The six trigonometric ratios of an acute angle, finding all six from one of them, the exact values at 30, 45 and 60 degrees, solving a right triangle with a calculator, and indirect measurement using angles of elevation and depression.
Subject: Algebra 2 · 65 slides · symbolic lesson
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Title
Algebra 2 · Chapter 13 — Trigonometric Ratios and Functions
Use Trigonometry with Right Triangles
Objectives
Five outcomes. Six ratios, and what you can measure with them.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 852-857 — the lesson these objectives are drawn from
Warm-up
The Pythagorean theorem relates the three sides of a right triangle.
Discussion prompt
A right triangle has legs 5 and 12. Find the hypotenuse. Then ask a new question: what is the RATIO of the leg of length 12 to that hypotenuse?
Hint: Five squared plus 12 squared.
Answer:
\[ c = \sqrt{5^2+12^2} = \sqrt{169} = 13 \]
\[ \frac{12}{13} \approx 0.923 \]
That ratio is the whole subject. It depends only on the ANGLE, not on how big the triangle is — double every side and the ratio is unchanged. Six such ratios can be formed, and each has a name.
Concept
For an acute angle of a right triangle the three sides can be paired six ways, giving sine, cosine, tangent and their reciprocals cosecant, secant and cotangent. Each depends only on the angle, so a ratio and a side determine the triangle.
sine — For an acute angle of a right triangle, the ratio of the side opposite that angle to the hypotenuse. Its reciprocal is the cosecant.
\[ \sin\theta = \frac{\text{opp}}{\text{hyp}}, \; \cos\theta = \frac{\text{adj}}{\text{hyp}}, \; \tan\theta = \frac{\text{opp}}{\text{adj}} \]
Opposite and adjacent are named relative to the chosen angle, so switching to the other acute angle swaps them. The hypotenuse is always the side across from the right angle.
Figure (svg): Two columns comparing the three primary ratios with their reciprocals
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 852-852
Section
Section 1
Concept
Name the sides opposite, adjacent and hypotenuse relative to the chosen angle. The six ratios are the six ways of putting one over another, and the last three are the reciprocals of the first three.
\[ \csc\theta = \frac{1}{\sin\theta}, \; \sec\theta = \frac{1}{\cos\theta}, \; \cot\theta = \frac{1}{\tan\theta} \]
Only sine, cosine and tangent carry new information. The other three exist because some formulas are far tidier written with them.
Figure (svg): A right triangle with its three sides named and the six ratios listed
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 852-852 — Right Triangle Definitions of Trigonometric Functions
Picture it
Example 1: the 5, 12, 13 triangle.
Figure (svg): A right triangle with its three sides named and the six ratios listed
The top three ratios have the hypotenuse or the adjacent side underneath; the bottom three are those same fractions flipped.
Worked example
Example 1.
\[ \text{The legs are } 5 \text{ (adjacent) and } 12 \text{ (opposite). Find all six ratios of } \theta. \]
Find the hypotenuse
Why: Twenty-five plus 144 is 169.
\[ h y p = 13 \]
Write the three primary ratios
Why: Opposite over hypotenuse, and so on.
\[ \frac{12}{13}, \frac{5}{13}, \frac{12}{5} \]
Flip each one
Why: The reciprocals.
\[ \frac{13}{12}, \frac{13}{5}, \frac{5}{12} \]
Label them
Why: In the standard order.
Figure (svg): A right triangle with its three sides named and the six ratios listed
\[ \sin\theta = \tfrac{12}{13}, \; \cos\theta = \tfrac{5}{13}, \; \tan\theta = \tfrac{12}{5} \]
Verify: check that sine and cosine are under one
Why: Both 12 over 13 and 5 over 13 are less than 1, as they must be: a leg is always shorter than the hypotenuse. Tangent has no such limit, since it compares two legs — here it is 2.4. A sine above 1 is always an arithmetic error.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 852-852
Fill the middle
Example 1.
Fill in the blanks
\text13 = \sqrt___ = \sqrt___ = ___
Why: Thirteen. The 5, 12, 13 triple appears constantly in trigonometry problems, so it is worth recognising on sight.
