Connects the circle definition back to the right triangle ratios. Shows that the two agree because the triangles are similar, defines all six functions as side ratios, solves right triangles from one side and one angle, and applies the results to angles of elevation and depression, where the measurement convention causes most of the errors.
Subject: Precalculus · 65 slides · symbolic lesson
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Title
Precalculus · Chapter 5 — Trigonometric Functions
§5.4 Right Triangle Trigonometry, pp. 693-706
Objectives
Five things, each one you can check yourself on paper.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 693-706 — the pages these objectives are drawn from
Warm-up
You met sine as a side ratio before this chapter, and §5.2 defined it as a coordinate. Those had better be the same number.
Discussion prompt
The sine of 30 degrees is one half by the circle definition. Check it against a right triangle containing a 30 degree angle.
Hint: Half an equilateral triangle has angles of 30, 60 and 90.
Answer:
Cut an equilateral triangle of side 2 in half. The result has a hypotenuse of 2, a shorter leg of 1, and angles of 30, 60 and 90 degrees.
The side opposite the 30 degree angle is the short one, so the ratio of opposite to hypotenuse is one over two — one half, exactly matching the circle definition.
They agree because the triangle drawn under a terminal side on the unit circle is similar to any other right triangle with the same angle. Similar triangles have equal ratios, so the two definitions cannot disagree.
Concept
A right triangle with a given acute angle is similar to the one formed under that angle's terminal side on the unit circle. Similar triangles have equal side ratios, so the triangle definition and the coordinate definition give the same values.
\[ \sin\theta=\frac{\text{opposite}}{\text{hypotenuse}}, \quad \cos\theta=\frac{\text{adjacent}}{\text{hypotenuse}}, \quad \tan\theta=\frac{\text{opposite}}{\text{adjacent}} \]
The triangle definition is restricted to acute angles, so it is the special case and the circle definition is the general one. But for the applications in this section every angle is acute, and the ratios are the more convenient tool.
Figure (svg): A right triangle with the sides labelled opposite, adjacent and hypotenuse relative to a marked angle, with the three primary ratios written beside it
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 693-697
Section
Section 1
Concept
Relative to a chosen acute angle, the three sides are the opposite, the adjacent and the hypotenuse. Each trigonometric function is one ratio of two of them.
The fourth point is the one that causes errors. A right triangle has two acute angles, and the leg opposite one is adjacent to the other — so every ratio changes when the reference angle changes. Marking the chosen angle before labelling anything prevents it.
Figure (svg): A right triangle with the sides labelled opposite, adjacent and hypotenuse relative to a marked angle, with the three primary ratios written beside it
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 693-698
Picture it
Each ratio names the two sides it compares.
Figure (svg): A right triangle with the sides labelled opposite, adjacent and hypotenuse relative to a marked angle, with the three primary ratios written beside it
The warning at the bottom is the operative part. The hypotenuse is fixed by the right angle, but which leg is opposite and which adjacent depends entirely on the angle marked.
Worked example
Label the sides relative to the marked angle first.
\[ \text{A right triangle has legs } 3 \text{ and } 4 \text{ and hypotenuse } 5. \text{ Find the six ratios for the angle opposite the } 3. \]
Label relative to the chosen angle
Why: The 3 is opposite it.
\[ o p p 3, a d j 4, h y p 5 \]
Write the three primary ratios
Why: Reading off the definitions.
\[ \sin 3 / 5, \cos 4 / 5, \tan 3 / 4 \]
Take the reciprocals
Why: For the other three.
\[ \csc 5 / 3, \sec 5 / 4, \cot 4 / 3 \]
Check the Pythagorean identity
Why: Nine plus sixteen over twenty-five.
\[ \text{equals } 1 \]
Figure (svg): A right triangle with the sides labelled opposite, adjacent and hypotenuse relative to a marked angle, with the three primary ratios written beside it
\[ \sin=\tfrac35, \cos=\tfrac45, \tan=\tfrac34, \csc=\tfrac53, \sec=\tfrac54, \cot=\tfrac43 \]
Verify: check against the other angle
Why: For the other acute angle the labels swap: the 4 becomes opposite and the 3 adjacent, giving a sine of four fifths and a cosine of three fifths. The two angles' sines and cosines are exchanged, which is the cofunction relationship this section takes up shortly.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 694-696
Matching
Each names the two it compares.
