MGU Mathematics Series — Trigonometry and Precalculus

Part II: Trigonometric Foundations

Law of Sines · Law of Cosines · Navigation · Surveying · Physics

Chapter 7 of 14

MGU Mathematics Series · Trigonometry and Precalculus · Library Textbook Edition

Chapter 7:
Triangle Trigonometry
and Applications

Right triangles, the Law of Sines, the Law of Cosines, the ambiguous case, area formulas, bearings, surveying, and force resolution — trigonometry as a measurement system

■ Math Level: Trigonometry and Precalculus ■

■ 14 Sections • Part II of 5 ■

Follows Chapter 6: Identities and Equations · Prepares students for Chapter 12: Parametric, Polar, and Vectors

Chapter 7 — Front Matter

Purpose and Overview

Chapter 7 connects the unit-circle view of trigonometry back to triangle measurement. Students who have studied radian measure, the unit circle, trigonometric functions, their graphs, and their identities are now ready to see how those ideas root themselves in the physical world of shape, distance, and angle. This chapter reviews right-triangle trigonometry, introduces the powerful tools of the Law of Sines and the Law of Cosines, develops area formulas from trigonometric relationships, and places all of this in real contexts: navigation, surveying, engineering, and physics.

This chapter serves as the bridge between Geometry and the function-based trigonometry of Precalculus. Students should leave understanding that trigonometry is a measurement system — one that allows unknown lengths and angles to be found when direct measurement is difficult or impossible.

Opening Question

A forest ranger in a fire lookout tower spots smoke in two directions. From Tower A, the smoke is observed at a bearing of N 42° E. From Tower B, located 15 miles due east of Tower A, the same smoke is observed at a bearing of N 28° W. How far is the fire from each tower?

This kind of problem cannot be solved with a ruler or a protractor alone. It requires trigonometry — specifically, the tools developed in this chapter. By the time you finish Chapter 7, you will have everything you need to solve this problem and dozens like it.

14Sections

14Objectives

13Key Terms

15+Worked Examples

Learning Objectives

1Review right-triangle trigonometric ratios and solve right triangles completely.
2Apply angles of elevation and depression to find heights and distances.
3Recognize and work with oblique triangles.
4State the Law of Sines and apply it to the AAS, ASA, and SSA cases.
5Analyze the ambiguous case (SSA) and determine the number of possible triangles.
6State the Law of Cosines and apply it to the SAS and SSS cases.
7Choose the correct law for a given triangle situation.
8Use the trigonometric area formula and Heron's Formula.
9Read and use compass bearings to solve navigation problems.
10Solve surveying and indirect measurement problems using trigonometry.
11Apply triangle trigonometry to force resolution in physics and engineering.
12Recognize and avoid common triangle trigonometry mistakes.
13Connect triangle trigonometry to polar coordinates, parametric equations, and vectors.
14Identify the triangle as the fundamental unit of indirect measurement in a technological world.

Chapter 7 — Contents

Sections at a Glance

7.1Right-Triangle Trigonometry ReviewSix trigonometric ratios; Pythagorean Theorem; 45-45-90 and 30-60-90 triangles; exact values
7.2Solving Right TrianglesThree procedures for given information; finding all sides and angles; checking with Pythagorean Theorem
7.3Angles of Elevation and DepressionDefinitions; alternate interior angle relationship; height and distance problems
7.4Oblique TrianglesDefinition; standard labeling convention; four cases: AAS, ASA, SSA, SAS, SSS
7.5The Law of SinesStatement; geometric proof using altitude; AAS and ASA worked examples
7.6The Ambiguous CaseSSA analysis; height test; zero, one, and two triangle cases; swing-compass diagram
7.7The Law of CosinesStatement; coordinate-system derivation; connection to Pythagorean Theorem; SAS and SSS examples
7.8Choosing Between Law of Sines and Law of CosinesDecision table; matched-pair strategy; four identification examples
7.9Area of a Triangle Using TrigonometryTrigonometric area formula; derivation; Heron's Formula; SAS and SSS area examples
7.10Bearings and NavigationTrue bearing vs. compass bearing; coordinate approach; two-bearing and opening-question examples
7.11Surveying and Indirect MeasurementRiver width; two-angle height measurement; setting up triangles from field data
7.12Trigonometry in Physics and EngineeringForce components; inclined plane; resultant forces; non-perpendicular force combination
7.13Common Triangle Trigonometry MistakesEight recurring errors; explanations; corrections; prevention strategies
7.14Preparing for Polar and Parametric ThinkingConnection to polar coordinates; parametric circles; vectors; dot product and Law of Cosines

Section 7.1

Right-Triangle Trigonometry Review

Before studying triangles without right angles, it is worth reviewing the foundation. Right-triangle trigonometry connects the six trigonometric functions to ratios of sides. Consider a right triangle with an acute angle θ at one vertex. Label the three sides relative to θ: the opposite side is the leg across from θ, the adjacent side is the leg next to θ (not the hypotenuse), and the hypotenuse is the side opposite the right angle.