Worked example
Guided Practice 1 and 2.
\[ \text{Find all six ratios for legs } 3 \text{ and } 4, \text{ and for hypotenuse } 17 \text{ with a leg } 15. \]
First: find the hypotenuse
Why: Nine plus 16 is 25.
\[ h y p = 5 \]
First: write the ratios
Why: Taking 4 as opposite and 3 as adjacent.
\[ \frac{4}{5}, \frac{3}{5}, \frac{4}{3}\text{ and their flips} \]
Second: find the missing leg
Why: Two eighty-nine minus 225 is 64.
\[ l e g = 8 \]
Second: write the ratios
Why: Taking 8 as opposite and 15 as adjacent.
\[ \frac{8}{17}, \frac{15}{17}, \frac{8}{15}\text{ and their flips} \]
Figure (svg): The solution to Worked example two more triangles shown as a ladder of expressions, one row per algebraic move
\[ \tfrac{4}{5}, \tfrac{3}{5}, \tfrac{4}{3}; \qquad \tfrac{8}{17}, \tfrac{15}{17}, \tfrac{8}{15} \]
Verify: notice both are whole-number triples
Why: Three, four, five and eight, fifteen, seventeen are Pythagorean triples, which is why the missing sides came out whole. Had the leg been 14 instead of 15 the third side would have been root 93, and nothing else would have changed about the method.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 853-853
Trap
\[ \text{legs } 5 \text{ and } 12, \; \text{hyp } 13 \]
Call the horizontal leg adjacent whatever angle is asked about
Why: The labels are attached to the drawing rather than to the angle.
\[ \sin\theta = \tfrac{12}{13} \text{ for BOTH acute angles} \quad \text{(wrong)} \]
Opposite and adjacent are defined relative to the chosen angle. The other acute angle has them the other way round.
\[ \sin\theta = \tfrac{12}{13}; \qquad \sin(\text{other angle}) = \tfrac{5}{13} \]
Re-label the sides for each angle you consider
Why: Only the hypotenuse keeps its name.
\[ \cos\theta = \sin(90^\circ-\theta) \]
This swap is exactly why sine of 30 equals cosine of 60: the two acute angles see the same triangle from opposite corners.
Matching
All six, from the angle's point of view.
Match the pairs
Why: Secant is the reciprocal of cosine, not of sine, which is the pairing most often mixed up. The names give no hint: co-secant pairs with sine, and secant with cosine.
Sorting
Think about which side is on top.
Sort into buckets
Sort each ratio by whether it can be greater than 1 for an acute angle.
This is a free error check: a sine or cosine above 1 is impossible for a right triangle, so any calculation producing one has gone wrong.
Prediction
Commit before reasoning.
Predict first
Double every side of the 5, 12, 13 triangle. What happens to sine of theta?
Correct: It is unchanged, because both parts of the ratio doubled.
\[ \frac{24}{26} = \frac{12}{13} \]
Why: The sides become 10, 24 and 26, and 24 over 26 reduces to 12 over 13 — exactly what it was. Every right triangle with the same acute angle is similar to every other, so corresponding side ratios are equal. That invariance is the entire reason these ratios are functions of the ANGLE alone, and it is what lets one table of values serve every triangle in the world.
Section
Section 2
Concept
A given ratio names two of the three sides. Draw a right triangle with those side lengths, use the Pythagorean theorem for the third, and then every other ratio can simply be read off.
\[ \sin\theta = \tfrac{3}{8} \;\Longrightarrow\; \text{opp} = 3, \; \text{hyp} = 8 \]
Any triangle with those proportions will do, since the ratios depend only on the angle. Choosing the numbers in the given fraction is simplest.
Figure (svg): One known ratio giving the third side by the Pythagorean theorem
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 853-853 — Example 2
Picture it
Example 2: sine of theta is three eighths.
Figure (svg): One known ratio giving the third side by the Pythagorean theorem
Sixty-four minus 9 is 55, so the adjacent side is root 55 and every remaining ratio follows immediately.