Match the pairs
Why: The first three are the primary ratios and the last is a reciprocal — the secant inverts the cosine, so it is the hypotenuse over the adjacent. Note that the tangent is the only one of the four not involving the hypotenuse, which is why it is the ratio for slope questions.
Worked example
The Pythagorean theorem supplies whichever side is missing.
\[ \text{A right triangle has hypotenuse } 13 \text{ and one leg } 5. \text{ Find the sine of the angle opposite the other leg.} \]
Find the missing leg
Why: Pythagoras: 169 minus 25.
\[ 144,\text{ so the leg is } 12 \]
Identify the sides for the chosen angle
Why: The 12 is opposite it.
\[ o p p 12, h y p 13 \]
Form the ratio
Why: Opposite over hypotenuse.
\[ \frac{12}{13} \]
Check
Why: The other angle's sine is 5 over 13.
Figure (svg): The solution to Worked example find a missing side first shown as a ladder of expressions, one row per legal move
\[ \sin\theta=\tfrac{12}{13} \]
Verify: check the identity
Why: Twelve thirteenths squared plus five thirteenths squared is 144 plus 25 over 169, which is 1 — so the sine and cosine of this angle are consistent. Finding the third side before choosing ratios is nearly always the right first move.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 696-698
Trap
\[ \text{a leg of } 3 \text{ is 'the opposite side', whichever angle is asked about} \]
Label the sides once and reuse the labels
Why: The triangle's sides are named and used for both acute angles.
The same labelling is applied to a question about the other acute angle.
Opposite and adjacent are relative to the chosen angle. The leg opposite one acute angle is adjacent to the other, so the labels swap when the angle does.
Only the hypotenuse is fixed, because it is defined by the right angle rather than by either acute one.
Mark the angle first, then label. Relabelling for each angle takes two seconds and removes the error entirely, whereas reusing labels produces a sine where a cosine belongs.
Sorting
For a right triangle with the lower-left acute angle marked.
Sort into buckets
Sort each side by its role relative to that angle.
Faded example
A right triangle has opposite 8, adjacent 15 and hypotenuse 17.
Fill in the blanks
\sin = \frac1715}, \qquad \tan = \frac______}
Why: The sine compares the opposite with the hypotenuse and the tangent compares it with the adjacent. Checking Pythagoras: 64 plus 225 is 289, which is 17 squared — so the triangle is valid and the ratios are consistent.
Socratic
Two of the three labels change when the reference angle does.
Discussion prompt
Explain why the hypotenuse is the exception.
Hint: What defines each of the three labels?
Answer:
The hypotenuse is defined by the right angle — it is the side across from it — and the right angle does not move when you switch between the two acute angles.
Opposite and adjacent are defined relative to the acute angle you chose. Switch angles and the leg that was across from one is now touching the other, so the two labels exchange.
That is why the sine and cosine of the two acute angles are swapped versions of each other, which is exactly the cofunction relationship. The whole phenomenon comes from which angle the labels are measured against.
Section
Section 2
Concept
Any right triangle with a given acute angle is similar to the one under that angle's terminal side on the unit circle, so their corresponding side ratios are identical.
The scaling argument also explains why the ratios do not depend on the triangle's size. A large and a small right triangle with the same angle give the same sine, which is what makes the trigonometric functions functions of the angle alone.
Figure (svg): A small right triangle on the unit circle and a larger similar triangle beside it, showing that the ratios of corresponding sides are equal so the two definitions agree
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 697-700
Picture it
The small one sits on the unit circle and the large one is any triangle with the same angle.
Figure (svg): A small right triangle on the unit circle and a larger similar triangle beside it, showing that the ratios of corresponding sides are equal so the two definitions agree
Their sides are proportional, so their ratios are equal. That is the whole reason the two definitions never disagree, and it is a fact about similarity rather than about trigonometry.