The Six Trigonometric Ratios

sin θ = opposite / hypotenuse cos θ = adjacent / hypotenuse tan θ = opposite / adjacent

csc θ = hypotenuse / opposite sec θ = hypotenuse / adjacent cot θ = adjacent / opposite

The Pythagorean Theorem

In every right triangle with legs a and b and hypotenuse c:

a² + b² = c²

Special Right Triangles

Triangle Type	Side Ratios	Key Exact Values
45–45–90	1 : 1 : √2	sin 45° = cos 45° = √2/2, tan 45° = 1
30–60–90	1 : √3 : 2	sin 30° = 1/2, cos 30° = √3/2, sin 60° = √3/2, cos 60° = 1/2

Example 7.1.1

Finding a missing side in a right triangle

ProblemA right triangle has a hypotenuse of 13 and an acute angle of θ = 35°. Find the side opposite θ.
ThinkThe opposite side and hypotenuse are related by sine. Set up the ratio and solve for the unknown.
Representsin θ = opposite / hypotenuse
Solvesin 35° = opposite / 13 opposite = 13 sin 35° ≈ 13(0.5736) ≈ 7.46
CheckThe opposite side (7.46) is shorter than the hypotenuse (13) ✓. Since 35° < 45°, the opposite side should be less than the adjacent. The adjacent = 13 cos 35° ≈ 10.65 > 7.46 ✓

Answer: The side opposite the 35° angle is approximately 7.46 units.

Example 7.1.2

Finding an angle in a right triangle

ProblemIn a right triangle, the leg opposite angle A has length 8 and the leg adjacent to A has length 15. Find angle A.
Representtan A = opposite / adjacent = 8 / 15
SolveA = tan⁻¹(8/15) ≈ tan⁻¹(0.5333) ≈ 28.07°
CheckHypotenuse = √(8² + 15²) = √289 = 17. Verify: sin 28.07° ≈ 8/17 ≈ 0.471 ✓

Answer: Angle A ≈ 28.1°

Calculator Mode

When solving for an angle using an inverse trigonometric function, make sure your calculator is in degree mode for problems using degrees, or radian mode for problems using radians. A wrong mode produces an answer that looks completely reasonable but is entirely incorrect.

Section 7.2

Solving Right Triangles

To solve a right triangle means to find all three sides and all three angles. You need at least two pieces of information beyond the right angle — either two sides, or one side and one acute angle. The three most common procedures are summarized below.

Given	Find	Procedure
Acute angle A and hypotenuse c	a, b, and B	a = c sin A; b = c cos A; B = 90° − A
Acute angle A and opposite side a	c, b, and B	c = a / sin A; b = a / tan A; B = 90° − A
Two legs a and b	c, A, and B	c = √(a² + b²); A = tan⁻¹(a/b); B = 90° − A

Example 7.2.1

Solving a right triangle completely

ProblemIn right triangle ABC, angle C = 90°, angle A = 52°, and hypotenuse c = 24. Solve the triangle.
RepresentUse sine for the opposite side, cosine for the adjacent side, and the angle sum for the third angle.
Solvea = c sin A = 24 sin 52° ≈ 24(0.7880) ≈ 18.91 b = c cos A = 24 cos 52° ≈ 24(0.6157) ≈ 14.78 B = 90° − 52° = 38°
CheckPythagorean Theorem: 18.91² + 14.78² ≈ 357.6 + 218.5 = 576.1 ≈ 24² = 576 ✓

Answer: a ≈ 18.91, b ≈ 14.78, B = 38°

Section 7.3

Angles of Elevation and Depression

Many practical problems involve looking upward or downward at an object from a horizontal reference line.

Angle of elevation: The angle formed between a horizontal line and a line of sight directed upward toward an object above.

Angle of depression: The angle formed between a horizontal line and a line of sight directed downward toward an object below.

A key geometric fact: the angle of elevation from the ground to an elevated observer equals the angle of depression from the elevated observer back to the ground. These are alternate interior angles formed by a transversal crossing two parallel horizontal lines — they are equal.

Diagram Instruction

Draw a horizontal reference line. From a point on that line, draw a ray going upward — the angle between the horizontal and the ray is the angle of elevation. From a raised point, draw a ray going downward — the angle between the horizontal and the downward ray is the angle of depression. Label the horizontal clearly in every problem.

Example 7.3.1

Finding the height of a building

ProblemFrom a point 80 feet from the base of a building, the angle of elevation to the top is 62°. How tall is the building?
ThinkHeight is opposite; distance from base is adjacent. Use tangent.
Solvetan 62° = h / 80 h = 80 tan 62° ≈ 80(1.8807) ≈ 150.5 feet
CheckThe angle exceeds 45°, so the height should exceed the base distance. 150.5 > 80 ✓

Answer: The building is approximately 150.5 feet tall.

Example 7.3.2

Using an angle of depression

ProblemA lighthouse keeper 120 feet above sea level spots a boat at an angle of depression of 18°. How far is the boat from the base of the lighthouse?
ThinkThe angle of depression from the top equals the angle of elevation from the boat (alternate interior angles). The height is opposite; the horizontal distance is adjacent.
Solvetan 18° = 120 / d d = 120 / tan 18° ≈ 120 / 0.3249 ≈ 369.3 feet
CheckA shallow angle of depression means the boat is far away. 369 feet for a 120-foot height is reasonable ✓

Answer: The boat is approximately 369 feet from the base of the lighthouse.

Student Warning

The angle of depression is at the top of the triangle — the elevated viewpoint — not at the base. Many errors come from placing the angle incorrectly in the diagram. Draw and label the horizontal reference line at the elevated position before setting up any ratio.

Section 7.4

Oblique Triangles

A triangle with no right angle is called an oblique triangle. Oblique triangles require new tools because the right-triangle ratios alone are not enough. Throughout this chapter we use a standard labeling system: vertices are labeled A, B, and C; the angle at each vertex carries the same letter; and the side opposite each vertex has the corresponding lowercase letter — side a is opposite angle A, side b is opposite angle B, and side c is opposite angle C.

To solve any triangle you need at least three pieces of information, and at least one must be a side length. Four standard cases arise, each calling for a different tool.

