6.8 — Are Relationships Predictable?
1. In class: A Develop Understanding Task
Name the sides relative to an acute angle
Every right triangle has a hypotenuse: the longest side, directly opposite the 90° angle. To describe the other two sides, choose one of the triangle's acute angles as your reference angle.
- The opposite side is the side across from the reference angle; it does not touch that angle.
- The adjacent side touches the reference angle but is not the hypotenuse.
- The hypotenuse is determined by the 90° angle and never changes when you switch the reference angle.
In the 3–4–5 triangle, the right angle is at C. Relative to angle A, side BC is opposite (length 3), AC is adjacent (length 4), and AB is the hypotenuse (length 5). Thus
\[\begin{aligned}\frac{\text{opposite}}{\text{hypotenuse}}&=\frac35,\\[4pt]\frac{\text{adjacent}}{\text{hypotenuse}}&=\frac45,\\[4pt]\frac{\text{opposite}}{\text{adjacent}}&=\frac34.\end{aligned}\]Relative to angle B, opposite and adjacent switch: AC (length 4) is opposite B and BC (length 3) is adjacent B. AB remains the hypotenuse.
Why the ratios are predictable
Imagine drawing two right triangles that both have a 60° angle. Each has one angle of 90° and another of 60°, so the third angle must be 30°. The triangles are similar by AA: they have the same angles, even if their lengths differ.
Similar triangles have proportional corresponding sides. If a larger copy multiplies every length by a factor of 2, then
\[\frac{2\cdot\text{opposite}}{2\cdot\text{hypotenuse}}=\frac{\text{opposite}}{\text{hypotenuse}}.\]The factor of 2 cancels. The same cancellation works for adjacent/hypotenuse and opposite/adjacent. Therefore the side ratios are determined by the chosen acute angle, not by the size of the triangle.
One right triangle has, relative to its reference angle, an opposite side of 6, an adjacent side of 8, and a hypotenuse of 10. A similar triangle is three times as large: its corresponding lengths are 18, 24, and 30.
\[\begin{aligned}\frac6{10}&=\frac{18}{30}=\frac35,\\[5pt]\frac8{10}&=\frac{24}{30}=\frac45,\\[5pt]\frac68&=\frac{18}{24}=\frac34.\end{aligned}\]The side measurements change; the three ratios do not.
Measurements made with a ruler may not match exactly because of drawing and measuring error. But the equality is mathematically exact for truly similar triangles.
The two 60° triangles in the main task
The packet's first triangle has a hypotenuse of 6 and a leg of 3. Use the Pythagorean theorem to find the other leg:
\[\begin{aligned}3^2+b^2&=6^2,\\b^2&=36-9=27,\\b&=\boxed{3\sqrt3}.\end{aligned}\]In the second triangle, the corresponding side lengths are twice as large: 6, \(6\sqrt3\), and 12.
For reference angle A in the smaller triangle, the opposite side is \(3\sqrt3\), the adjacent side is 3, and the hypotenuse is 6.
\[\begin{aligned}\frac{\text{opp}}{\text{hyp}}&=\frac{3\sqrt3}{6}=\frac{\sqrt3}{2},\\[4pt]\frac{\text{adj}}{\text{hyp}}&=\frac36=\frac12,\\[4pt]\frac{\text{opp}}{\text{adj}}&=\frac{3\sqrt3}{3}=\sqrt3.\end{aligned}\]For reference angle D in the larger triangle, the same three ratios are
\[\begin{aligned}\frac{6\sqrt3}{12}&=\frac{\sqrt3}{2},\\[4pt]\frac6{12}&=\frac12,\\[4pt]\frac{6\sqrt3}{6}&=\sqrt3.\end{aligned}\]Both reference angles are 60° because both triangles are 30°–60°–90° right triangles. Every right triangle with a 60° reference angle gives these ratios, however large or small.
2. Ready review — Find a missing side and angle
Use two familiar facts:
\[\boxed{a^2+b^2=c^2}\]Here \(a,b\) are the legs and \(c\) is the hypotenuse. Also, in every right triangle:
\[\boxed{\text{Acute angles sum to }90^\circ}\]First identify which side is the hypotenuse. Its square belongs alone on the right side of the Pythagorean equation. Use the positive square root for a length. For the remaining acute angle, subtract the given acute angle from 90°.
