Right Triangle Slides Geometry & Trigonometry
All 6 Trig Functions
Right Triangle Trigonometry
Mastering the six trigonometric ratios, calculator reciprocals, and real-world models.
1
All 6 Trig Ratios
2
Calculator Reciprocals
3
Real-World Modeling
The Six Ratios: Primary & Reciprocal
Definitions
SINE & COSECANT Opposite / Hypotenuse Pair
\( \sin\theta = \frac{\text{Opp}}{\text{Hyp}} \)
\( \csc\theta = \frac{\text{Hyp}}{\text{Opp}} = \frac{1}{\sin\theta} \)
COSINE & SECANT Adjacent / Hypotenuse Pair
\( \cos\theta = \frac{\text{Adj}}{\text{Hyp}} \)
\( \sec\theta = \frac{\text{Hyp}}{\text{Adj}} = \frac{1}{\cos\theta} \)
TANGENT & COTANGENT Opposite / Adjacent Pair
\( \tan\theta = \frac{\text{Opp}}{\text{Adj}} \)
\( \cot\theta = \frac{\text{Adj}}{\text{Opp}} = \frac{1}{\tan\theta} \)
Memory Trick: Every reciprocal pair has exactly ONE “co-”! Sine ↔ Co secant, Co sine ↔ Secant, Tangent ↔ Co tangent.
Evaluating All 6 Ratios: \(\triangle ABC\)
5-12-13 Triangle
A C B 12 (Adj) 5 (Opp) 13 (Hyp) Relative to Angle A
\(\sin A\)
\(\frac{5}{13}\)
\(\csc A\)
\(\frac{13}{5}\)
\(\cos A\)
\(\frac{12}{13}\)
\(\sec A\)
\(\frac{13}{12}\)
\(\tan A\)
\(\frac{5}{12}\)
\(\cot A\)
\(\frac{12}{5}\)
Notice: Reciprocals simply flip the fraction numerator and denominator!
Exact values remain simplified improper fractions (e.g. \(13/5\), not mixed numbers). \(\csc A = \frac{1}{\sin A} = \frac{13}{5}\)
CFU #1
Check for Understanding: Rotated Ratios
Fill in Guided Notes Box
α 20 (Adj) 15 (Opp) 25 (Hyp) Target Angle: \(\alpha\)
\(\sin\alpha = \frac{15}{25} = \mathbf{\frac{3}{5}}\) \(\csc\alpha = \mathbf{\frac{5}{3}}\)
\(\cos\alpha = \frac{20}{25} = \mathbf{\frac{4}{5}}\) \(\sec\alpha = \mathbf{\frac{5}{4}}\)
\(\tan\alpha = \frac{15}{20} = \mathbf{\frac{3}{4}}\) \(\cot\alpha = \mathbf{\frac{4}{3}}\)
Check: Once you simplify the primary ratio, simply flip it for the reciprocal ratio!
All 6 ratios are exact fractions derived strictly from the side lengths.
Calculator Reciprocals: No “SEC” Key!
Calculator Protocol
The Inverse Button Trap
There are NO dedicated csc, sec, or cot keys on standard calculators!
FATAL ERROR: Pressing \(\sin^{-1}, \cos^{-1}, \tan^{-1}\)
Those are INVERSE functions to find angles, NOT reciprocals!
Correct: Use division \(\frac{1}{\cos(\theta)}\) OR the \([x^{-1}]\) key!
How to Evaluate on Your Calculator:
\(\sec(40^\circ) = 1 / \cos(40^\circ)\) \(\sec(40^\circ)\)
\(\approx 1.3054\)
\(\csc(55^\circ) = 1 / \sin(55^\circ)\) \(\csc(55^\circ)\)
\(\approx 1.2208\)
\(\cot(28^\circ) = 1 / \tan(28^\circ)\) \(\cot(28^\circ)\)
\(\approx 1.8807\)
Remember: Make sure you are in DEGREE mode first! (tan(45) = 1)
CFU #2
Check for Understanding: Spot the Error
Whiteboard / Notes Check
Student A (Taylor) CORRECT
sec(60°) = 1 / cos(60°) = 2.0
Taylor correctly calculated the reciprocal of \(\cos(60^\circ) = 0.5\). The reciprocal of \(1/2\) is \(2\).
Student B (Morgan) ERROR!
sec(60°) = cos⁻¹(60) = ERROR / DOMAIN
Morgan pressed the inverse cosine button instead of dividing by \(\cos(60^\circ)\)!
Teacher Diagnostic: Never use the “2nd” or “Shift” button for reciprocal trig functions!