Case	Given Information	Tool to Use
AAS or ASA	Two angles and one side	Law of Sines
SSA	Two sides and a non-included angle	Law of Sines — watch for ambiguous case
SAS	Two sides and the included angle	Law of Cosines
SSS	Three sides, no angles	Law of Cosines

Note on AAA

AAA (three angles, no sides) does not uniquely determine a triangle. It determines only the shape — similar triangles all share the same three angles but can be any size. You must always know at least one side length.

Section 7.5

The Law of Sines

The Law of Sines is the first major tool for oblique triangles. It relates every angle in a triangle to the side directly opposite it.

Law of Sines

For any triangle with angles A, B, C and opposite sides a, b, c:

a / sin A = b / sin B = c / sin C

Equivalently: sin A / a = sin B / b = sin C / c

Use the first form to find a missing side. Use the second form to find a missing angle.

Why the Law of Sines Works

Draw an oblique triangle ABC. Drop a perpendicular from C to side AB and call its height h. In the two right triangles formed: h = b sin A and h = a sin B. Since both equal h, set them equal: b sin A = a sin B, which gives a / sin A = b / sin B. A similar argument with a perpendicular from A completes the full proportion. The Law of Sines follows directly from right-triangle trigonometry applied to an altitude.

Example 7.5.1

AAS case — two angles and a non-included side

ProblemIn triangle ABC, A = 40°, B = 75°, and a = 22. Find sides b and c.
ThinkFind the third angle first. Then apply the Law of Sines using the known side a opposite angle A.
SolveC = 180° − 40° − 75° = 65° b = (22 sin 75°) / sin 40° = (22)(0.9659) / 0.6428 ≈ 33.1 c = (22 sin 65°) / sin 40° = (22)(0.9063) / 0.6428 ≈ 31.0
CheckThe largest angle is B = 75°, so the longest side should be b. b ≈ 33.1 > c ≈ 31.0 > a = 22 ✓

Answer: b ≈ 33.1, c ≈ 31.0, C = 65°

Example 7.5.2

ASA case — two angles and the included side

ProblemTwo angles of a triangle are A = 55° and C = 82°, and the side between them is b = 14.5. Find the remaining sides.
SolveB = 180° − 55° − 82° = 43° a = (14.5 sin 55°) / sin 43° = (14.5)(0.8192) / 0.6820 ≈ 17.4 c = (14.5 sin 82°) / sin 43° = (14.5)(0.9903) / 0.6820 ≈ 21.1

Answer: a ≈ 17.4, c ≈ 21.1

Section 7.6

The Ambiguous Case

The ambiguous case arises when you are given two sides and a non-included angle — the SSA configuration. Unlike other triangle cases, SSA does not always determine a unique triangle. Depending on the given values, there may be two triangles, one triangle, or no triangle at all.

This ambiguity occurs because a given side opposite a given angle might swing to two possible positions, one position, or not reach the other side at all. Imagine fixing angle A and side b at one vertex, then swinging side a like a compass arc from the end of b — the arc can cross the base line twice, once, or not at all.

Condition	Number of Triangles
Angle A is obtuse or 90°, and a > b	One triangle
Angle A is obtuse or 90°, and a ≤ b	No triangle
Angle A is acute, and a < h = b sin A	No triangle (side a too short)
Angle A is acute, and a = h = b sin A	One right triangle
Angle A is acute, and h < a < b	Two triangles
Angle A is acute, and a ≥ b	One triangle

Example 7.6.1

SSA — two triangles exist

ProblemGiven A = 35°, a = 10, b = 14. Find all possible triangles.
ThinkAngle A is acute. Compute h = b sin A and compare to a.
Solveh = 14 sin 35° ≈ 14(0.5736) ≈ 8.03 Since h ≈ 8.03 < a = 10 < b = 14 → TWO triangles. sin B = (b sin A) / a = (14)(0.5736) / 10 ≈ 0.8030 B₁ = sin⁻¹(0.8030) ≈ 53.5° Triangle 1: B₁ = 53.5°, C₁ = 180° − 35° − 53.5° = 91.5° c₁ = (10 sin 91.5°) / sin 35° ≈ (10)(0.9997) / 0.5736 ≈ 17.4 B₂ = 180° − 53.5° = 126.5° (check: 35° + 126.5° = 161.5° < 180° ✓) C₂ = 180° − 35° − 126.5° = 18.5° c₂ = (10 sin 18.5°) / sin 35° ≈ (10)(0.3173) / 0.5736 ≈ 5.5

Triangle 1: B ≈ 53.5°, C ≈ 91.5°, c ≈ 17.4 | Triangle 2: B ≈ 126.5°, C ≈ 18.5°, c ≈ 5.5

Example 7.6.2

SSA — no triangle exists

ProblemGiven A = 50°, a = 6, b = 9. How many triangles are possible?
Solveh = b sin A = 9 sin 50° ≈ 9(0.7660) ≈ 6.89 Since a = 6 < h ≈ 6.89, side a does not reach the base.

Answer: No triangle is possible with these measurements.

Student Warning — The Ambiguous Case

The most frequent mistake is finding one value of B with sin⁻¹ and stopping there. Because sin⁻¹ always returns a value between 0° and 90°, you must always check whether 180° minus that value also gives a valid triangle. Ask: "Does A + B₂ < 180°?" If yes, a second triangle exists. Never skip this check in SSA problems.

Section 7.7

The Law of Cosines

When you know two sides and the included angle (SAS), or when you know all three sides (SSS), the Law of Sines is insufficient — there is no matched angle-side pair available. The Law of Cosines fills that gap.