Worked example
Compare the two definitions at one angle.
\[ \text{Check that both definitions give } \sin 45^\circ = \tfrac{\sqrt2}{2}. \]
Use the circle definition
Why: The point is on the diagonal.
\[ \text{vertical coordinate } \sqrt{2} / 2 \]
Build a triangle with a 45 degree angle
Why: Half a square, legs 1 and 1.
\[ \text{hypotenuse } \sqrt{2} \]
Form the triangle ratio
Why: Opposite over hypotenuse.
\[ 1 / \sqrt{2} \]
Rationalise
Why: Multiply top and bottom by root 2.
\[ \sqrt{2} / 2 \]
Figure (svg): A small right triangle on the unit circle and a larger similar triangle beside it, showing that the ratios of corresponding sides are equal so the two definitions agree
\[ \tfrac{\sqrt2}{2} \text{ by either definition} \]
Verify: note why they had to agree
Why: The triangle with legs 1 and 1 is similar to the one on the unit circle with legs root 2 over 2 each — the second is the first scaled by one over root 2. Similar triangles have equal ratios, so agreement was guaranteed rather than coincidental.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 698-699
Prediction
A right triangle is enlarged by a factor of 3, keeping its shape.
Predict first
What happens to its sine?
Correct: It is unchanged, since a ratio survives scaling.
Why: Both the opposite side and the hypotenuse triple, so their ratio is unaffected. This is what makes the sine a function of the angle rather than of the triangle, and it is why one table of values serves every triangle with a given angle.
Worked example
Scaling a triangle leaves every ratio unchanged.
\[ \text{Compare the sine of the same angle in a } 3\text{-}4\text{-}5 \text{ and a } 6\text{-}8\text{-}10 \text{ triangle.} \]
Take the first triangle's ratio
Why: Opposite 3 over hypotenuse 5.
\[ \frac{3}{5} \]
Take the second's
Why: Opposite 6 over hypotenuse 10.
\[ \frac{6}{10} \]
Simplify
Why: Both reduce to the same value.
\[ \frac{3}{5} \]
Explain
Why: The second is the first scaled by 2.
Figure (svg): The solution to Worked example size does not matter shown as a ladder of expressions, one row per legal move
\[ \tfrac35 \text{ in both} \]
Verify: state the general consequence
Why: Every ratio is unchanged by scaling, so the six functions depend on the angle alone and not on the triangle's size. That is what makes them functions of an angle, which is a stronger statement than it first appears and is exactly what the similarity guarantees.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 699-700
Error analysis
A student computes a sine from a large triangle.
Annotate
On: \( \text{a bigger triangle has bigger sides, so its sine is bigger} \)
Every trigonometric value is a ratio, and ratios survive scaling. This is why a table of sines can exist at all: one value per angle, valid for every triangle containing it.
Sorting
Only changes to the angle can.
Sort into buckets
Sort each change to a right triangle.
Faded example
For a 30-60-90 triangle with hypotenuse 2 and short leg 1.
Fill in the blanks
\sin 30^\circ = \frac21/2} = ___, \text___
Why: The side opposite the 30 degree angle is the short leg of 1 and the hypotenuse is 2, giving one half — exactly the vertical coordinate of the point at 30 degrees on the unit circle. The two definitions agree because the triangles are similar.
Explain it to yourself
Both definitions give the same numbers, but one is more general.
Discussion prompt
Explain which is the special case and why.
Hint: Which angles can each handle?
Answer:
The triangle definition is the special case, because it only works for acute angles — a right triangle cannot contain an obtuse one, and it certainly cannot contain a negative angle.
The circle definition is general, assigning values to every angle. And on acute angles it agrees with the triangle version, by the similarity argument.
So nothing is lost by adopting the circle definition, and a great deal is gained: periodicity, negative angles and the whole of Chapter 6's graphs all require it. The triangle version survives because it is more convenient for the applications, where the angles are acute anyway.
