Trigonometric functions are defined using a right triangle.

Mathematics · FE Reference Handbook section

Mathematics

8 formulas

10 exam-style examples

~60 min

All Mathematics lectures

Learning objectives

What you must be able to do before leaving this section.

This chapter section covers Trigonometric functions are defined using a right triangle. within Mathematics. Read it the way you would read a textbook chapter: the theory first so the relations mean something, then every equation with its use and its trap, then 10 fully worked examples with the arithmetic shown line by line, and finally a self-check you should be able to answer without notes.

Explain, in your own words, what trigonometric functions are defined using a right triangle. describes physically and when it applies.
State every one of the 8 relations the handbook lists here and name each symbol with its unit.
Select the correct relation from the wording of an exam stem within 20 seconds.
Carry a complete solution from givens to a "most nearly" answer with the correct unit.
Recognise the distractors generated by the unit trap: radians vs degrees — set the calculator before the first trig entry.

Lecture

Why this section exists. Trigonometric functions are defined using a right triangle. is the part of Mathematics that lets you connect an algebraic or calculus expression that must be evaluated exactly to a number you can defend. Before any equation is useful you must be able to picture the physical situation it describes; the schematic below is that picture.

How the theory is built. The handbook prints results, not derivations. Each relation in this section comes from one governing principle applied to the idealised system: state the principle, impose the stated assumptions, and the printed equation follows. Knowing which assumption each relation rests on is what lets you reject a wrong answer choice in seconds.

How it is examined. Items from this page are written as short symbolic manipulations with one numeric evaluation at the end. Roughly two thirds are direct substitution, one third require one intermediate quantity from a neighbouring relation, and a small number are conceptual — testing whether you know the assumption, not the arithmetic.

The habit that earns the points. Unit discipline. radians vs degrees — set the calculator before the first trig entry. Every relation below is dimensionally consistent only when that rule is honoured, and the distractor set is deliberately built from candidates who ignored it. Write the unit next to every number you substitute, every time.

How to study this page. Read the theory, then cover the formula cards and try to reproduce each relation from its description. Then work the examples with the solution hidden, revealing one line at a time. Finish with the self-check questions; if you cannot answer one, return to the matching formula card.

Three engineers in hard hats and safety vests reviewing drawings on a truck tailgate. — Photo 1. Where this shows up in practice: trigonometric functions are defined using a right triangle..
Capstone Studio instructional photograph

Mathematics — Trigonometric functions are defined using a right triangle.: reference schematic for orienting the symbols used in this section.

Theory, developed

Read this before the equations — it is what makes them memorable.

The physical situation

Every item from this section describes an algebraic or calculus expression that must be evaluated exactly. Sketch it before you compute — a labelled sketch with the givens on it converts a wordy stem into a solvable problem and exposes the quantity the examiner left out on purpose.

The governing principle

The 8 relations on this page are consequences of one principle applied to that idealised system. Identify which quantity is conserved, balanced, or defined, and the correct equation follows without memorisation.

Assumptions and limits of validity

Each printed relation carries silent assumptions — linearity, steady state, uniformity, small deformation, or standard conditions, depending on the subject. Conceptual exam items are written by violating exactly one of these, so read the sentence above the equation as carefully as the equation itself.

Solution procedure you should automate

1) Read the last sentence of the stem to identify the requested quantity. 2) Locate the relation on this page whose left-hand side is that quantity. 3) Tabulate the givens with units and mark the missing symbol. 4) If a symbol is missing, find the one relation that produces it. 5) Rearrange symbolically, substitute once, evaluate, and round only at the end.

Notation used in this section

sin θ	Quantity produced by "sin θ = y/r, cos θ = x/r" — read its definition and unit from the handbook line directly above the equation.
tan θ	Quantity produced by "tan θ = y/x, cot θ = x/y" — read its definition and unit from the handbook line directly above the equation.
csc θ	Quantity produced by "csc θ = r/y, sec θ = r/x" — read its definition and unit from the handbook line directly above the equation.
θ	Quantity produced by "θ" — read its definition and unit from the handbook line directly above the equation.
a2	Quantity produced by "a2 = b2 + c2 – 2bc cos A" — read its definition and unit from the handbook line directly above the equation.
b2	Quantity produced by "b2 = a2 + c2 – 2ac cos B" — read its definition and unit from the handbook line directly above the equation.
c2	Quantity produced by "c2 = a2 + b2 – 2ab cos C" — read its definition and unit from the handbook line directly above the equation.