Law of Cosines — Three Forms

a² = b² + c² − 2bc cos A

b² = a² + c² − 2ac cos B

c² = a² + b² − 2ab cos C

To find an angle when all three sides are known, solve for cosine:

cos A = (b² + c² − a²) / (2bc)

cos B = (a² + c² − b²) / (2ac)

cos C = (a² + b² − c²) / (2ab)

Connection to the Pythagorean Theorem

When angle A = 90°, cos A = 0, and the Law of Cosines reduces to a² = b² + c². The Pythagorean Theorem is a special case of the Law of Cosines. The derivation places the triangle in a coordinate system and applies the distance formula; the Pythagorean identity sin²θ + cos²θ = 1 appears naturally, connecting the Law of Cosines directly to the unit circle.

Example 7.7.1

SAS case — finding a side

ProblemIn triangle ABC, b = 15, c = 20, and A = 48°. Find side a.
Representa² = b² + c² − 2bc cos A
Solvea² = 15² + 20² − 2(15)(20) cos 48° = 225 + 400 − 600(0.6691) = 625 − 401.5 = 223.5 a ≈ √223.5 ≈ 14.95
CheckThe triangle inequality requires |c − b| < a < c + b, i.e., 5 < a < 35. Indeed 5 < 14.95 < 35 ✓

Answer: a ≈ 15.0

Example 7.7.2

SSS case — finding all angles

ProblemA triangle has sides a = 8, b = 11, c = 13. Find all three angles.
ThinkFind the largest angle first (opposite the longest side) to reduce error risk. The largest angle is C, opposite c = 13.
Solvecos C = (a² + b² − c²) / (2ab) = (64 + 121 − 169) / (2·8·11) = 16 / 176 ≈ 0.0909 C = cos⁻¹(0.0909) ≈ 84.8° cos B = (a² + c² − b²) / (2ac) = (64 + 169 − 121) / (2·8·13) = 112 / 208 ≈ 0.5385 B = cos⁻¹(0.5385) ≈ 57.4° A = 180° − 84.8° − 57.4° = 37.8°
CheckAngle sum: 37.8° + 57.4° + 84.8° = 180° ✓ Largest angle C is opposite longest side c ✓

Answer: A ≈ 37.8°, B ≈ 57.4°, C ≈ 84.8°

Advantage Over Law of Sines for Angles

When using the Law of Cosines to find an angle, the result is always between 0° and 180° because cos⁻¹ returns a unique value in that range. This means the ambiguous case does not arise when you use the Law of Cosines to find angles.

Section 7.8

Choosing Between Law of Sines and Law of Cosines

Given Information	Which Law	Why
AAS — two angles, non-included side	Law of Sines	Already have a matched pair (angle + opposite side)
ASA — two angles, included side	Law of Sines	Find third angle first; then a matched pair exists
SSA — two sides, non-included angle	Law of Sines	Check for ambiguous case carefully
SAS — two sides, included angle	Law of Cosines	No matched pair yet; included angle is between two sides
SSS — three sides, no angles	Law of Cosines	No angles known at all

Memory Guide

Start with the Law of Sines when you have a matched pair — one angle and the side directly opposite it. Start with the Law of Cosines when you do not have a matched pair. In SAS, the known angle is between two known sides, not opposite a known side. In SSS, no angles are known at all.

Example 7.8.1

Identifying the correct law — four situations

(a)A = 62°, B = 45°, c = 18 → AAS. Find C = 73°, then use Law of Sines.
(b)a = 7, b = 9, C = 110° → SAS. Use Law of Cosines to find c first.
(c)a = 5, b = 8, c = 10 → SSS. Use Law of Cosines to find all angles.
(d)A = 30°, a = 6, b = 10 → SSA. Use Law of Sines, but check for the ambiguous case.

Section 7.9

Area of a Triangle Using Trigonometry

The familiar formula Area = ½bh requires knowing the base and perpendicular height. When the height is not known directly, trigonometry provides it. Drop a perpendicular of height h from vertex C to side c. Since sin A = h/b, we have h = b sin A. Substituting into ½ · c · h gives the trigonometric area formula.

Trigonometric Area Formula

For any triangle with two sides b and c and the included angle A:

Area = (1/2) bc sin A

Similarly: Area = (1/2) ab sin C or Area = (1/2) ac sin B

When all three sides are known, use Heron's Formula. Let s = (a + b + c)/2 (the semi-perimeter). Then:

Area = √(s(s−a)(s−b)(s−c))

Example 7.9.1

Area with SAS

ProblemFind the area of a triangle with sides b = 12 and c = 17 and included angle A = 66°.
SolveArea = (1/2)(12)(17) sin 66° ≈ (1/2)(204)(0.9135) ≈ 93.2 square units

Answer: Area ≈ 93.2 square units

Example 7.9.2

Area using Heron's Formula

ProblemFind the area of a triangle with sides a = 9, b = 12, c = 15.
ThinkNote: 9² + 12² = 81 + 144 = 225 = 15². This is a right triangle. Heron's Formula will confirm the answer.
Solves = (9 + 12 + 15) / 2 = 18 Area = √(18 · 9 · 6 · 3) = √2916 = 54
CheckRight-triangle formula: (1/2)(9)(12) = 54 ✓

Answer: Area = 54 square units

Section 7.10

Bearings and Navigation

Navigation problems require knowing direction as well as distance. Two bearing systems are commonly used:

True bearing: Measured clockwise from north, 0° to 360°. Due east = 090°; due south = 180°; due west = 270°.

Compass bearing: Given as a reference direction (N or S), followed by an acute angle, followed by a second direction (E or W). Example: N 42° E means 42° east of north.

Diagram Instruction

Draw a vertical north-south line through your starting point. Angles are measured from north (up) toward east (right) or west (left), or from south (down) toward east or west. Label the reference direction and the acute angle at every point in the diagram before setting up equations.