Worked example
Example 2.
\[ \text{If } \theta \text{ is acute and } \sin\theta = \tfrac{3}{8}, \text{ find } \tan\theta. \]
Draw the triangle
Why: Opposite 3, hypotenuse 8.
Find the third side
Why: Sixty-four minus 9.
\[ a d j = \sqrt{55} \]
Write the tangent
Why: Opposite over adjacent.
\[ 3\text{ over } \sqrt{55} \]
Rationalise
Why: Multiply top and bottom by root 55.
\[ 3 \sqrt{55}\text{ over } 55 \]
Figure (svg): One known ratio giving the third side by the Pythagorean theorem
\[ \tan\theta = \frac{3}{\sqrt{55}} = \frac{3\sqrt{55}}{55} \]
Verify: sanity-check the size
Why: Root 55 is about 7.42, so the tangent is about 0.405 — slightly larger than the sine of 0.375, which is right because the adjacent side 7.42 is slightly shorter than the hypotenuse 8. Tangent always exceeds sine for an acute angle, for exactly that reason.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 853-853
Fill the middle
Example 2, Step 1.
Fill in the blanks
x = \sqrt55 = \sqrt___ = \sqrt___}
Why: Fifty-five, which has no perfect-square factor, so root 55 is already simplified. It is about 7.42.
Worked example
Guided Practice 4, and Guided Practice 3.
\[ \text{If } \cos\theta = \tfrac{7}{10}, \text{ find } \sin\theta. \text{ Also find all six ratios for a } 5,5,5\sqrt{2} \text{ triangle.} \]
First: draw the triangle
Why: Adjacent 7, hypotenuse 10.
\[ o p p = \sqrt{51} \]
First: write the sine
Why: Opposite over hypotenuse.
\[ \sqrt{51}\text{ over } 10 \]
Second: identify the triangle
Why: Two legs of 5, hypotenuse 5 root 2.
\[ a 45 ^\circ\text{ triangle} \]
Second: write the ratios
Why: Five over 5 root 2, and so on.
\[ \sqrt{2}\text{ over } 2, \sqrt{2}\text{ over } 2, 1 \]
Figure (svg): The solution to Worked example find sine from cosine shown as a ladder of expressions, one row per algebraic move
\[ \sin\theta = \frac{\sqrt{51}}{10}; \qquad \tfrac{\sqrt2}{2}, \tfrac{\sqrt2}{2}, 1 \]
Verify: check the first with the Pythagorean identity
Why: Sine squared plus cosine squared should be 1: fifty-one hundredths plus forty-nine hundredths is exactly 1. That identity, which Lesson 13.3 will state formally, is already available here as a check on any pair of values.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 853-853
Error analysis
A student is told sine of theta is 3/8 and looks for the adjacent side.
Annotate
On: \( \text{adj} = 8-3 = 5 \)
The check is quick: 5 squared plus 3 squared is 34, not 64. Squaring the proposed sides and testing the theorem catches this instantly.
Fill the middle
Guided Practice 4.
Fill in the blanks
\cos\theta = \tfrac51___ \;\Longrightarrow\; \sin^2\theta = 1-\tfrac______ = \tfrac___}___
Why: Fifty-one, so the sine is root 51 over 10, about 0.714. Notice this is slightly larger than the cosine of 0.7, so the angle is a little over 45 degrees.
Matching
Draw the triangle, then read.
Match the pairs
Why: The last two describe the same triangle from the same angle, which is why they hand each other back and forth. Once any one ratio is known, the triangle is determined up to size and every other ratio follows.
Prediction
Commit before reasoning.
Predict first
Given sin theta = 3/8, why is it legitimate to draw a triangle with opposite exactly 3 and hypotenuse exactly 8?
Correct: Because every triangle with that ratio is similar, so all its other ratios are the same.
\[ \frac{6}{16} = \frac{3}{8}; \quad \text{same angle, same ratios} \]
Why: A triangle with opposite 6 and hypotenuse 16 has the same angle and the same six ratios; so does one with opposite 1.5 and hypotenuse 4. Since the answer wanted is a ratio, the size cancels out and the most convenient representative can be drawn. Picking the numerator and denominator of the given fraction is simply the least arithmetic.