Handbook notes for this section

Definitions and conditions exactly as the handbook states them.

sin A sin B sin C
Law of Cosines
Brink, R.W., A First Year of College Mathematics, D. Appleton-Century Co., Inc., Englewood Cliffs, NJ, 1937.

Core formulas for this FE topic

Definitions, applicability, units, assumptions and worked examples for each relation.

Worked exam-style examples

The four ways this section is written on the real exam — thoughts first, then equations, then substitution.

Example 1

Closing a traverse leg with the law of cosines

Two traverse legs of 120 m and 90 m meet at an included angle of 68°. What is the closing distance between the far ends?

Given

a = 120 m
b = 90 m
C = 68°

Find

Closing length c

Start with the thinking

Two sides and the included angle → law of cosines.
Check that c falls between |a − b| and a + b.

Step-by-step solution

Law of cosines
$c^{2} = a^{2} + b^{2} - 2 ab cos C$
Substitute
$c^{2} = 12 0^{2} + 9 0^{2} - 2 (120) (90) cos 6 8^{\circ}$
Terms
$14, 400 + 8, 100 - 21, 600 (0.3746) = 22, 500 - 8, 091$
Result
$c^{2} = 14, 409, soc = 120.0 m$
Range check
$∣120 - 90∣ = 30 < 120.0 < 210 ✓$

Answer: c ≈ 120 m

Why the other options are there

150 m (Pythagoras used, angle ignored)
170 m (cosine term added instead of subtracted)

Reference: FE Reference Handbook — Mathematics — Trigonometry

Example 2

Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle.

Two sides of a triangle measure 138 ft and 71 ft with an included angle of 96°. Most nearly, what is the third side and the enclosed area?

Given

b = 138 ft
c = 71 ft
A = 96°

Find

Side a and the area

Start with the thinking

Two sides plus the included angle is the law-of-cosines case.
Set the calculator to degrees before the first trig entry.

Figure for Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle.

Step-by-step solution

Law of cosines — a² = b² + c² − 2bc·cos A
Substituting
$a^{2} = 13 8^{2} + 7 1^{2} - 2 (138) (71) cos 9 6^{\circ}$
Evaluate
$a^{2} = 24, 085 - - 2, 048 = 26, 133$
Side
$a = 161.7 f t$
Area — A = ½bc·sin A
Substituting
$A = 0.5 (138) (71) sin 9 6^{\circ} = 4, 872 f t^{2}$

Answer: a ≈ 161.7 ft, area ≈ 4,872 ft²

Why the other options are there

209.0 ft (sides added)
155.2 ft (Pythagoras used with a non-right angle)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 3

Verifying trigonometric identities numerically — Trigonometric functions are defined using a right triangle.

For θ = 43°, evaluate sin θ and cos θ, verify the Pythagorean identity sin²θ + cos²θ = 1 to three decimals, and compute sin 2θ using the double-angle identity.

Given

θ = 43°
θ = 0.7505 rad

Find

sin θ, cos θ, sin²θ + cos²θ, and sin 2θ

Start with the thinking

Identities are exact relations, so a numerical check must return 1.000 within rounding.
The double-angle identity avoids re-entering 2θ into the calculator.

Step-by-step solution

Values
$sin 4 3^{\circ} = 0.6820, cos 4 3^{\circ} = 0.7314$
Formula
$sin^{2} θ + cos^{2} θ = 1$
Substituting
$(0.6820)^{2} + (0.7314)^{2} = 0.4651 + 0.5349 = 1.0000$
Formula
$sin 2 θ = 2 sin θ cos θ$
Substituting
$sin 8 6^{\circ} = 2 (0.6820) (0.7314) = 0.9976$

Answer: sin²θ + cos²θ = 1.000 (identity verified); sin 2θ = 0.9976

Why the other options are there

sin 2θ = 1.3640 (doubled the sine instead of the angle)
1.4134 (added sine and cosine)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 4

Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (2)

Two sides of a triangle measure 88 ft and 67 ft with an included angle of 104°. Most nearly, what is the third side and the enclosed area?

Given

b = 88 ft
c = 67 ft
A = 104°

Find

Side a and the area

Start with the thinking

Two sides plus the included angle is the law-of-cosines case.
Set the calculator to degrees before the first trig entry.