Section
Section 3
Concept
Given one acute angle and one side, every other part of a right triangle can be found. The technique is choosing, for each unknown, the ratio relating it to a known quantity.
The fourth point is the same advice §2.1's worked examples gave about chaining computed values, and it matters for the same reason: a rounded intermediate value carries its error into everything computed from it, while the given data is exact.
Figure (svg): A right triangle with one angle and one side known, showing which ratio connects the known pair to each unknown side
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 700-703
Picture it
The known pair and the wanted side decide which ratio to use.
Figure (svg): A right triangle with one angle and one side known, showing which ratio connects the known pair to each unknown side
There is no searching involved. Each unknown appears in exactly one ratio alongside the known side, and that is the one to use.
Worked example
One ratio per unknown side.
\[ \text{A right triangle has a } 35^\circ \text{ angle and hypotenuse } 12. \text{ Find both legs.} \]
Find the opposite leg
Why: Sine relates opposite to hypotenuse.
\[ o p p = 12 \sin 35 \]
Compute
Why: Twelve times about 0.5736.
\[ \text{about } 6.88 \]
Find the adjacent leg
Why: Cosine relates adjacent to hypotenuse.
\[ a d j = 12 \cos 35 \]
Compute
Why: Twelve times about 0.8192.
\[ \text{about } 9.83 \]
Figure (svg): A right triangle with one angle and one side known, showing which ratio connects the known pair to each unknown side
\[ \text{opp}\approx 6.88, \quad \text{adj}\approx 9.83 \]
Verify: check with Pythagoras
Why: Six point eight eight squared plus 9.83 squared is about 47.3 plus 96.6, which is 143.9 — and 12 squared is 144, agreeing to rounding. Note both legs were computed from the given hypotenuse rather than one from the other, so neither inherits the other's rounding error.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 701-702
Sorting
Pick the one mentioning both the known and the wanted side.
Sort into buckets
Given an angle and the hypotenuse, sort each unknown.
Worked example
The tangent relates the two legs.
\[ \text{A right triangle has a } 28^\circ \text{ angle whose adjacent leg is } 20. \text{ Find the opposite leg and the hypotenuse.} \]
Find the opposite leg
Why: Tangent relates the two legs.
\[ o p p = 20 \tan 28 \]
Compute
Why: Twenty times about 0.5317.
\[ \text{about } 10.6 \]
Find the hypotenuse
Why: Cosine relates adjacent to hypotenuse.
\[ h y p = 20 / \cos 28 \]
Compute
Why: Twenty over about 0.8829.
\[ \text{about } 22.7 \]
Figure (svg): The solution to Worked example solve from an angle and a leg shown as a ladder of expressions, one row per legal move
\[ \text{opp}\approx 10.6, \quad \text{hyp}\approx 22.7 \]
Verify: check the hypotenuse is the longest
Why: Twenty two point seven exceeds both 20 and 10.6, as a hypotenuse must. And Pythagoras gives 400 plus 112 which is 512, whose root is about 22.6 — agreeing to rounding. Confirming the hypotenuse is longest is a fast check that the ratios were not inverted.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 702-703
Trap
\[ \sin 35^\circ=\frac{\text{opp}}{12} \;\Longrightarrow\; \text{opp}=\frac{12}{\sin 35^\circ}\approx 20.9 \]
Rearrange the equation to isolate the unknown
Why: The equation is correct and the unknown is isolated.
The hypotenuse is divided by the sine rather than multiplied.
Multiplying is what isolates the numerator. Opposite over 12 equals the sine, so the opposite is 12 times the sine, about 6.88.
The wrong answer of 20.9 exceeds the hypotenuse of 12, which is impossible — a leg is always shorter than the hypotenuse.
Check that every leg is shorter than the hypotenuse. That single comparison catches an inverted ratio immediately, and it costs nothing.
Faded example
An angle of 40 degrees has an opposite leg of 9. Find the hypotenuse.