Example 7.10.1

The opening question — two-bearing problem

ProblemFrom Tower A, smoke is observed at N 42° E. From Tower B, 15 miles due east of A, the same smoke is observed at N 28° W. How far is the fire from each tower?
ThinkLet the fire be F. Triangle AFB has known side AB = 15. Identify the interior angles at A and B.
SolveInterior angle at A: N 42° E means 42° east of north. Side AB runs due east (90° from north). Angle at A = 90° − 42° = 48° Interior angle at B: N 28° W means 28° west of north. Side BA runs due west (90° from north). Angle at B = 90° − 28° = 62° Angle at F = 180° − 48° − 62° = 70° Law of Sines (AB = 15 is opposite angle F): AF = (15 sin 62°) / sin 70° ≈ (15)(0.8829) / 0.9397 ≈ 14.1 miles BF = (15 sin 48°) / sin 70° ≈ (15)(0.7431) / 0.9397 ≈ 11.9 miles

Answer: The fire is approximately 14.1 miles from Tower A and 11.9 miles from Tower B.

Example 7.10.2

Two-leg navigation problem

ProblemShip A sails 35 km in the direction N 68° E, then 50 km in the direction S 20° E. Find the straight-line distance from port to the final position.
ThinkResolve each leg into x (east) and y (north) components. Add components, then apply the distance formula.
SolveLeg 1 (N 68° E → 68° from north toward east): x₁ = 35 sin 68° ≈ 32.45 km y₁ = 35 cos 68° ≈ 13.11 km Leg 2 (S 20° E → south and slightly east): x₂ = x₁ + 50 sin 20° ≈ 32.45 + 17.10 = 49.55 km y₂ = y₁ − 50 cos 20° ≈ 13.11 − 46.98 = −33.87 km d = √(49.55² + 33.87²) ≈ √(2455.2 + 1147.2) ≈ √3602.4 ≈ 60.0 km

Answer: The ship is approximately 60.0 km from port.

Section 7.11

Surveying and Indirect Measurement

Surveying uses trigonometry to measure land features when direct measurement is not possible. The key principle is always the same: identify a triangle (or system of triangles) whose missing measurements can be found using the laws of sines and cosines.

Example 7.11.1

Width of a river

ProblemFrom point A on one bank, a stake C is driven directly across the river. A surveyor walks 200 feet to point B, then measures: angle at A = 82°, angle at B = 55°. Find the width (perpendicular distance from AB to C).
SolveAngle at C = 180° − 82° − 55° = 43° AC = (200 sin 55°) / sin 43° ≈ (200)(0.8192) / 0.6820 ≈ 240.1 ft Width h = AC sin A = 240.1 sin 82° ≈ 240.1(0.9903) ≈ 237.8 ft

Answer: The river is approximately 237.8 feet wide.

Example 7.11.2

Two-angle height measurement

ProblemFrom level ground, the angle of elevation to a hilltop is 28°. After walking 300 feet toward the hill, the elevation angle is 43°. Find the height of the hill.
ThinkLet h = height and d = distance from the nearer point to the base. Form two right-triangle equations.
SolveFrom far point: tan 28° = h / (d + 300) From near point: tan 43° = h / d → d = h / tan 43° Substitute: tan 28° = h / (h/tan 43° + 300) tan 28° · (h/tan 43° + 300) = h 300 tan 28° = h(1 − tan 28°/tan 43°) h = 300 tan 28° / (1 − tan 28°/tan 43°) = 300(0.5317) / (1 − 0.5317/0.9325) ≈ 159.5 / 0.4297 ≈ 371 feet

Answer: The hill is approximately 371 feet tall.

Section 7.12

Trigonometry in Physics and Engineering

Trigonometry is essential in physics and engineering because many quantities — forces, velocities, displacements — have both magnitude and direction. When a force acts at an angle, finding its horizontal and vertical components using trigonometry is a fundamental step. For a force F acting at angle θ above the horizontal:

Force Component Formulas

F_x = F cos θ (horizontal component)

F_y = F sin θ (vertical component)

This decomposition is the foundation of vector analysis, developed formally in Chapter 12.

Example 7.12.1

Inclined plane

ProblemA 150-pound block rests on a ramp inclined at 22°. Find the component of weight parallel to the ramp (tending to cause sliding) and perpendicular to the ramp (pressing into the surface).
SolveParallel (down the slope): W_∥ = 150 sin 22° ≈ 150(0.3746) ≈ 56.2 lb Perpendicular (into the surface): W_⊥ = 150 cos 22° ≈ 150(0.9272) ≈ 139.1 lb
Check√(56.2² + 139.1²) ≈ √(3158 + 19349) ≈ √22507 ≈ 150 lb ✓

Parallel: ≈ 56.2 lb (tends to slide the block) | Perpendicular: ≈ 139.1 lb (presses into the ramp)

Example 7.12.2

Combining two non-perpendicular forces

ProblemRope 1 pulls with 120 N at 30° above horizontal. Rope 2 pulls with 95 N at 55° above horizontal (same side). Find the combined force.
SolveRope 1: F₁ₓ = 120 cos 30° ≈ 103.9 N, F₁ᵧ = 120 sin 30° = 60.0 N Rope 2: F₂ₓ = 95 cos 55° ≈ 54.5 N, F₂ᵧ = 95 sin 55° ≈ 77.8 N Totals: Rₓ = 103.9 + 54.5 = 158.4 N Rᵧ = 60.0 + 77.8 = 137.8 N |R| = √(158.4² + 137.8²) ≈ √(25090 + 18989) ≈ √44079 ≈ 209.9 N θ = tan⁻¹(137.8 / 158.4) ≈ tan⁻¹(0.870) ≈ 41.0°

Answer: Combined force ≈ 210 N at approximately 41° above horizontal.

