴𝐵𝐶𝐷 is a trapezoid-shaped piece of land where 𝐴𝐷 is parallel to 𝐵𝐶 and 𝐴𝐵𝐵𝐶. Find 𝑚𝐶 given that 𝐴𝐷=20m, 𝐵𝐶=35m, and 𝐷𝐶=25m. Give the answer to the nearest second.

Answer

We can start by highlighting 𝐶 on the diagram:

We can drop a perpendicular line from 𝐷 onto 𝐵𝐶 at a point 𝐸.

We see that triangle 𝐷𝐸𝐶 is a right triangle at 𝐸. We find the length of 𝐸𝐶 by noting that 𝐴𝐷=𝐵𝐸=20m. We can use this along with the fact that 𝐵𝐶=𝐵𝐸+𝐸𝐶=35m to find 𝐸𝐶. Substituting 𝐵𝐸=20m into the equation gives us 20+𝐸𝐶=35.

We then subtract 20 from both sides of the equation to get 𝐸𝐶=15.m

We can add this onto our diagram.

We can determine 𝑚𝐶 by noting that it is an angle in a right triangle with two known side lengths, so we can find the measure of this angle by using trigonometry.

To do this, we start by labeling the sides of right triangle 𝐷𝐸𝐶 based on their positions relative to 𝐶. We see that 𝐸𝐶 is the side adjacent to 𝐶 in the right triangle 𝐷𝐸𝐶 and 𝐷𝐶 is the hypotenuse of the right triangle since it is opposite the right angle.

We can now want to determine the relevant trigonometric ratio. To do this, we will use the acronym SOH CAH TOA to help us recall which ratios each trigonometric function gives.

We know the lengths of the side adjacent to the angle and the hypotenuse, so we will use the cosine function cosAHcos𝐶==1525𝐶=35.

We can use our calculator to evaluate this expression. We use the following inputsshiftcos35=

However, the exact buttons needed may be different on other calculators.

This gives us 𝑚𝐶=53748.37.

Hence, to the nearest second, 𝑚𝐶=53748.

Example 5: Solving Story Problems with Trigonometry

A 5 m ladder is leaned against a perpendicular 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 the nearest second.

Answer

Our first step in solving a question like this is to draw a diagram of the situation.

In this diagram, relative to the angle 𝑥, we have labeled the sides whose lengths we know. Here we know the length of the adjacent and the hypotenuse so we need to use the cosine ratio to find the measure of the unknown angle. We know that cosAH𝑥=.

If we substitute in the lengths A and H, we get cos𝑥=25.

We can find 𝑥 by using our calculators to evaluate cos25, an example of the sequence of buttons that we might need to press is shiftcos25=. However, the exact buttons will vary from calculator to calculator.

Calculating this, we find that 𝑥=662519.

We will finish by considering one final story problem.

Example 6: Solving Story Problems with Trigonometry

The height of a ski slope is 16 metres and the length is 20 metres. Find the measure of 𝜃 giving the answer to the nearest second.

Answer

In this question, we are fortunate to have been given an associated diagram, which means we do not need to draw this ourselves. Our first step is to label the sides relative to the angle theta.

Here, we know the lengths of the opposite and the hypotenuse and, therefore, need to use the sine ratio to find the measure of the unknown angle. Recall that sinOH𝜃=. If we substitute in the lengths O and H, we get sin𝜃=1620.

We can find 𝜃 by using our calculators to evaluate sin1620, an example of the sequence of buttons that we might need to press is shiftsin1620=. However, the exact buttons will vary from calculator to calculator.

Calculating this, we find that 𝜃=53748.

Key Points

  • When working with right triangles, we use the terms opposite, adjacent, and hypotenuse to refer to the sides of the triangle. The hypotenuse is always opposite the right angle and is the longest side. The opposite and the adjacent are labeled in relation to a given angle often denoted 𝜃. The adjacent is the side next to the angle 𝜃 which is not the hypotenuse. As for the opposite, it is the last side of the triangle. It is called the opposite since it is opposite the given angle.
  • Recall the acronym “SOH CAH TOA,” where O stands for the opposite, A stands for the adjacent, H stands for the hypotenuse, and 𝜃 is the angle. The trigonometric ratios are sinOHcosAHandtanOA𝜃=,𝜃=,𝜃=.
  • We can find the measure of an angle given the side lengths using the Shift button and then the relevant trigonometric ratio on the calculator followed by the ratio of the known lengths.

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