Fill in the blanks
\sin 40^\circ = \fracsin 4014.0 \;\Longrightarrow\; h = \frac______} \approx ___
Why: The unknown is in the denominator here, so isolating it means dividing rather than multiplying. Nine over about 0.643 gives about 14.0, which is longer than the leg of 9 as a hypotenuse must be.
Prediction
A right triangle has been solved.
Predict first
Which side must be the longest?
Correct: The hypotenuse, always.
Why: The hypotenuse is opposite the right angle, which is the largest angle in the triangle, and the largest side is always opposite the largest angle. Checking this after solving catches an inverted ratio, since dividing instead of multiplying typically produces a leg longer than the hypotenuse.
Step zero
You are given one acute angle and one side of a right triangle.
Discussion prompt
What do you write down before doing any trigonometry?
Hint: Is anything available for free?
Answer:
The third angle, which is 90 minus the given acute one. It needs no trigonometry and is often what a question asks for.
Then, for each unknown side, decide which ratio mentions it and the known side. That choice is the only real decision in the problem, and making it explicitly prevents a search.
Also worth noting up front: compute every unknown from the given side rather than from one you calculated, so no rounding error propagates. That is §2.1's advice, and it matters more here because several quantities are usually wanted.
Section
Section 4
Concept
The two acute angles of a right triangle are complementary, and switching between them exchanges the opposite and adjacent sides — so each function of one equals its cofunction of the other.
\[ \sin\theta=\cos(90^\circ-\theta), \qquad \tan\theta=\cot(90^\circ-\theta) \]
The relationship halves any table of values. Knowing the sines from 0 to 45 degrees gives the cosines from 45 to 90 for free, which is exactly how the old printed tables were compressed onto half the pages they would otherwise have needed.
Figure (svg): A right triangle with the sides labelled opposite, adjacent and hypotenuse relative to a marked angle, with the three primary ratios written beside it
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 698-701
Picture it
Opposite and adjacent are relative to the chosen angle.
Figure (svg): A right triangle with the sides labelled opposite, adjacent and hypotenuse relative to a marked angle, with the three primary ratios written beside it
Switching to the other acute angle exchanges those two labels while leaving the hypotenuse alone. That single exchange produces every cofunction relationship.
Worked example
Find the complement and switch the function.
\[ \text{Express } \sin 25^\circ \text{ as a cosine.} \]
Find the complement
Why: Ninety minus 25.
\[ 65 ^\circ \]
Switch to the cofunction
Why: Sine becomes cosine.
\[ \cos 65 \]
State the equality
Why: The two are equal.
\[ \sin 25 = \cos 65 \]
Check numerically
Why: Both are about 0.4226.
Figure (svg): The solution to Worked example use the relationship shown as a ladder of expressions, one row per legal move
\[ \sin 25^\circ=\cos 65^\circ \]
Verify: see it in a triangle
Why: In a right triangle with a 25 degree angle, the other acute angle is 65. The leg opposite the 25 is adjacent to the 65, so the same ratio is a sine for one and a cosine for the other. The relationship is a relabelling rather than a computation.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 699-700
Faded example
Express the tangent of 15 degrees as a cotangent.
Fill in the blanks
\tan 15^\circ = \cot(90 - 15)^\circ = \cot 75^\circ
Why: The complement of 15 is 75, and the tangent's cofunction is the cotangent. Both equal about 0.268. The relationship exchanges each function with its co- partner while replacing the angle by its complement.
Worked example
The relationship turns one equation into another.
\[ \text{Solve } \sin(2x)=\cos(30^\circ) \text{ for an acute } 2x. \]
Rewrite the right side as a sine
Why: The cofunction of the complement.
\[ \cos 30 = \sin 60 \]
Equate the angles
Why: Both sides are now sines of acute angles.
\[ 2 x = 60 \]
Solve
Why: Divide by 2.
\[ x = 30 \]
Check
Why: Sine of 60 equals cosine of 30.
\[ \text{both } \sqrt{3} / 2 \]
Figure (svg): The solution to Worked example solve using a cofunction shown as a ladder of expressions, one row per legal move
\[ x=30^\circ \]
Verify: confirm the equating step is valid
Why: Both sides are sines of acute angles, and the sine is one-to-one on acute angles — it increases steadily from 0 to 1. So equal sines force equal angles, which licenses the step. Without that restriction the equation would have other solutions, as §7.5 will explore.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 700-701
Error analysis
A student applies the cofunction relationship.