MGU Library Connection

This section previews vectors, developed formally in Chapter 12: Parametric Equations, Polar Coordinates, and Vectors.
Force decomposition and vector addition appear extensively in MGU Physics I.
Computer graphics courses use identical techniques to animate motion and simulate forces.

Section 7.13

Common Triangle Trigonometry Mistakes

Mistake 1 — Law of Sines without a matched pair

Applying the Law of Sines to an SAS problem where no complete angle-side pair exists yet.

Correction: In SAS and SSS, start with the Law of Cosines. The Law of Sines requires one known angle and the side directly opposite it.

Mistake 2 — Stopping after one triangle in the ambiguous case

Using sin⁻¹ to find angle B and writing the solution without checking whether B₂ = 180° − B also works.

Correction: In every SSA problem, compute B₂ = 180° − B₁ and check whether A + B₂ < 180°. If yes, a second valid triangle exists.

Mistake 3 — Calculator in the wrong mode

Leaving the calculator in radian mode while solving a degree problem, or vice versa.

Correction: Check calculator mode before every computation. Most chapter 7 problems use degrees.

Mistake 4 — Incorrect triangle labeling

Pairing side b with angle A, or otherwise mismatching sides and their opposite angles in the Law of Sines.

Correction: Before applying either law, clearly label all six parts using the a–A, b–B, c–C convention. Sketch the triangle even for abstract problems.

Mistake 5 — Using sin⁻¹ for obtuse angles in SSS

Finding all angles in an SSS triangle using the Law of Sines and getting an acute angle where an obtuse angle belongs.

Correction: In SSS problems, use the Law of Cosines to find angles. cos⁻¹ returns values up to 180° and handles obtuse angles correctly.

Mistake 6 — Assuming all triangles are solvable

Proceeding to solve without checking that the given values can actually form a triangle.

Correction: The triangle inequality requires each side to be less than the sum of the other two. In SSA, compute h = b sin A and compare to a before proceeding.

Mistake 7 — Not checking angle sums

Finishing a solution without verifying that A + B + C = 180°.

Correction: Always add the three angles as a final check. This step catches most computational errors at no additional effort.

Mistake 8 — Misplacing the angle of depression

Placing the angle of depression at the base of the right triangle instead of at the elevated viewpoint.

Correction: The angle of depression is measured downward from the horizontal at the elevated position. Draw and label the horizontal reference line at the top of the diagram, not the bottom.

Section 7.14

Preparing for Polar and Parametric Thinking

Chapter 7 has shown that trigonometry is a system for measuring triangles. As you move into the next part of this textbook, trigonometry will appear in a broader context: describing position, direction, motion, and curves.

Connection to polar coordinates: In Chapter 12, every point in the plane is described by a distance r and an angle θ. The connection to triangle trigonometry is immediate — x = r cos θ and y = r sin θ are the same component relationships used throughout this chapter.

Connection to parametric equations: A circle of radius r can be described parametrically as x = r cos t, y = r sin t. The motion of any object on a circular path — a satellite in orbit, a wheel turning — is described by exactly these equations.

Connection to vectors: The force problems of Section 7.12 previewed vectors. When you decomposed a force into horizontal and vertical components, you were writing a vector in component form: ⟨F cos θ, F sin θ⟩. Chapter 12 develops vectors formally.

Connection to the dot product: The Law of Cosines and the vector dot product are the same idea expressed differently. If u and v are vectors with angle θ between them, then u · v = |u||v| cos θ. Recognizing this connection will make Chapter 12 and Calculus III feel natural.

Looking Ahead — Part III

Advanced Algebraic Functions

You have now completed Part II. You have built trigonometry systematically: angles as rotations (Chapter 3), the unit circle (Chapter 4), periodic graphs (Chapter 5), identities and equations (Chapter 6), and triangle measurement in the physical world (Chapter 7). In Part III, the book will deepen the algebraic function work begun in Part I — polynomial and rational functions, exponential and logarithmic functions, sequences, and series. Those chapters will sharpen your algebraic skills and extend your modeling toolkit. Then, in Part IV, everything converges: Chapter 12 will bring polar coordinates, parametric equations, and vectors, and there, every triangle ratio you practiced in this chapter will become the formal language of position, motion, and force.