Figure for Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (2)

Step-by-step solution

Law of cosines — a² = b² + c² − 2bc·cos A
Substituting
$a^{2} = 8 8^{2} + 6 7^{2} - 2 (88) (67) cos 10 4^{\circ}$
Evaluate
$a^{2} = 12, 233 - - 2, 853 = 15, 086$
Side
$a = 122.8 f t$
Area — A = ½bc·sin A
Substituting
$A = 0.5 (88) (67) sin 10 4^{\circ} = 2, 860 f t^{2}$

Answer: a ≈ 122.8 ft, area ≈ 2,860 ft²

Why the other options are there

155.0 ft (sides added)
110.6 ft (Pythagoras used with a non-right angle)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 5

Verifying trigonometric identities numerically — Trigonometric functions are defined using a right triangle. (2)

For θ = 19°, evaluate sin θ and cos θ, verify the Pythagorean identity sin²θ + cos²θ = 1 to three decimals, and compute sin 2θ using the double-angle identity.

Given

θ = 19°
θ = 0.3316 rad

Find

sin θ, cos θ, sin²θ + cos²θ, and sin 2θ

Start with the thinking

Identities are exact relations, so a numerical check must return 1.000 within rounding.
The double-angle identity avoids re-entering 2θ into the calculator.

Step-by-step solution

Values
$sin 1 9^{\circ} = 0.3256, cos 1 9^{\circ} = 0.9455$
Formula
$sin^{2} θ + cos^{2} θ = 1$
Substituting
$(0.3256)^{2} + (0.9455)^{2} = 0.1060 + 0.8940 = 1.0000$
Formula
$sin 2 θ = 2 sin θ cos θ$
Substituting
$sin 3 8^{\circ} = 2 (0.3256) (0.9455) = 0.6157$

Answer: sin²θ + cos²θ = 1.000 (identity verified); sin 2θ = 0.6157

Why the other options are there

sin 2θ = 0.6511 (doubled the sine instead of the angle)
1.2711 (added sine and cosine)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 6

Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (3)

Two sides of a triangle measure 90 ft and 85 ft with an included angle of 100°. Most nearly, what is the third side and the enclosed area?

Given

b = 90 ft
c = 85 ft
A = 100°

Find

Side a and the area

Start with the thinking

Two sides plus the included angle is the law-of-cosines case.
Set the calculator to degrees before the first trig entry.

Figure for Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (3)

Step-by-step solution

Law of cosines — a² = b² + c² − 2bc·cos A
Substituting
$a^{2} = 9 0^{2} + 8 5^{2} - 2 (90) (85) cos 10 0^{\circ}$
Evaluate
$a^{2} = 15, 325 - - 2, 657 = 17, 982$
Side
$a = 134.1 f t$
Area — A = ½bc·sin A
Substituting
$A = 0.5 (90) (85) sin 10 0^{\circ} = 3, 767 f t^{2}$

Answer: a ≈ 134.1 ft, area ≈ 3,767 ft²

Why the other options are there

175.0 ft (sides added)
123.8 ft (Pythagoras used with a non-right angle)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 7

Verifying trigonometric identities numerically — Trigonometric functions are defined using a right triangle. (3)

For θ = 61°, evaluate sin θ and cos θ, verify the Pythagorean identity sin²θ + cos²θ = 1 to three decimals, and compute sin 2θ using the double-angle identity.

Given

θ = 61°
θ = 1.0647 rad

Find

sin θ, cos θ, sin²θ + cos²θ, and sin 2θ

Start with the thinking

Identities are exact relations, so a numerical check must return 1.000 within rounding.
The double-angle identity avoids re-entering 2θ into the calculator.

Step-by-step solution

Values
$sin 6 1^{\circ} = 0.8746, cos 6 1^{\circ} = 0.4848$
Formula
$sin^{2} θ + cos^{2} θ = 1$
Substituting
$(0.8746)^{2} + (0.4848)^{2} = 0.7650 + 0.2350 = 1.0000$
Formula
$sin 2 θ = 2 sin θ cos θ$
Substituting
$sin 12 2^{\circ} = 2 (0.8746) (0.4848) = 0.8480$

Answer: sin²θ + cos²θ = 1.000 (identity verified); sin 2θ = 0.8480

Why the other options are there

sin 2θ = 1.7492 (doubled the sine instead of the angle)
1.3594 (added sine and cosine)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 8

Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (4)

Two sides of a triangle measure 130 ft and 134 ft with an included angle of 44°. Most nearly, what is the third side and the enclosed area?