Annotate
On: \( \sin 20^\circ=\cos 160^\circ \)
The relationship comes from the two acute angles of a right triangle, which sum to 90. A quick sign check catches the supplement version, since the two sides come out with opposite signs.
Matching
The pairs are the ones sharing a co- prefix.
Match the pairs
Why: Each function pairs with the one whose name adds or removes the co- prefix. Note this is the OPPOSITE pairing from the reciprocal relationship, where the secant goes with the cosine — so secant and cosecant are cofunctions of each other but reciprocals of cosine and sine respectively.
Prediction
The cofunction relationship holds for all six functions.
Predict first
Where does it come from?
Correct: The two acute angles of a right triangle sum to 90 degrees.
Why: Because they sum to 90, they are complementary, and switching between them exchanges which leg is opposite and which is adjacent. Every cofunction relationship is that single exchange, applied to the six ratios in turn.
Explain it to yourself
Cosine, cotangent and cosecant all carry a co- prefix.
Discussion prompt
Explain what the prefix means and why it was chosen.
Hint: What does the cosine of an angle equal?
Answer:
The co- stands for complement. The cosine of an angle is the sine of the complement, which is where the name comes from — it was originally written as the complement's sine.
The same holds for the other two: the cotangent is the tangent of the complement, and the cosecant is the secant of it.
So the prefix is genuinely informative rather than decorative. It is unfortunate only in that it does not match the reciprocal pairing — the secant's reciprocal partner is the cosine, not the cosecant — which is the source of the section's one memorisation burden.
Section
Section 5
Concept
An angle of elevation is measured upward from a horizontal line of sight and an angle of depression downward from one. Both use the horizontal as the reference, never the vertical.
The fourth point is worth exploiting. A problem stating an angle of depression from the top of a cliff can be redrawn with the same angle as an elevation at the bottom, which often puts the angle inside the triangle where the ratios are easier to apply.
Figure (svg): Two application diagrams: an angle of elevation from an observer up to a distant top, and an angle of depression from a height down to a point below
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 703-706
Picture it
Both measured from a horizontal line, in opposite directions.
Figure (svg): Two application diagrams: an angle of elevation from an observer up to a distant top, and an angle of depression from a height down to a point below
The warning is the operative part. Measuring from the vertical gives the complementary angle, which produces a sine where a cosine belongs and an answer that looks entirely plausible.
Worked example
Draw the triangle, then choose the ratio.
\[ \text{From } 50 \text{ m away, a tower's top has an elevation of } 32^\circ. \text{ How tall is it?} \]
Draw the triangle
Why: Horizontal distance, vertical height, line of sight.
Label relative to the angle
Why: The height is opposite, the distance adjacent.
Choose the ratio
Why: Tangent relates the two legs.
\[ \tan 32 = \frac{h}{50} \]
Solve
Why: Multiply by 50.
\[ h = 50 \tan 32\text{ about } 31.2 \]
Figure (svg): Two application diagrams: an angle of elevation from an observer up to a distant top, and an angle of depression from a height down to a point below
\[ h=50\tan 32^\circ\approx 31.2 \text{ m} \]
Verify: sanity-check the size
Why: An elevation under 45 degrees means the height is less than the distance, and 31 is indeed less than 50. Had the answer exceeded 50, the ratio would have been inverted. That comparison against 45 degrees is a fast check on any elevation problem.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 704-705
Sorting
Both elevation and depression use the horizontal.
Sort into buckets
Sort each description.