3-Step Solving Strategy
Method
01
Label the Triangle
Circle reference angle \(\theta\). Label sides as Opposite , Adjacent , and Hypotenuse .
Identify given side and target unknown side.
02
Select the Ratio
Choose the trig ratio (\(\sin, \cos, \tan\) or reciprocal) linking the known side and unknown side.
Opp & Hyp \(\rightarrow \sin\)
03
Set Up & Solve
Isolate the unknown variable. Multiply or swap and divide across the equals sign.
Variable in numerator \(\rightarrow\) Multiply.
Reciprocal note: Using \(\sec(34^\circ) = \frac{x}{15}\) gives \(x = 15 \cdot \sec(34^\circ)\), avoiding division! \(x = 15 \cdot (1.2062) \approx 18.1\)
CFU #3
Check for Understanding: Missing Side
Fill in Guided Notes Box
48° 30 h = ? Given: \(\theta = 48^\circ\), Adj = \(30\) | Target: Opp = \(h\)
1. Ratio Selected: Tangent (\(\tan\)) or \(\cot\)
2. Equation Setup: \(\tan(48^\circ) = \frac{h}{30}\)
3. Solve: \(h = 30 \cdot \tan(48^\circ)\)
4. Result: \(h \approx 33.3\)
Reciprocal Alternative: \(\cot(48^\circ) = \frac{30}{h} \implies h = \frac{30}{\cot(48^\circ)} = 30 \cdot \tan(48^\circ) \approx 33.3\).
Both primary and reciprocal approaches yield identical geometric answers. Consistent & Verified
Real-World Angles: Elevation & Depression
Crucial Distinction
Observer Target Angle of Elevation Angle of Depression Horizontal Line of Sight
Angle of Elevation
Measured UPWARD from the horizontal line of sight to an elevated object.
Angle of Depression
Measured DOWNWARD from the horizontal line of sight to an object below.
Geometric Fact: Because horizontal lines are parallel, Angle of Elevation = Angle of Depression (Alternate Interior Angles).
Common Trap: Never measure the angle against the vertical wall or cliff! Always from the horizontal!
Guided Challenge: Height of a Lighthouse
Modeling
The Scenario
A marine surveyor stands 120 feet away from the base of a lighthouse. The angle of elevation from ground level to the top is 34°.
Find the height (\(h\)) of the lighthouse.
34° 120 ft h = ?
Step 1: Identify Given & Unknown
Angle = \(34^\circ\) | Adjacent = \(120\text{ ft}\) | Opposite = \(h\)
Step 2: Choose the Trig Ratio
Opposite & Adjacent \(\rightarrow\) Tangent (\(\tan\))
Step 3: Equation & Calculator Evaluation
\(\tan(34^\circ) = \frac{h}{120}\)
\(h = 120 \cdot \tan(34^\circ) = 120 \cdot (0.6745) \approx \mathbf{80.94\text{ ft}}\)
Ready for individual & partner practice! Answer: The lighthouse is approximately 80.9 feet tall.
Lesson Summary & Next Steps
Wrap-Up
All 6 Trig Ratios
Primary: \(\sin, \cos, \tan\). Reciprocals: \(\csc = \frac{1}{\sin}, \sec = \frac{1}{\cos}, \cot = \frac{1}{\tan}\).
DEG Mode & 1 / x
Always check DEG mode. Use division \(\frac{1}{\cos(\theta)}\) for secant — never use the \(\cos^{-1}\) inverse key!
Real-World Models
Draw the triangle, identify the horizontal line of sight, and solve for height or ground distance.
Guided Notes complete → Practice Worksheet → Exit Ticket
10-min Exit Ticket at end
Right Triangle Practice Worksheet The 6 Trig Ratios Primary & Reciprocal Functions
Right Triangle Trigonometry: Practice
Name:
Date: Period:
6 RATIOS: \(\sin = \frac{O}{H} \leftrightarrow \csc = \frac{H}{O}\) | \(\cos = \frac{A}{H} \leftrightarrow \sec = \frac{H}{A}\) | \(\tan = \frac{O}{A} \leftrightarrow \cot = \frac{A}{O}\)
Calculator: \(\sec\theta = 1/\cos\theta\)
1 Evaluating All Six Trig Ratios (Exact Fractions)
Simplify all fractions completely.