Given

b = 130 ft
c = 134 ft
A = 44°

Find

Side a and the area

Start with the thinking

Two sides plus the included angle is the law-of-cosines case.
Set the calculator to degrees before the first trig entry.

Figure for Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (4)

Step-by-step solution

Law of cosines — a² = b² + c² − 2bc·cos A
Substituting
$a^{2} = 13 0^{2} + 13 4^{2} - 2 (130) (134) cos 4 4^{\circ}$
Evaluate
$a^{2} = 34, 856 - 25, 062 = 9, 794$
Side
$a = 98.97 f t$
Area — A = ½bc·sin A
Substituting
$A = 0.5 (130) (134) sin 4 4^{\circ} = 6, 050 f t^{2}$

Answer: a ≈ 98.97 ft, area ≈ 6,050 ft²

Why the other options are there

264.0 ft (sides added)
186.7 ft (Pythagoras used with a non-right angle)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 9

Verifying trigonometric identities numerically — Trigonometric functions are defined using a right triangle. (4)

For θ = 52°, evaluate sin θ and cos θ, verify the Pythagorean identity sin²θ + cos²θ = 1 to three decimals, and compute sin 2θ using the double-angle identity.

Given

θ = 52°
θ = 0.9076 rad

Find

sin θ, cos θ, sin²θ + cos²θ, and sin 2θ

Start with the thinking

Identities are exact relations, so a numerical check must return 1.000 within rounding.
The double-angle identity avoids re-entering 2θ into the calculator.

Step-by-step solution

Values
$sin 5 2^{\circ} = 0.7880, cos 5 2^{\circ} = 0.6157$
Formula
$sin^{2} θ + cos^{2} θ = 1$
Substituting
$(0.7880)^{2} + (0.6157)^{2} = 0.6210 + 0.3790 = 1.0000$
Formula
$sin 2 θ = 2 sin θ cos θ$
Substituting
$sin 10 4^{\circ} = 2 (0.7880) (0.6157) = 0.9703$

Answer: sin²θ + cos²θ = 1.000 (identity verified); sin 2θ = 0.9703

Why the other options are there

sin 2θ = 1.5760 (doubled the sine instead of the angle)
1.4037 (added sine and cosine)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Example 10

Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (5)

Two sides of a triangle measure 95 ft and 60 ft with an included angle of 45°. Most nearly, what is the third side and the enclosed area?

Given

b = 95 ft
c = 60 ft
A = 45°

Find

Side a and the area

Start with the thinking

Two sides plus the included angle is the law-of-cosines case.
Set the calculator to degrees before the first trig entry.

Figure for Law of cosines applied to a traverse leg — Trigonometric functions are defined using a right triangle. (5)

Step-by-step solution

Law of cosines — a² = b² + c² − 2bc·cos A
Substituting
$a^{2} = 9 5^{2} + 6 0^{2} - 2 (95) (60) cos 4 5^{\circ}$
Evaluate
$a^{2} = 12, 625 - 8, 061 = 4, 564$
Side
$a = 67.56 f t$
Area — A = ½bc·sin A
Substituting
$A = 0.5 (95) (60) sin 4 5^{\circ} = 2, 015 f t^{2}$

Answer: a ≈ 67.56 ft, area ≈ 2,015 ft²

Why the other options are there

155.0 ft (sides added)
112.4 ft (Pythagoras used with a non-right angle)

Reference: FE Reference Handbook — Mathematics → Trigonometric functions are defined using a right triangle.

Self-check

Answer these without notes before moving on.

Without looking, state the relation on this page whose left-hand side is the quantity most often requested, and name every symbol in it.
Which assumption, if violated, makes the main relation of this section invalid?
Given an algebraic or calculus expression that must be evaluated exactly, what is the first quantity you would compute, and why that one first?
Which unit conversion in this subject most often produces a wrong answer choice, and what is its numerical factor?
Rework Example 1 above from the givens alone, without reading the solution lines.

Chapter summary

Trigonometric functions are defined using a right triangle. contains 8 relations; you must be able to find this page in under 15 seconds.
Exam style: short symbolic manipulations with one numeric evaluation at the end.
Unit rule: radians vs degrees — set the calculator before the first trig entry.
Work the 10 examples until the solution path, not the answer, is automatic.

Common traps in this section

radians vs degrees — set the calculator before the first trig entry
Answering the intermediate quantity instead of the quantity requested.
Rounding intermediate values before the final step.
Using a relation from an adjacent handbook section that shares a symbol.
Skipping the sketch — most lost points on this page start with a misread geometry.