Section
Section 3
Concept
The 30-60-90 triangle with sides 1, root 3, 2 and the 45-45-90 triangle with sides 1, 1, root 2 give the exact values of all six ratios at 30, 45 and 60 degrees.
\[ \sin 30^\circ = \tfrac{1}{2}, \; \sin 45^\circ = \tfrac{\sqrt2}{2}, \; \sin 60^\circ = \tfrac{\sqrt3}{2} \]
These are exact values, not decimal approximations, and they are the ones expected in answers unless a decimal is requested.
Figure (svg): The two special right triangles and the values they generate
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 853-853 — Trigonometric Values for Special Angles
Picture it
Both special triangles, with the values they generate.
Figure (svg): The two special right triangles and the values they generate
Sine of 30 equals cosine of 60, and sine of 60 equals cosine of 30, because those two angles are the two acute corners of one triangle.
Worked example
Example 3.
\[ \text{A right triangle has hypotenuse } 8 \text{ and a } 30^\circ \text{ angle. Find the adjacent leg } x. \]
Choose the ratio
Why: Adjacent over hypotenuse is cosine.
\[ \cos 30 = \frac{x}{8} \]
Substitute the exact value
Why: Cosine of 30 is root 3 over 2.
\[ \sqrt{3}\text{ over } 2 = \frac{x}{8} \]
Solve for x
Why: Multiply both sides by 8.
\[ x = 4 \sqrt{3} \]
Approximate if wanted
Why: Four times 1.732.
\[ \text{about } 6.93 \]
Figure (svg): The two special right triangles and the values they generate
\[ x = 4\sqrt3 \approx 6.93 \]
Verify: check the size against the hypotenuse
Why: The leg 6.93 is shorter than the hypotenuse 8, as it must be, and it is the LONGER leg — correct, because it lies opposite the 60 degree angle. The shorter leg here would be exactly 4, half the hypotenuse, which is the signature of a 30-60-90 triangle.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 854-854
Fill the middle
Example 3.
Fill in the blanks
\frac4___ = \frac______ \;\Longrightarrow\; x = 8\!\left(\frac______\right) = ___\sqrt3
Why: Four root 3, about 6.93. Eight halved is 4, and the radical is carried through untouched.
Worked example
Guided Practice 5.
\[ \text{Solve } \triangle ABC \text{ with } C = 90^\circ, \; B = 45^\circ, \; c = 5. \]
Find the missing angle
Why: Ninety minus 45.
\[ A = 45 ^\circ \]
Recognise the triangle
Why: Both acute angles equal.
Find a leg
Why: Five times sine of 45.
\[ a = 5 \sqrt{2}\text{ over } 2 \]
Find the other leg
Why: The legs are equal.
\[ b = a,\text{ about } 3.54 \]
Figure (svg): The solution to Worked example a 45 degree triangle shown as a ladder of expressions, one row per algebraic move
\[ a = b = \frac{5\sqrt2}{2} \approx 3.54 \]
Verify: check with the Pythagorean theorem
Why: Two legs of 3.54 give 12.5 plus 12.5, which is 25, the square of the hypotenuse 5. The exact form 5 root 2 over 2 squared twice is 25 over 2 twice, or exactly 25 — the decimal rounding hides that the identity is exact.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 854-854
Trap
\[ \cos 30^\circ = 0.866 \]
Use the decimal throughout
Why: The calculator value is entered and carried.
\[ x = 8(0.866) = 6.928 \quad \text{(inexact)} \]
The answer is right to three places but is not the exact value the question wants, and the rounding compounds in any later step.
\[ \cos 30^\circ = \frac{\sqrt3}{2} \]
Use the exact value from the special table
Why: Keep the radical until the end.
\[ x = 8\!\left(\frac{\sqrt3}{2}\right) = 4\sqrt3 \]
Four root 3 is exact and can be approximated at the very last step if a decimal is wanted.
Matching
From the special triangles.
