SL 3.3 Solve contextual two- and three-dimensional trigonometry problems | International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

Geometry and Trigonometry

IB Mathematics: Analysis and Approaches SL — study topic SL 3.3

Contextual trigonometry turns measurements and angles into a triangle model. Begin by identifying what is known, what is required, and which points and lines form the relevant triangle. In a two-dimensional diagram, the triangle may be visible directly. In a three-dimensional situation, it is often helpful to first find a horizontal distance, then use that distance with a vertical height in a right triangle. A sketch is a model, not necessarily a scale drawing: label known lengths, angles, and units, and state any assumption needed to interpret the situation.

What you will learn

Choose a suitable trigonometric method for a contextual triangle and explain why it applies.
Use the sine rule, cosine rule, and right-triangle trigonometry to find unknown lengths or angles.
Represent a three-dimensional situation with a horizontal projection and a right triangle.
Use a calculator to check numerical work while retaining a clear mathematical method.
Give answers with appropriate units, accuracy, and interpretation.

1. Review: choose the triangle and the method

A triangle’s interior angles add to

180^\circ

. In a right triangle, the hypotenuse is opposite the

90^\circ

angle. Relative to a chosen acute angle

\theta

, the opposite side is across from

\theta

and the adjacent side touches it without being the hypotenuse. These labels depend on which angle you choose.

For a right triangle, use sine, cosine, or tangent according to the sides involved. The calculator must be in degree mode when the context gives angles in degrees. Check that the answer makes sense: a distance is positive, and an angle in a triangle is between

0^\circ

and

180^\circ

.

For a non-right triangle, use the cosine rule when you know two sides and their included angle, or all three sides and need an angle. Use the sine rule when you know an opposite side-angle pair and another side or angle. The included angle is the angle between the two known sides.

\sin\theta=\frac{\text{opposite}}{\text{hypotenuse}},\quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}},\quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}}

Draw and label a triangle before substituting values.
Use right-triangle ratios only when the triangle has a right angle.
Identify the included angle carefully when using the cosine rule.

2. Solve two-dimensional contexts

In a context such as two routes leaving a junction, the given distances and the angle between them may form two sides and an included angle. The cosine rule gives the third side directly. For an unknown angle, rearrange the rule and use the inverse cosine function on the calculator. Keep the angle’s position in the diagram clear.

For a line of sight to a tall object, a horizontal ground distance and an angle of elevation form a right triangle. The angle of elevation is measured upward from the horizontal. If the observer’s eye is above the ground, distinguish the height above eye level from the total height above ground.

A calculator is useful for evaluating trigonometric values and inverse trigonometric functions. It does not decide which sides belong in a model. Set the angle unit first, preserve unrounded values during working, and round only the final result to the requested precision.

a^2=b^2+c^2-2bc\cos A,\qquad \frac{a}{\sin A}=\frac{b}{\sin B}=\frac{c}{\sin C}

Include units in intermediate quantities and in the final contextual answer.
A bearing or direction must be interpreted from the diagram; do not assume it is the triangle’s included angle without checking.
State a sensible final accuracy, such as the nearest tenth of a metre.

3. Model three-dimensional situations

A three-dimensional diagram can be difficult to calculate from directly. Look for a horizontal projection: the ground distance from an observation point to the point directly below the object. If the ground position is described by perpendicular east-west and north-south distances, these form a right triangle on the ground. Pythagoras’ theorem gives the horizontal distance.

The horizontal distance and vertical height then form a second right triangle with the line of sight. For an angle of elevation, the vertical height is opposite the angle and the horizontal distance is adjacent. The line of sight is the hypotenuse. This separates a three-dimensional situation into manageable two-dimensional calculations.

A sketch should show which distances are horizontal and which are vertical. A numerical or calculator check can confirm the arithmetic, while the diagram explains why the chosen ratio applies. Check that a calculated line of sight is longer than either of its perpendicular components.

d=\sqrt{x^2+y^2},\qquad \tan\theta=\frac{h}{d}

Find the horizontal projection before using the elevation angle.
Use Pythagoras’ theorem only for a right triangle.
Do not confuse the horizontal ground distance with the direct three-dimensional distance.

4. Check, interpret, and communicate

A complete solution connects the context to a labeled diagram, the diagram to an equation, and the numerical result back to the context. Explain which angle is used and why the selected rule applies. A calculator screen alone is not a mathematical explanation.

Use a graphing calculator or ordinary scientific calculator to evaluate the final expression and, where useful, check a result by substituting it back into the original relationship. In a non-right triangle, a plotted sketch may help you see whether the calculated angle is acute or obtuse, but the labeled geometry and equation determine the answer.

Check rounding and units. If the question asks for a distance, report a length rather than an angle; if it asks for an angle, include degrees when that is the context’s unit. Avoid rounding an intermediate side too early, as this can change the final answer.

Show the model and the equation before giving a calculator result.
Use a contextually meaningful final statement.
Check size, units, and rounding.

Worked example

Two routes from a junction

Two straight paths leave a junction. One is

8.0

km long and the other is

11.0

km long. The angle between them is

62^\circ

. Find the straight-line distance between their endpoints, to the nearest

0.1

km.

Model the situation
The paths and the straight line between their endpoints form a triangle. The two known sides meet at the given $62^\circ$ angle, so this is the included angle. The cosine rule is appropriate.
Substitute into the cosine rule
Let $x$ be the distance between the endpoints. Substitute the two path lengths and the included angle.
$x^2=8.0^2+11.0^2-2(8.0)(11.0)\cos 62^\circ$
Evaluate and round
In degree mode, the calculation gives $x\approx10.118$ . Take the positive square root because distance is positive, then round to the nearest tenth.
$x\approx10.1\text{ km}$

Answer: The endpoints are approximately

10.1

km apart.