Worked example
Redraw it as an elevation to put the angle inside the triangle.
\[ \text{From a } 40 \text{ m cliff, a boat has a depression angle of } 25^\circ. \text{ How far out is it?} \]
Draw the horizontal at the top
Why: The depression is measured down from it.
\[ 25 ^\circ\text{ down} \]
Use alternate angles
Why: The elevation from the boat equals it.
\[ 25 ^\circ\text{ at the boat} \]
Label relative to that angle
Why: The cliff is opposite, the distance adjacent.
\[ o p p 40, a d j = d \]
Choose and solve
Why: Tangent relates the legs.
\[ d = 40 / \tan 25\text{ about } 85.8 \]
Figure (svg): The solution to Worked example an angle of depression shown as a ladder of expressions, one row per legal move
\[ d=\frac{40}{\tan 25^\circ}\approx 85.8 \text{ m} \]
Verify: check the shallow angle
Why: A depression of only 25 degrees means a distant boat, so the horizontal distance should exceed the cliff's height comfortably — and 86 does exceed 40. Redrawing the depression as an elevation at the boat put the angle inside the triangle, which made the labelling straightforward.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 705-706
Trap
\[ \text{a } 32^\circ \text{ elevation, so the angle at the top of the tower is } 32^\circ \]
Place the given angle in the triangle
Why: The angle is marked at a vertex of the right triangle.
It is placed at the top, between the vertical and the line of sight.
An elevation is measured from the horizontal, so the 32 degrees sits at the observer, between the ground and the line of sight.
The angle at the top is the complement, 58 degrees. Using 32 there would give a height of 50 over the tangent of 32, about 80 metres rather than 31.
Draw the horizontal line explicitly and mark the angle against it. The two placements differ by a complement, so the wrong one produces a plausible-looking answer that is badly wrong.
Faded example
A kite string 60 m long makes an elevation of 48 degrees. Find the kite's height.
Fill in the blanks
\sin 48^\circ = \frac6044.6} \;\Longrightarrow\; h = 60\sin 48^\circ \approx ___
Why: The string is the hypotenuse and the height is opposite the elevation angle, so the sine relates them. Sixty times about 0.743 gives about 44.6 metres, which is less than the string's length as a leg must be.
Prediction
An observer on a cliff looks down at a boat, and someone on the boat looks up at the observer.
Predict first
How do the two angles compare?
Correct: They are equal, by alternate angles.
Why: The two horizontal lines are parallel and the line of sight is a transversal, so the depression from above and the elevation from below are alternate angles and therefore equal. This is what lets a depression problem be redrawn as an elevation one, which usually simplifies the labelling.
Real world
Elevation and depression angles are the basis of a great deal of surveying.
Discussion prompt
How does a surveyor measure the height of a building without climbing it?
Hint: What can be measured on the ground, and what instrument gives the angle?
Answer:
Measure the horizontal distance to the base, then use an instrument to read the angle of elevation to the top. Both are done from the ground.
The height is then that distance times the tangent of the angle, which is this section's first worked example. Adding the instrument's own height above the ground completes it.
The same method scales enormously. Triangulation built the first accurate maps and measured mountains and the Earth itself, all from angles measured at ground level and one carefully surveyed baseline — which is why trigonometry was a navigator's and a surveyor's subject long before it was a student's.
Comparison
Fill the blanks from memory. They agree where both apply, and one applies far more widely.
Comparison matrix
| right triangle | unit circle | |
|---|---|---|
| defined for | acute angles only | every angle |
| sine is | opposite over hypotenuse | the vertical coordinate |
| handles negatives | no | yes |
| shows periodicity | no | yes |
| best for | applied problems with acute angles | graphs, identities and everything general |
The last row is why both survive. The circle definition is the real one and the triangle version is the convenient special case for the applications, which is most of what this section is about.
Pattern
Six steps, and the first two are where the errors are.
Steps 1 and 2 account for most wrong answers in applied problems. An angle placed at the wrong vertex differs from the right one by a complement, and the resulting answer is plausible enough to survive unexamined.
OpenStax Algebra and Trigonometry 2e, §7.2 Right Triangle Trigonometry §7.2
Check
Name the two sides each ratio compares.
Check your understanding
Which ratio is the tangent?
Answer: A
Why: The tangent compares the two legs and is the only primary ratio not involving the hypotenuse. That is why it is the ratio for slope and gradient questions, where only a rise and a run are available.