Match the pairs
Why: The sines rise 1/2, then about 0.707, then about 0.866 as the angle grows — sine always increases across the acute angles, while cosine always decreases. Tangent of 45 is exactly 1 because the two legs are equal.
Ranking
Smallest first.
Put in order
Why: The values are 0.5, 0.707, 0.866, 1 and 1.732. Sine stays below 1 for every acute angle while tangent passes through 1 at 45 degrees and grows without bound as the angle approaches 90.
Comparison
Fill the blanks. Both are built from a familiar shape.
Comparison matrix
| Question | 30-60-90 | 45-45-90 |
|---|---|---|
| Side lengths | 1, root 3, 2 | 1, 1, root 2 |
| Comes from | half an equilateral triangle | half a square |
| sin of the smaller angle | 1/2 | root 2 over 2 |
| tan of the smaller angle | root 3 over 3 | 1 |
Both triangles come from cutting a familiar figure in half, which is why their side lengths are so clean and worth memorising rather than re-deriving each time.
Section
Section 4
Concept
Solving a triangle means finding all unknown side lengths and angle measures. In a right triangle, one acute angle and one side are enough, and a calculator in degree mode handles angles other than the special ones.
\[ B = 90^\circ-A; \quad \tan A = \frac{a}{b} \]
A capital letter names an angle and the matching lowercase letter names the side opposite it, which is the convention for the rest of this chapter.
Figure (svg): A right triangle being solved from one angle and one side
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 854-854 — Solving a triangle
Picture it
Example 4: angle A is 28 degrees and side b is 15.
Figure (svg): A right triangle being solved from one angle and one side
The missing angle comes from the complement, and each missing side from one equation pairing it with the side already known.
Worked example
Example 4.
\[ \text{Solve } \triangle ABC \text{ with } C = 90^\circ, \; A = 28^\circ, \; b = 15. \]
Find angle B
Why: The acute angles are complementary.
\[ B = 62 ^\circ \]
Set up for side a
Why: Opposite over adjacent is tangent.
\[ \tan 28 = \frac{a}{15} \]
Solve for a
Why: Fifteen times tangent of 28.
\[ a = 7.98 \]
Solve for c
Why: Fifteen over cosine of 28.
\[ c = 17.0 \]
Figure (svg): A right triangle being solved from one angle and one side
\[ B = 62^\circ, \; a \approx 7.98, \; c \approx 17.0 \]
Verify: check with the Pythagorean theorem
Why: Seven point nine eight squared plus 15 squared is 63.7 plus 225, or 288.7, whose root is 16.99 — matching c to three places. Checking a solved triangle against the theorem catches a wrong ratio choice immediately.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 854-854
Fill the middle
Example 4.
Fill in the blanks
A = 28^\circ \;\Longrightarrow\; B = 90^\circ-28^\circ = 62^\circ
Why: Sixty-two degrees. The two acute angles of a right triangle always sum to 90, since all three sum to 180.
Worked example
Guided Practice 6 and 7.
\[ \text{Solve } \triangle ABC \text{ for } A = 32^\circ, b = 10; \text{ and for } A = 71^\circ, c = 20. \]
First: find B and a
Why: Ninety minus 32; then 10 times tangent 32.
\[ B = 58, a = 6.25 \]
First: find c
Why: Ten over cosine of 32.
\[ c = 11.79 \]
Second: find B and a
Why: Ninety minus 71; then 20 times sine of 71.
\[ B = 19, a = 18.91 \]
Second: find b
Why: Twenty times cosine of 71.
\[ b = 6.51 \]
Figure (svg): The solution to Worked example two more triangles shown as a ladder of expressions, one row per algebraic move
\[ 58^\circ, 6.25, 11.79; \qquad 19^\circ, 18.91, 6.51 \]
Verify: notice which ratio each case needed
Why: When the given side was a LEG, tangent and cosine were used; when it was the HYPOTENUSE, sine and cosine were. The rule is to pick the ratio containing both the unknown you want and the side you were handed, which never requires more than one equation.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 854-854
Error analysis
A student computes 15 times the tangent of 28 degrees.