A right triangle has a hypotenuse of 17, one leg of 8, and an acute angle of 28°. Find the other leg and the other acute angle.
\[\begin{aligned}b^2&=17^2-8^2,\\&=289-64=225,\\b&=\boxed{15}.\end{aligned}\] \[90^\circ-28^\circ=\boxed{62^\circ}.\]These are two separate steps: finding the leg uses lengths; finding the other acute angle uses the triangle angle sum.
A right triangle has legs of lengths 7 and 9. Find its hypotenuse.
\[\begin{aligned}c^2&=7^2+9^2=130,\\c&=\sqrt{130}\approx\boxed{11.40}.\end{aligned}\]Do not subtract when looking for the hypotenuse; its square is the sum of the squared legs.
3. Set review — Write ratios for both acute angles
The side ratios explored in class now receive their trigonometric names.
Three named ratios: sine, cosine, and tangent
These ratios are important enough to have names. For an acute angle \(\theta\) in a right triangle:
| Name | Definition | Mnemonic |
|---|---|---|
| Sine, \(\sin\theta\) | \(\dfrac{\text{opposite}}{\text{hypotenuse}}\) | SOH |
| Cosine, \(\cos\theta\) | \(\dfrac{\text{adjacent}}{\text{hypotenuse}}\) | CAH |
| Tangent, \(\tan\theta\) | \(\dfrac{\text{opposite}}{\text{adjacent}}\) | TOA |
SOHCAHTOA is commonly pronounced “soh-kah-TOH-uh”, almost like a single made-up word. It is not a mathematical term; it is simply a way to remember which sides go in each ratio. Break it into three chunks:
SOH: Sine = Opposite / Hypotenuse.
CAH: Cosine = Adjacent / Hypotenuse.
TOA: Tangent = Opposite / Adjacent.
Use it in this order: Identify the reference angle, find the hypotenuse across from the right angle, and label the opposite and adjacent legs relative to the reference angle. Then use the mnemonic to choose the ratio. The mnemonic does not tell you which leg is opposite or adjacent.
Each trigonometric ratio is a number, not a length. A larger similar triangle has the same ratio.
In a right triangle, the side opposite angle \(T\) is 5, the adjacent leg is 12, and the hypotenuse is 13. Then
\[\begin{aligned}\sin T&=\frac5{13},\\[4pt]\cos T&=\frac{12}{13},\\[4pt]\tan T&=\frac5{12}.\end{aligned}\]The letters in the triangle's name do not determine the ratio. You must use the angle indicated in the question.
Each Set diagram places its right angle at C. The other two vertices are A and B. Once you have the three side lengths, work from the same picture twice: first use angle A as the reference, then use angle B. The hypotenuse \(AB\) stays the same; the legs switch which one is opposite and which is adjacent.
Use the labeled triangle in the In class section, with right angle C and lengths \(AC=4\), \(BC=3\), \(AB=5\).
| Ratio | Using angle A | Using angle B |
|---|---|---|
| Sine | \(\sin A=\frac35\) | \(\sin B=\frac45\) |
| Cosine | \(\cos A=\frac45\) | \(\cos B=\frac35\) |
| Tangent | \(\tan A=\frac34\) | \(\tan B=\frac43\) |
For angle A, BC is opposite and AC is adjacent. For angle B, AC is opposite and BC is adjacent. The two angles are not interchangeable.
Right triangle PQR has a right angle at R. Suppose \(PR=5\) and \(QR=12\). The hypotenuse is
\[PQ=\sqrt{5^2+12^2}=\sqrt{169}=13.\]Using angle P, the opposite leg is \(QR=12\) and the adjacent leg is \(PR=5\), so
\[\begin{aligned}\sin P&=\frac{12}{13},\\[4pt]\cos P&=\frac5{13},\\[4pt]\tan P&=\frac{12}{5}.\end{aligned}\]Using angle Q instead gives \(\sin Q=5/13\), \(\cos Q=12/13\), and \(\tan Q=5/12\).
Some displayed hypotenuse lengths are rounded: for example, problem 7 shows 8.06 alongside legs 7 and 4, while their exact Pythagorean hypotenuse is \(\sqrt{65}\approx8.0623\). In problem 8, legs 8.5 and 3 give \(\sqrt{81.25}\approx9.0139\), displayed as 9.01. The printed decimals are approximations, not exact equalities to the square-root expressions. If you calculate from the legs, you may keep an exact square root until the last step and then round ratios as needed.