Triangle A: 8-15-17 (Target: \(\angle A\)) Right \(\angle C\)
A C B 15 8 17
\(\sin A =\)
\(\csc A =\)
\(\cos A =\)
\(\sec A =\)
\(\tan A =\)
\(\cot A =\)
Triangle B: 9-12-? (Target: \(\angle X\)) Hyp = 15
X Z Y 12 9 15
\(\sin X =\)
\(\csc X =\)
\(\cos X =\)
\(\sec X =\)
\(\tan X =\)
\(\cot X =\)
2 Calculator Reciprocal Lab (Degree Mode)
Evaluate to 4 decimal places.
A. Primary Ratios
\(\sin(38^\circ) \approx\)
\(\cos(72^\circ) \approx\)
\(\tan(15^\circ) \approx\)
B. Reciprocal Ratios
\(\csc(38^\circ) \approx\)
\(\sec(72^\circ) \approx\)
\(\cot(15^\circ) \approx\)
C. Solve for \(x\) (1 dec.)
\(x = 14 \cdot \sec(40^\circ)\)
\(x = 9 \cdot \csc(52^\circ)\)
\(x = 20 \cdot \cot(33^\circ)\)
3 Solving for Unknown Sides
Round to tenths.
32° x 14
Problem 1: Find \(x\)
Ratio setup:
\(x \approx\)
55° y 22
Problem 2: Find \(y\)
Ratio setup:
\(y \approx\)
The 6 Trig Functions — Practice Worksheet Page 1 of 2 Turn over for Real-World Modeling →
Part 4: Applied Modeling
Real-World Trigonometry & Indirect Measurement
Protocol: Sketch → Ratio → Solve with Units
! Modeling Reminder: You may use either primary ratios or their reciprocals to solve!
\(\cot\theta = 1/\tan\theta\)
1 The Pine Tree Height (Angle of Elevation)
Standard Modeling
A forestry technician stands 65 feet from the base of a tall ponderosa pine. Using a clinometer, the technician measures the angle of elevation to the top of the tree to be \(54^\circ\) . Disregarding eye height, how tall is the tree? Round to the nearest tenth of a foot.
Draw & Label Triangle:
Show Equation & Calculation:
Tree Height =
2 Coastal Search & Rescue (Angle of Depression)
Use \(\tan\) or \(\cot\)
A lifeguard sits in an observation tower with their eye level 28 feet above the water. They spot a swimmer in distress at an angle of depression of \(18^\circ\) . What is the direct horizontal distance from the base of the tower to the swimmer? Round to tenths.
Draw Triangle (relative to horizontal):
Show Equation & Calculation:
Horizontal Distance =
3 Surveyor's Challenge: Eye-Level Adjustment
Multi-Step Extension
An architect whose eyes are 5.5 feet above ground level stands 90 feet away from a school building. The angle of elevation to the roofline is \(28^\circ\) . What is the total height of the school building? (Hint: Find the height above eye level, then add 5.5 ft. )
Draw Schematic (include eye height):
Show Both Steps (Trig Height + 5.5 ft):
Total Building Height =
Self-Check: How confident are you with both primary and reciprocal trig ratios?
Still shaky Getting there Got it mastered!
Trig Mastery Exit Ticket Formative Check 10 Minutes • All 6 Trig Ratios
Trigonometry Exit Ticket
Name:
Date: Period:
Directions: Complete independently. Show clear algebraic setups and exact fractions where appropriate.
Score: ____ / 10
1 Evaluate Primary & Reciprocal Ratios (4 pts) Exact fraction form
θ 24 7 25
\(\sin\theta\)
\(\csc\theta\)
\(\cos\theta\)
\(\sec\theta\)
2 Calculator Reciprocal Evaluation (3 pts) Round to 4 decimal places
A. Evaluate \(\sec(50^\circ)\):
Method: \(1 / \cos(50^\circ)\)
\(\sec(50^\circ) \approx\)
B. Evaluate \(\cot(25^\circ)\):
Method: \(1 / \tan(25^\circ)\)
\(\cot(25^\circ) \approx\)
C. Solve for \(x\) (tenth):
\(x = 16 \cdot \cos(37^\circ)\)
\(x \approx\)
3 Real-World Modeling: Kite Flight (3 pts) Applied Trig
Maya is flying a kite at the park. She has let out 160 feet of string, and the string is pulled taut at an angle of elevation of \(52^\circ\) with the ground. Assuming her hands hold the spool at ground level, find the vertical height of the kite. Round to the nearest tenth of a foot.