Annotate
On: \( 15\tan 28 = 15(-0.271) = -4.07 \)
A negative side length in a right triangle is impossible, so the sign alone flags the error. Checking that the tangent of 45 returns exactly 1 confirms the mode in two keystrokes.
Sorting
Pair the unknown with the side you know.
Sort into buckets
In a right triangle with acute angle A, sort each situation by the ratio it needs.
Note that the same ratio serves whichever of its two sides is missing. Cosine handles both hypotenuse-from-adjacent and adjacent-from-hypotenuse; only the algebra afterwards differs.
Matching
The right angle is always at C.
Match the pairs
Why: The last is a 30-60-90 triangle, so its answers are exact: b is 7 root 3 and c is exactly 14, twice the shorter leg. The others needed a calculator because their angles are not special.
Prediction
Commit before reasoning.
Predict first
In a RIGHT triangle, are one side and one acute angle always enough to solve it completely?
Correct: Yes — the right angle is a third known part, so one side and one angle give all six parts.
\[ 90^\circ + A + \text{one side} \;\Longrightarrow\; \text{solved} \]
Why: A triangle has six parts and needs three to be determined, at least one of them a side. In a right triangle the 90 degree angle is free, so one more angle and one side complete the set. Any general triangle would need three genuine pieces, which is what Lessons 13.5 and 13.6 will supply through the laws of sines and cosines.
Section
Section 5
Concept
An angle measured from a known position, together with a known distance, gives a length that cannot be measured directly. The angle above the horizontal is the angle of elevation; the angle below it is the angle of depression, and the two are equal.
\[ \sin 48^\circ = \frac{h}{300} \;\Longrightarrow\; h = 300\sin 48^\circ \]
Drawing the diagram first, with the right angle and the known parts marked, is what turns a description into an equation.
Figure (svg): An angle of elevation used to find a height indirectly
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 855-855 — Use indirect measurement
Picture it
Example 6: a 300 foot rope at 48 degrees.
Figure (svg): An angle of elevation used to find a height indirectly
The rope is the hypotenuse and the height is opposite the angle, so a single sine equation gives 223 feet.
Worked example
Example 6, and Guided Practice 10.
\[ \text{A } 300 \text{ foot rope makes a } 48^\circ \text{ angle with the water. Find the height, and again at } 38^\circ. \]
Draw and label
Why: Rope as hypotenuse, height opposite the angle.
\[ \sin 48 = \frac{h}{300} \]
Solve
Why: Three hundred times sine of 48.
\[ h = 223\text{ feet} \]
Repeat at 38 degrees
Why: Three hundred times sine of 38.
\[ h = 185\text{ feet} \]
Compare
Why: Ten degrees lower.
\[ 38\text{ feet less height} \]
Figure (svg): An angle of elevation used to find a height indirectly
\[ h \approx 223; \qquad h \approx 185 \]
Verify: check the height against the rope
Why: Two hundred twenty-three feet is well under the 300 foot rope, as it must be — the height is a leg and the rope the hypotenuse. At 90 degrees the parasailer would be directly overhead at the full 300 feet, and at 0 degrees at water level, so 223 at 48 degrees sits sensibly between.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 855-855
Fill the middle
Example 6.
Fill in the blanks
h = 300\sin 48^\circ = 300(0.7431) \approx 223
Why: About 223 feet, comfortably less than the 300 foot rope because the height is a leg rather than the hypotenuse.
Worked example
Example 5, and Guided Practice 9.
\[ \text{From Powell Point the angle to Widforss Point is } 76^\circ \text{ and the near side is } 2 \text{ miles. Find the width, and the far distance.} \]
Identify the sides
Why: The 2 miles is adjacent, the width opposite.
\[ \tan 76 = \frac{x}{2} \]
Solve for the width
Why: Two times tangent of 76.
\[ x = 8.0\text{ miles} \]
Set up for the third side
Why: Two over cosine of 76.