4. Go review — Factor quadratics and find intercepts
A quadratic in the form \(f(x)=x^2+bx+c\) can often be factored into \((x+m)(x+n)\). To find \(m\) and \(n\), look for two numbers that multiply to \(c\) and add to \(b\).
When a function is in factored form, its x-intercepts occur where the output is zero, so set each factor equal to zero. Its y-intercept occurs where \(x=0\), so substitute 0 into the original formula.
Factor \(f(x)=x^2+11x+24\) and identify all intercepts.
Factor: 3 and 8 add to 11 and multiply to 24.
\[f(x)=(x+3)(x+8).\]x-intercepts: \(x+3=0\) or \(x+8=0\), so \(\boxed{(-3,0),\ (-8,0)}\).
y-intercept: \(f(0)=24\), so \(\boxed{(0,24)}\).
Factor \(g(x)=x^2-x-42\). Find its intercepts.
The numbers must multiply to -42 and add to -1. They are -7 and 6.
\[g(x)=(x-7)(x+6).\]x-intercepts: \(\boxed{(7,0),\ (-6,0)}\). y-intercept: \(\boxed{(0,-42)}\).
Check: \((-7)(6)=-42\) and \(-7+6=-1\).
Special case: difference of squares
When the middle x-term is missing and the constant is a negative square, use \(a^2-b^2=(a-b)(a+b)\):
For \(h(x)=x^2-100\), factor and find intercepts.
\[h(x)=(x-10)(x+10).\]x-intercepts: \(\boxed{(-10,0),\ (10,0)}\); y-intercept: \(\boxed{(0,-100)}\).
Special case: a perfect square
Sometimes both factors are the same. This gives one distinct x-intercept, not two different points.
For \(k(x)=x^2-10x+25\), use -5 and -5 as the two numbers.
\[k(x)=(x-5)^2.\]x-intercept: \(\boxed{(5,0)}\) (a repeated root). y-intercept: \(\boxed{(0,25)}\). The graph touches the x-axis at x = 5 instead of crossing it.
Additional Learning Resources
| Resource and direct link | What it covers / why use it | Scope / What to skip |
|---|---|---|
| Khan Academy — Trigonometric ratios in right triangles | A clear explanation of opposite, adjacent, hypotenuse and how to write sine, cosine, and tangent, matching the main task and Set. | Close match. Use the whole article for this section. |
| Professor Leonard — Introduction to Right Triangle Trigonometry (Precalculus — Trigonometry 30) | Full-classroom explanation of the meaning of sine, cosine, and tangent in right triangles. | Some extra material. Watch the explanation of the three basic ratios; skip reciprocal functions and cofunction identities, which are beyond today's required material. Exact timestamps were not verified. |
| Khan Academy — Factoring simple quadratics review | Written review of factoring expressions with leading coefficient 1, useful for Go 11–19. | Close match. Focus on products of two binomials and the sum/product method. |
| Khan Academy — Interpret quadratic models: factored form | Illustrates how roots from factors become x-intercepts. | Some extra material. Watch the factored-form/intercept explanation; the applied modeling context is optional. |
These online resources require internet access. The lesson itself and its equations are bundled for offline use.
Teacher review — Substantive packet issues
No substantive mathematical errors identified in Section 6.8. Two Set diagrams (7 and 8) show hypotenuse lengths rounded to two decimals rather than exact radicals; a brief clarification is provided above so students do not mistake rounded quantities for exact identities.
The main task starts by having students measure a hand-drawn 60° triangle. The equality of ratios is exact only for the ideal geometric triangles; small disagreements from ruler measurements are expected and should not be treated as counterexamples.
Section 6.8 — Packet-to-lesson coverage
| Curriculum component | Original problems / mathematical objective | Supplement section |
|---|---|---|
| Main task, p. 42 | Draw a 60° right triangle; identify three sides and calculate three ratios; Problem 1 compares measurements | In class |
| Main task, p. 43 | Problem 2: find missing sides and three ratios for ABC, DEF | In class |
| Main task, pp. 43–44 | Problems 3–6: reason about common angles and invariant side ratios | In class |
| Ready, p. 45 | Problems 1–6: missing leg or hypotenuse and remaining acute angle | Section 2 |
| Set, p. 46 | Problems 7–10: sin, cos, tan for acute angles A and B; Pythagorean review | Section 3 |
| Go, p. 47 | Problems 11–19: factor quadratics; x- and y-intercepts; repeated roots; difference of squares | Section 4 + Go practice A–D |