Draw & Label Triangle:
Equation & Calculator Steps:
Kite Height =
Self-Reflection
One concept I feel 100% confident with today:
One question or doubt I still have:
Right Triangle Teacher Guide Educator Facilitation Guide One-Period Lesson Plan (50–55 Minutes)
The Six Trig Ratios: Teacher Guide
Standards: CCSS HSG-SRT.C.6, C.7, C.8
Key Learning Targets
Evaluate all six trigonometric ratios (\(\sin, \cos, \tan, \csc, \sec, \cot\)).
Configure calculators in DEGREE mode and evaluate reciprocals via \(1/\text{trig}\).
Avoid the inverse button trap (\(\cos^{-1}\) vs. \(\sec\)).
Model elevation & depression scenarios to find real-world distances.
Bundle Architecture
Slides (11 slides): 6 ratios, reciprocal keys, 3 CFUs.Guided Notes (2 pages): All 6 ratios + 4 active CFUs.Practice Worksheet (2 pages): 6-ratio evaluations & modeling.Exit Ticket (1 page): 10-minute check including reciprocals.
Instructional Pacing & CFU Routine 50-Minute Standard Period
00–14 min Direct + CFU 1
All 6 Trig Ratios
Slides 1–4. Teach primary and reciprocal pairs (memory trick: one "co-" per pair). Pause at Slide 4 for CFU #1 (Rotated 6-Ratio Check) .
14–22 min Lab + CFU 2
Calculator Reciprocals
Slides 5–6. Model evaluating \(\sec(40^\circ) = 1/\cos(40^\circ)\). Run CFU #2 (Spot the Error) to eliminate the inverse key error.
22–38 min CFU 3, 4 & Pairs
Solving & Real Modeling
Slides 7–10. Demonstrate how \(\sec\) can simplify equations where \(x\) is in the hypotenuse. Complete CFU #3 & #4 , then practice in pairs.
38–50 min Formative
Exit Ticket & Closure
Slide 11. Distribute independent 10-minute Exit Ticket. Evaluates primary ratios, reciprocals, and real-world modeling.
Top Student Misconceptions & Fixes
1. Inverse Button Trap: Pressing \(\cos^{-1}\) instead of \(1/\cos\). Fix: Emphasize \(\cos^{-1}\) finds angles; reciprocals flip ratios.2. Pair Mismatches: Pairing Sine with Secant. Fix: "Every pair gets one 'co-'" rule: Sine ↔ Cosecant; Cosine ↔ Secant.3. Depression Placement: Measuring depression against the vertical wall. Fix: Always draw a dashed horizontal sightline first!
High-Leverage Questioning Prompts
“Does your pair have one 'co-'?” Fast self-check for students pairing \(\sin\) with \(\sec\) or \(\csc\).“How do you tell your calculator to flip a ratio?” Reinforces \(1 / \text{function}\) or the \([x^{-1}]\) key.“Can \(\sec\theta\) or \(\csc\theta\) ever be less than 1?” No! Because hypotenuse is the longest side, \(\text{Hyp}/\text{Leg} \ge 1\).
Targeted Differentiation Strategies
Support / Scaffolding:
Have students write the 3 primary ratios first, then draw double arrows to flip them into reciprocals.
Post a calculator cheat sheet on the board: \(\sec\theta = 1 \div \cos(\theta)\).
Extension / Challenge:
Derive the Pythagorean identities: \(\sin^2\theta + \cos^2\theta = 1\) and \(1 + \tan^2\theta = \sec^2\theta\).
Explore the quotient identity for cotangent: \(\cot\theta = \frac{\cos\theta}{\sin\theta}\).
The Six Trig Ratios — Educator Guide Page 1 of 3 Turn over for Guided Notes Master Solutions →
Educator Facilitation
Guided Notes: Master Answer Key & CFUs
All 6 Ratios Included
Page 1: 6 Ratios & CFU 1
Section 1 & 2 Blanks:
Hypotenuse: longest side; opposite 90° angle.
Opposite Leg: Across from reference angle \(\theta\) .
\(\sin\theta = \mathbf{\frac{\text{Opp}}{\text{Hyp}}}\) • \(\csc\theta = \mathbf{\frac{\text{Hyp}}{\text{Opp}}}\)
\(\cos\theta = \mathbf{\frac{\text{Adj}}{\text{Hyp}}}\) • \(\sec\theta = \mathbf{\frac{\text{Hyp}}{\text{Adj}}}\)
\(\tan\theta = \mathbf{\frac{\text{Opp}}{\text{Adj}}}\) • \(\cot\theta = \mathbf{\frac{\text{Adj}}{\text{Opp}}}\)
CFU #1 Solution (Rotated Triangle \(\alpha\)):
Sides: Opp = 15, Adj = 20, Hyp = 25
\(\sin\alpha = \frac{15}{25} = \mathbf{\frac{3}{5}}\)
\(\csc\alpha = \mathbf{\frac{5}{3}}\)
\(\cos\alpha = \frac{20}{25} = \mathbf{\frac{4}{5}}\)
\(\sec\alpha = \mathbf{\frac{5}{4}}\)
\(\tan\alpha = \frac{15}{20} = \mathbf{\frac{3}{4}}\)
\(\cot\alpha = \mathbf{\frac{4}{3}}\)
Page 1 (Cont.): Calculator Reciprocals & CFU 2
Section 3 Blanks (Calculator):
\(\sec(\theta) = \mathbf{\frac{1}{\cos(\theta)}}\) OR \([\cos(\theta)]^{-1}\).