Solve
Why: Two over 0.242.
\[ \text{about } 8.3\text{ miles} \]
Figure (svg): The solution to Worked example the width of a canyon shown as a ladder of expressions, one row per algebraic move
\[ x \approx 8.0; \qquad \approx 8.3 \]
Verify: check the hypotenuse is the longest side
Why: Eight point three exceeds both 8.0 and 2, as the hypotenuse must. The two distances 8.0 and 8.3 are close because the 76 degree angle is steep, making the triangle long and thin — a two-mile walk converted into an eight-mile measurement.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 855-855
Trap
\[ \text{a } 48^\circ \text{ angle of elevation} \]
Measure the angle from the vertical
Why: The angle is placed at the top of the triangle rather than at the observer.
\[ h = 300\cos 48^\circ = 201 \quad \text{(wrong)} \]
An angle of elevation is measured from the HORIZONTAL. Using cosine puts the height adjacent to the angle instead of opposite it.
\[ \sin 48^\circ = \frac{h}{300} \]
Draw the horizontal and mark the angle above it
Why: Then read which side is opposite.
\[ h = 300\sin 48^\circ \approx 223 \]
The two answers differ by more than twenty feet, so drawing the diagram is not optional bookkeeping.
Fill the middle
Example 5.
Fill in the blanks
x = 2\tan 76^\circ = 2(4.011) \approx 8.0
Why: About 8.0 miles. A steep angle makes the tangent large, which is how a two-mile baseline measures an eight-mile distance.
Comparison
Fill the blanks. Looking up, and looking down.
Comparison matrix
| Question | Angle of elevation | Angle of depression |
|---|---|---|
| Measured from | the horizontal, looking up | the horizontal, looking down |
| Who measures it | the person below | the person above |
| Their measures | equal | equal |
| Why equal | they are alternate interior angles between parallel horizontals | the same reason |
Because the two horizontals are parallel, the angles are alternate interior angles and therefore equal. That equality lets you take the measurement from whichever end is easier to reach.
Prediction
Commit before reasoning.
Predict first
At 48 degrees the parasailer is 223 feet up. At 96 degrees, would the height be about 446 feet?
Correct: No — the height cannot exceed the 300 foot rope, and 96 degrees is not even a valid elevation here.
\[ 300\sin 90^\circ = 300 \text{ (the maximum)} \]
Why: Height is 300 times the sine of the angle, and sine never exceeds 1, so the height caps at 300 feet no matter what. Sine is not proportional to the angle: going from 30 to 60 degrees raises the sine from 0.5 only to 0.866, not to 1. This non-proportionality is exactly why a table or calculator is needed rather than simple scaling, and it is the first hint of the wave shape Lesson 14.1 will draw.
Comparison
Fill the blanks. Which sides each one connects.
Comparison matrix
| Question | Sine | Cosine | Tangent |
|---|---|---|---|
| Definition | opp/hyp | adj/hyp | opp/adj |
| Reciprocal | cosecant | secant | cotangent |
| Value at 45 degrees | root 2 over 2 | root 2 over 2 | 1 |
| Can exceed 1? | no | no | yes |
Sine and cosine are bounded because the hypotenuse is the longest side. Tangent compares two legs and so has no ceiling, growing without limit as the angle nears 90 degrees.
Pattern
Label, choose, solve.
Check the calculator is in degree mode. The tangent of 45 should return exactly 1.
OpenStax Algebra and Trigonometry 2e, §7.2 Right Triangle Trigonometry §7.2
Check
Find the hypotenuse first.
Check your understanding
In a right triangle the leg opposite theta is 12 and the adjacent leg is 5. What is cos theta?
Answer: A
Why: The hypotenuse is 13, and cosine is adjacent over hypotenuse.
Check
Draw the triangle the ratio describes.
Check your understanding
If theta is acute and sin theta = 3/8, what is tan theta?
Answer: A
Why: The adjacent side is root 55, so tan = 3/root 55 = 3 root 55 over 55.
Check
Pair the unknown with a side you know.
Check your understanding
In right triangle ABC with C = 90 degrees, A = 28 degrees and b = 15, what is side a to two decimal places?
Answer: A
Why: a is opposite A and b is adjacent, so tan 28 = a/15 and a = 15 tan 28.