Check
Complement, not supplement.
Check your understanding
The sine of 35 degrees equals the cosine of which angle?
Answer: A
Why: The cofunction relationship uses the complement, which is 90 minus 35, or 55. Both values are about 0.574. The relationship comes from the two acute angles of a right triangle summing to 90.
Check
Multiply or divide?
Check your understanding
An angle of 40 degrees has hypotenuse 10. What is the opposite side?
Answer: A
Why: The sine is the opposite over the hypotenuse, so the opposite is the hypotenuse times the sine — about 6.4. Since a leg must be shorter than the hypotenuse, an answer under 10 is expected.
Real world
Right triangle trigonometry is the oldest applied mathematics there is, and it is still how distances are measured indirectly.
Discussion prompt
How is the distance to a nearby star measured, given that nobody can travel there?
Hint: What baseline is available, and what angle changes over six months?
Answer:
The Earth's orbit provides a baseline: observe the star six months apart and the observation points are about 300 million kilometres apart. The star's apparent position shifts slightly against the distant background.
That shift is an angle, and with the baseline known, the distance is one tangent calculation — exactly this section's method at an enormous scale. The technique is called parallax.
The angles involved are tiny, under one second of arc, which is why the method needed centuries of instrument improvement before it worked. But the mathematics is a right triangle, unchanged from the surveying problems in this section, which is a good illustration of how far a simple idea can be pushed.
Commit first
State your confidence along with your answer.
Predict first
Why do the triangle and circle definitions of sine agree?
Correct: Because the triangles involved are similar, so their ratios are equal.
Why: A right triangle with a given acute angle is similar to the one under that angle's terminal side on the unit circle, and similar triangles have proportional sides and therefore equal ratios. The agreement is guaranteed by similarity rather than being a coincidence, and it holds for every acute angle.
Explain it
The test of understanding is being able to say why, not just what.
Discussion prompt
Explain to a classmate why the sine of an angle does not depend on the size of the triangle.
Hint: What happens to both parts of the ratio when the triangle is scaled?
Answer:
The sine is a ratio of two sides, not a length. Scaling the triangle multiplies both the opposite side and the hypotenuse by the same factor.
So the numerator and denominator scale together and the ratio is unchanged. A triangle twice the size gives exactly the same sine.
That is what makes the sine a function of the angle alone, which is why a single table of values can serve every triangle. A good explanation adds that this is also why the circle definition agrees: the unit-circle triangle is just one particular scaling of every other.
Exit ticket
One honest answer, so the next chapter can start in the right place.
Predict first
Which idea from this lesson would you most want to see again?
Correct: Any of these is a legitimate answer; the useful one is the honest one.
Why: There is no correct choice here. The fourth causes the most wrong answers in applied problems, because a misplaced angle differs by a complement and produces a plausible result. The first is its underlying cause, since labelling relative to the wrong angle has the same effect.
Connect it up
One page, drawn from memory, closes the chapter.
Draw it
Draw a right triangle with one acute angle marked, labelling all three sides relative to it, and write the six ratios beside it. Then draw the same angle on a unit circle with its small triangle, and note why the two are similar. Finally sketch an elevation and a depression problem, marking the horizontal reference line in each and showing where the given angle sits.
If your two triangles are marked as similar and your horizontal reference lines are drawn explicitly, both of the section's error sources are addressed on the page.
Recap
Five things, closing a chapter that began with angles and ends with them measuring the world.
| if you remember one thing | it should be this |
|---|---|
| about labelling | opposite and adjacent depend on which angle you marked |
| about the two definitions | similar triangles have equal ratios, so they agree |
| about solving | pick the ratio mentioning what you know and what you want |
| about applications | elevation and depression are measured from the horizontal |
Chapter 6 graphs these functions as functions rather than as ratios, where the periodicity that the circle definition made possible becomes the whole subject.
OpenStax, Precalculus, §5.4 Right Triangle Trigonometry §5.4, pp. 693-706 — everything on these slides traces back here
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