Chapter 7 — Back Matter

Chapter Summary

Sec 7.1Right-triangle ratios define sine, cosine, and tangent as opposite/hypotenuse, adjacent/hypotenuse, and opposite/adjacent. The Pythagorean Theorem connects the three sides. Special triangles (45-45-90 and 30-60-90) give exact values.
Sec 7.2Solving a right triangle means finding all three sides and all three angles. With any two pieces of information (beyond the right angle), the remaining parts can be found using the six trigonometric ratios.
Sec 7.3Angles of elevation are measured upward from horizontal; angles of depression are measured downward. The angle of elevation from the ground equals the angle of depression from the elevated viewpoint (alternate interior angles).
Sec 7.4An oblique triangle has no right angle. The standard labeling pairs each side with the lowercase letter of the opposite vertex's angle. Four solvable cases exist: AAS, ASA, SSA, SAS, SSS.
Sec 7.5The Law of Sines: a/sin A = b/sin B = c/sin C. It applies to AAS and ASA cases. The proof uses a perpendicular altitude, connecting the law directly to right-triangle trigonometry.
Sec 7.6The ambiguous case (SSA) can produce two triangles, one triangle, or no triangle. Always check whether a second triangle exists by testing whether B₂ = 180° − B₁ is valid.
Sec 7.7The Law of Cosines: a² = b² + c² − 2bc cos A. It applies to SAS and SSS. When A = 90° it reduces to the Pythagorean Theorem. Use cos⁻¹ to find angles — no ambiguous case arises.
Sec 7.8Start with the Law of Sines when a matched angle-side pair exists. Start with the Law of Cosines when no matched pair exists (SAS or SSS).
Sec 7.9Area = (1/2)bc sin A for any triangle with sides b, c and included angle A. Heron's Formula, Area = √(s(s−a)(s−b)(s−c)), uses only the three side lengths.
Sec 7.10True bearings measure clockwise from north (0°–360°). Compass bearings give an acute angle east or west of north or south. Resolve each leg into components and apply triangle laws or the distance formula.
Sec 7.11Surveying problems identify triangles from field data and apply the Law of Sines or Cosines. Indirect measurement finds distances and heights that cannot be measured directly.
Sec 7.12Forces are resolved into horizontal and vertical components using F_x = F cos θ and F_y = F sin θ. Resultant forces are found by adding components, then computing magnitude and direction.
Sec 7.13Eight common mistakes: wrong law choice, missing the second triangle in SSA, wrong calculator mode, mismatched labels, using sin⁻¹ for obtuse angles, invalid triangles, unchecked angle sums, misplaced elevation/depression angles.
Sec 7.14Triangle trigonometry connects directly to polar coordinates (x = r cos θ, y = r sin θ), parametric circles, vector component form, and the dot product. The triangle is the fundamental unit of indirect measurement.

Key Terms Review

oblique triangle

A triangle that contains no right angle. Requires the Law of Sines or Law of Cosines to solve.

Law of Sines

a / sin A = b / sin B = c / sin C. Applies to AAS, ASA, and SSA configurations.

Law of Cosines

a² = b² + c² − 2bc cos A. Applies to SAS and SSS. Reduces to the Pythagorean Theorem when A = 90°.

ambiguous case

The SSA configuration where the given data may produce two triangles, one triangle, or no triangle.

angle of elevation

The angle measured upward from a horizontal line to a line of sight aimed at an object above.

angle of depression

The angle measured downward from a horizontal line to a line of sight aimed at an object below.

true bearing

Direction measured clockwise from north, expressed in degrees from 000° to 360°.

compass bearing

Direction expressed as N or S followed by an acute angle followed by E or W. Example: S 35° W.

Heron's Formula

Area = √(s(s−a)(s−b)(s−c)) where s = (a + b + c)/2. Computes triangle area from three sides alone.

included angle

The angle formed between two known sides of a triangle. The SAS case provides an included angle.

included side

The side between two known angles of a triangle. The ASA case provides an included side.

semi-perimeter

Half the perimeter of a triangle: s = (a + b + c)/2. Used in Heron's Formula.

component form

Expressing a force or vector as horizontal and vertical parts: ⟨F cos θ, F sin θ⟩. Foundational for vectors in Chapter 12.

Concept Review Questions

Answer each question in complete sentences without performing calculations.

What are the six trigonometric ratios for a right triangle, and how is each defined in terms of the three sides?
Why does the Law of Sines work for any triangle, not just right triangles? Describe the geometric idea behind the proof using an altitude.
What is the ambiguous case in triangle trigonometry, and when does it arise?
Why might the SSA case produce two valid triangles? Use a compass-arc description to explain.
State the Law of Cosines in all three forms. What famous theorem is a special case, and why?
How do you decide whether to use the Law of Sines or the Law of Cosines for a given triangle problem?
Why is it advisable to find the largest angle of a triangle first in SSS problems?
What does it mean to "solve a triangle completely"?
Explain the difference between angle of elevation and angle of depression, and state the geometric relationship between them.
What is the difference between a true bearing and a compass bearing?
In force problems, what does it mean to "resolve" a force into components?
How does triangle trigonometry connect to the polar coordinate system in Chapter 12?

Practice ARight Triangle Practice

Problems 1–6: Find the indicated part. Angle C = 90° throughout.

A = 41°, c = 18. Find a.
B = 67°, c = 30. Find b.
a = 7, c = 25. Find A and B.
a = 10, b = 24. Find c and A.
A = 28°, a = 9. Find b and c.
b = 16, B = 53°. Find a and c.

Problems 7–10: Elevation and depression.

From 60 ft from a flagpole's base, angle of elevation to the top is 48°. Height of the flagpole?
A bird 45 ft high has an angle of depression to a worm of 32°. Distance from base to worm?
Angle of depression from a 200-ft hilltop to a car below is 14°. Distance from hill base?
From the top of an 80-ft building, angle of elevation to a taller building across the street is 35°; angle of depression to that building's base is 22°. Find the height of the taller building.

Practice BLaw of Sines Practice

Problems 11–16: Solve each triangle. Round to the nearest tenth.

A = 38°, B = 72°, a = 15
A = 55°, C = 80°, c = 22
B = 44°, C = 61°, b = 18
A = 33°, B = 79°, c = 30
A = 65°, B = 48°, b = 12
B = 52°, C = 73°, a = 25

Problems 17–22: Ambiguous case (SSA). Determine the number of possible triangles, then solve each.

A = 40°, a = 12, b = 10
A = 25°, a = 7, b = 12
A = 60°, a = 18, b = 20
A = 110°, a = 25, b = 18
A = 30°, a = 6, b = 12
A = 45°, a = 14, b = 16

Practice CLaw of Cosines Practice

Round to the nearest tenth.

b = 10, c = 14, A = 50°. Find a.
a = 8, c = 11, B = 72°. Find b.
a = 5, b = 7, c = 9. Find all angles.
a = 12, b = 15, c = 18. Find all angles.
a = 20, b = 20, C = 100°. Find c.
b = 6, c = 8, A = 120°. Find a.