Real world
You stand 40 feet from the base of a tree. The angle of elevation to the top is 52 degrees, and your eyes are 5.5 feet above the ground.
Discussion prompt
How tall is the tree?
Hint: The triangle sits on your eye level, not on the ground.
Answer:
\[ \tan 52^\circ = \frac{h}{40} \;\Longrightarrow\; h = 40\tan 52^\circ \approx 51.2 \]
\[ \text{tree height} = 51.2+5.5 = 56.7 \text{ feet} \]
The tree is about 56.7 feet tall.
The 5.5 feet is the step everyone forgets. The triangle's horizontal leg runs at eye level, so the height it computes is the distance from EYE LEVEL to the treetop, and your own height has to be added back. Any indirect measurement taken from a standing observer needs the same correction, and the shorter the object the more the omission matters — for a six-foot fence post the error would be almost the whole answer.
Commit first
Answer, then rate your confidence honestly.
Predict first
A student computes a sine of 1.4 for an acute angle. Is that possible?
Correct: No — sine is opposite over hypotenuse, and a leg is always shorter than the hypotenuse, so sine is under 1.
\[ 0 < \sin\theta < 1; \quad \tan\theta \text{ unbounded} \]
Why: The hypotenuse is opposite the right angle and is therefore the longest side of the triangle, so any leg divided by it gives a fraction below 1. The same holds for cosine. Tangent has no such bound because it compares two legs, and secant and cosecant have the hypotenuse on top so they are always at least 1. Knowing which ratios are bounded gives four free error checks on every problem in this chapter.
Explain it
They know the Pythagorean theorem and have never met sine.
Discussion prompt
In four sentences or fewer, explain what sine of an angle means.
Hint: Talk about triangles of different sizes with the same angle.
Answer:
Draw any right triangle with your angle in it. Divide the side across from the angle by the longest side.
You get the same answer whatever size triangle you drew, because bigger triangles have both sides bigger in the same proportion. That number is the sine of the angle, and it depends only on the angle.
Exit ticket
Name the weakest spot before you close the deck.
Predict first
Which of these would you least want handed to you cold?
Correct: Whichever you picked is tonight's ten minutes, and each has a one-line fix.
Why: For opposite and adjacent, re-label from scratch for each angle you consider. For recovering ratios, draw the triangle the given ratio names and use the Pythagorean theorem. For choosing a ratio, name the side you want and the side you have, then pick the ratio containing both. For diagrams, draw the horizontal first and mark the angle against it.
Connect it up
Paper. Fifteen minutes.
Draw it
Build a right-triangle trigonometry page. Top left: draw one right triangle, label the three sides from one acute angle, write all six ratios, then re-label from the OTHER acute angle and write all six again, circling the pairs that swapped. Top right: draw both special triangles with their exact side lengths and fill in the full table of sine, cosine and tangent at 30, 45 and 60. Middle: take a single given ratio, draw the triangle it describes, find the missing side, and write out the remaining five ratios in exact form. Bottom left: solve a triangle completely from one angle and one side, then check your answer with the Pythagorean theorem. Bottom right: draw an indirect measurement of your own invention, marking the horizontal, the angle of elevation and the right angle, and solve it.
If any sine or cosine on your page came out above 1, that ratio is upside down: the hypotenuse belongs underneath.
Recap
Five things, and a way to measure what you cannot reach.
| If you see | Then |
|---|---|
| Two sides of a right triangle | The third follows from the Pythagorean theorem |
| One ratio given | Draw that triangle and read off the others |
| An angle of 30, 45 or 60 | Use the exact value, not a decimal |
| An unknown side and a known side | Pick the ratio containing both |
| An angle of elevation | Measure it from the horizontal |
| A sine or cosine above 1 | The ratio is upside down |
Lesson 13.2 frees the angle from the triangle: it can be placed on the coordinate plane, made larger than 90 degrees, and even measured in a unit other than degrees.
McDougal Littell Algebra 2 (Texas Edition), Ch. 13 Trigonometric Ratios and Functions — Lesson 13.1 Use Trigonometry with Right Triangles §13.1, pp. 852-857 — everything on these slides traces back here
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