Check: The distance is less than the sum of the two paths,

19.0

km, and greater than their difference,

3.0

km. This is consistent with a triangle.

Worked example

Height from an angle of elevation

A surveyor stands

34

m horizontally from a vertical tower. The angle of elevation from the surveyor’s eye to the top is

28^\circ

. The surveyor’s eye is

1.6

m above the ground. Find the tower’s height above ground to the nearest

0.1

m.

Separate the heights
The right triangle’s vertical side is the height of the tower above the surveyor’s eye, not the total height above ground. The horizontal side is $34$ m.
Use tangent
Relative to the angle of elevation, the unknown height above eye level is opposite and $34$ m is adjacent. Tangent relates these sides.
$\tan 28^\circ=\frac{h}{34}$
Add the eye height
Solving gives $h=34\tan 28^\circ\approx18.08$ m. Add the eye height to obtain the tower’s height above ground.
$H=34\tan 28^\circ+1.6\approx19.7\text{ m}$

Answer: The tower is approximately

19.7

m high.

Check: The height above eye level is about

18.1

m, so the total should be slightly more than that. Adding

1.6

m gives the stated result.

Worked example

A three-dimensional line of sight

A viewing point is on level ground. The point directly below a beacon is

18

m north and

10

m east of the viewer. The beacon is

24

m vertically above that ground point. Find the line-of-sight distance and its angle of elevation, to the nearest

0.1

m and nearest degree.

Find the horizontal distance
The north and east ground distances are perpendicular, so they form a right triangle. Let $d$ be the horizontal distance to the point directly below the beacon.
$d=\sqrt{18^2+10^2}=\sqrt{424}\approx20.59\text{ m}$
Find the direct distance
The horizontal distance and the $24$ m vertical height are perpendicular sides of a right triangle whose hypotenuse is the line of sight, $L$ .
$L=\sqrt{d^2+24^2}=\sqrt{1000}\approx31.6\text{ m}$
Find the elevation angle
For the angle of elevation $\theta$ , the vertical height is opposite and the horizontal distance is adjacent. Use inverse tangent and round to the nearest degree.
$\theta=\tan^{-1}\!\left(\frac{24}{d}\right)\approx49^\circ$

Answer: The line of sight is approximately

31.6

m, and its angle of elevation is approximately

49^\circ

.

Check: The direct distance exceeds both the vertical height and the horizontal distance, as a hypotenuse should. A graphing calculator in degree mode can check both numerical results.

Common mistakes and how to avoid them

Using the cosine rule with an angle that is not between the two known sides.

Correction: Mark the known sides and inspect the angle between them. If that is not the given angle, reconsider the diagram and the information available.

Using the full tower height as the opposite side when the angle is measured from eye level.

Correction: First calculate the height above eye level. Add the observer’s eye height only when finding the height above ground.

Using the three-dimensional direct distance as the adjacent side for an angle of elevation.

Correction: The adjacent side is the horizontal ground projection. Find it from the perpendicular ground distances before using tangent.

Rounding intermediate calculations too early or using radians for degree measurements.

Correction: Set the calculator to degrees when appropriate and retain extra digits until the final rounding.

Lesson summary

Choose the rule from the triangle and the known information: right-triangle ratios, sine rule, or cosine rule.
For a three-dimensional context, find the horizontal projection and then use a vertical right triangle.
Use a calculator to evaluate and check the model, not as a substitute for explaining it.
Report a sensible answer with units and requested accuracy.

Check your understanding

Question 1

A right triangle has an angle of elevation of

35^\circ

and a horizontal distance of

12

m. Which expression gives the vertical height?

$12\sin 35^\circ$
$12\tan 35^\circ$
$\frac{12}{\tan 35^\circ}$
$\frac{12}{\sin 35^\circ}$

Show answer and explanation

12\tan 35^\circ

The height is opposite the angle and the horizontal distance is adjacent, so tangent gives

\tan 35^\circ=h/12

.

Question 2

Two sides of a triangle are

5

cm and

9

cm, and the angle between them is

60^\circ

. Which rule directly finds the third side?

The cosine rule
The sine rule, without any further information
Pythagoras’ theorem, because two sides are known
The tangent ratio

Show answer and explanation

The cosine rule

Two sides and their included angle are the information used by the cosine rule. The triangle is not stated to be right-angled.

Question 3

A point is

6

m east and

8

m north of an observer on level ground. What is its horizontal distance from the observer?

$2$ m
$7$ m
$10$ m
$14$ m

Show answer and explanation

10

m

The ground directions are perpendicular, so the horizontal distance is

\sqrt{6^2+8^2}=10

m.

Key terms

Angle of elevation: The angle measured upward from a horizontal line to a line of sight.
Included angle: The angle between two specified sides of a triangle.
Horizontal projection: The horizontal ground distance from an observation point to the point directly below an elevated object.
Line of sight: The straight line from an observer to the point being viewed.

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SL 3.3 · Solve contextual two- and three-dimensional trigonometry problems

International Baccalaureate (IB) IB AA SL: Mathematics: Analysis and Approaches SL

IB Mathematics: Analysis and Approaches SL — study topic SL 3.3

What you will learn

1. Review: choose the triangle and the method

2. Solve two-dimensional contexts

3. Model three-dimensional situations

4. Check, interpret, and communicate

Two routes from a junction

Height from an angle of elevation

A three-dimensional line of sight

Common mistakes and how to avoid them

Lesson summary

Check your understanding

Question 1

Question 2

Question 3

Key terms

Continue through IB AA SL

About this lesson and its review