Practice DArea Practice

b = 9, c = 14, A = 42°
a = 20, c = 15, B = 78°
a = 6, b = 8, c = 10
a = 5, b = 7, c = 9
A = 60°, b = 12, c = 12
A = 30°, B = 45°, a = 10

Application Problems

Define all variables, set up the triangle, apply the correct law, and interpret your answer in context.

Navigation: A plane leaves an airport and flies 180 miles on a bearing of N 52° E. It then turns and flies 240 miles on a bearing of S 38° E. How far is the plane from the airport?
Lake survey: From point A on one shore, a baseline AC = 520 feet is measured along the shore to point C. From A, angle BAC = 74°. From C, angle BCA = 63°. Find the distance AB across the lake.
Hiking: A hiker walks 3.5 km due north, then turns and walks 5.2 km on a bearing of N 65° E. How far is the hiker from the start, and on what bearing?
Structural brace: A diagonal brace in a wall frame is 8.5 feet long and makes a 34° angle with the floor. Find the horizontal span it covers and its vertical rise.
Harbor survey: Three buoys A, B, C mark a triangular harbor section. AB = 820 m, BC = 640 m, CA = 580 m. Find the area of the triangular region and the angle at buoy B.
Rope tension: A sign weighing 250 N is suspended from two ropes. The left rope makes 35° with the horizontal; the right rope makes 50° with the horizontal. Find the tension in each rope.
Sailing race: Buoy A to Buoy B is 0.8 miles on a bearing of S 72° E. Buoy A to Buoy C is 1.1 miles on a bearing of S 18° W. Find the distance from B to C.
Forest fire (opening question): From Tower 1, smoke appears at 32° east of north. Tower 2 is 6.5 miles directly south of Tower 1. The ranger at Tower 2 spots the same smoke at 28° west of north. Find the distance from each tower to the fire.
Two towers: From a new tower site, Tower X (45 m tall) is observed at an elevation angle of 28° and Tower Y (60 m tall) is observed at an elevation angle of 34°. Find the distance from the new site to each tower's base.
Inclined plane: A 200-lb crate rests on a loading ramp inclined at 18°. Find the components of its weight parallel and perpendicular to the ramp, and verify using the Pythagorean Theorem.

Error Analysis

Each problem shows a student error. Identify the mistake and explain the correct approach. Do not simply give the correct numerical answer — explain what went wrong and why.

Problem 1

SSA: A = 55°, a = 9, b = 11. Student finds sin B = (11 sin 55°)/9 ≈ 1.001, writes "B ≈ 90°, one triangle found."

What is the actual error? What does sin B > 1 tell you about this configuration?

Problem 2

SSS: a = 6, b = 8, c = 11. Student attempts to apply the Law of Sines directly, writing sin C / 11 = sin A / 6, then stops because no angles are known.

Why can the Law of Sines not be used here? What should the student do instead?

Problem 3

Area problem: a = 7, b = 9, C = 40°. Student writes Area = (1/2)(7)(9) sin 40° but then says "Wait — C is opposite c, not between a and b. So I should use sin B instead."

Is the student's original setup correct or incorrect? Explain which angle appears in the area formula and why.

Problem 4

Angle of depression: 25° from a 300-m cliff. Student writes tan 25° = horizontal / 300, so horizontal = 300 tan 25° ≈ 139.9 m.

Identify the error in the ratio setup. What is the correct equation and correct answer?

Problem 5

SSA: A = 35°, a = 7, b = 10. Student finds sin B ≈ 0.8192, writes B ≈ 55.0°, C = 90°, and stops — declaring one triangle with C = 90°.

What step did the student omit? What should the full solution look like?

Chapter 7 Checkpoint

Complete without assistance — assesses all major skills and concepts from Chapter 7 · Round to the nearest tenth

Part A — Right Triangles and Elevation/Depression

In right triangle ABC with C = 90°, find b and c given A = 47° and a = 13.
From the top of a 90-foot observation deck, the angle of depression to a ground marker is 31°. How far is the marker from the base?

Part B — Law of Sines

Solve: A = 44°, B = 71°, a = 20.
SSA case: A = 40°, a = 10, b = 14. Determine how many triangles are possible and solve completely for each one.

Part C — Law of Cosines

Find the missing side: a = 11, b = 16, C = 58°.
Find all three angles of a triangle with a = 7, b = 9, c = 12.

Part D — Area

Find the area of a triangle with b = 18, c = 24, and included angle A = 37°.
Find the area of a triangle with a = 10, b = 13, c = 15 using Heron's Formula.

Part E — Applications

A ship leaves port and travels 42 km on a bearing of N 35° E, then 60 km on a bearing of S 55° E. How far is the ship from port?
From point P, a surveyor measures to points A and B on opposite sides of a canyon: PA = 250 m, PB = 310 m, and angle APB = 78°. Find the width AB across the canyon.

Part F — Choosing the Right Method

For each situation, state which law to use and give one reason why. Do not solve.

(a) A = 62°, a = 14, b = 11 (b) a = 9, b = 12, C = 80° (c) A = 50°, B = 65°, c = 18 (d) a = 5, b = 8, c = 10

Part G — Error Analysis

A student solves triangle ABC with A = 30°, a = 6, b = 12 and reports one triangle with B = 90°. Is this complete? What is missing?
A student uses the Law of Sines to find angles in an SSS triangle and obtains B = 35°, C = 50°, A = 95°. Explain why the student should have used the Law of Cosines instead and identify a potential error this approach could have introduced.

Answers may be placed in the answer key.