Practice: \(\sec(40^\circ) = \frac{1}{\cos(40^\circ)} \approx \mathbf{1.3054}\).
Practice: \(\csc(55^\circ) = \frac{1}{\sin(55^\circ)} \approx \mathbf{1.2208}\).
CFU #2 Solution (Spot the Error):
Taylor: Correct! Entered \(1 / \cos(60^\circ) = 2.0\).
Morgan: Error! Pressed \(\cos^{-1}(60)\). Inverse cosine has a domain of \([-1, 1]\), so entering 60 produces a domain error.
Fix: Never use 2nd/Shift keys for reciprocal functions!
Page 2: Solving Strategies & Real-World Modeling
CFU #3 Solution (Missing Side):
1. Primary: \(\tan(48^\circ)\) | Reciprocal: \(\cot(48^\circ)\)
2. Setup: \(\tan(48^\circ) = \frac{h}{30}\) OR \(\cot(48^\circ) = \frac{30}{h}\).
3. Calculation: \(h = 30 \cdot \tan(48^\circ) \approx \mathbf{33.3}\).
CFU #4 Solution (Elevation vs. Depression):
Angle 1: Elevation (look upward from horizontal).
Angle 2: Depression (look downward from horizontal).
Camera: \(22^\circ\) is between sightline & horizontal .
Distance: \(d = 40 \cdot \cot(22^\circ) = \frac{40}{\tan(22^\circ)} \approx \mathbf{99.0\text{ ft}}\).
The Six Trig Ratios — Educator Guide Page 2 of 3 Turn over for Practice Worksheet & Exit Ticket Keys →
Official Solutions & Scoring Keys
Worksheet & Exit Ticket Master Solutions
Classroom Keys & Rubrics
Worksheet Page 1: All 6 Ratios & Calculator Drills
Part 1: 6 Ratio Evaluations
Triangle A (8-15-17): \(\sin A = \mathbf{8/17}\), \(\csc A = \mathbf{17/8}\)
\(\cos A = \mathbf{15/17}\), \(\sec A = \mathbf{17/15}\)
\(\tan A = \mathbf{8/15}\), \(\cot A = \mathbf{15/8}\)
Triangle B (9-12-15): \(\sin X = \frac{9}{15} = \mathbf{3/5}\), \(\csc X = \mathbf{5/3}\)
\(\cos X = \frac{12}{15} = \mathbf{4/5}\), \(\sec X = \mathbf{5/4}\)
\(\tan X = \frac{9}{12} = \mathbf{3/4}\), \(\cot X = \mathbf{4/3}\)
Part 2 & Part 3: Calculator Lab & Missing Sides
Section A: \(\sin(38^\circ) \approx \mathbf{0.6157}\), \(\cos(72^\circ) \approx \mathbf{0.3090}\), \(\tan(15^\circ) \approx \mathbf{0.2679}\)Section B (Reciprocals): \(\csc(38^\circ) = \frac{1}{\sin(38^\circ)} \approx \mathbf{1.6243}\)
\(\sec(72^\circ) = \frac{1}{\cos(72^\circ)} \approx \mathbf{3.2361}\)
\(\cot(15^\circ) = \frac{1}{\tan(15^\circ)} \approx \mathbf{3.7321}\)
Section C (Algebra): \(x = 14\sec(40^\circ) \approx 14(1.3054) \approx \mathbf{18.3}\)
\(x = 9\csc(52^\circ) \approx 9(1.2690) \approx \mathbf{11.4}\)
\(x = 20\cot(33^\circ) \approx 20(1.5399) \approx \mathbf{30.8}\)
Part 3: Prob 1: \(x \approx \mathbf{7.4}\) | Prob 2: \(y \approx \mathbf{18.0}\)
Worksheet Page 2: Applied Real-World Modeling
Problem 1 (Pine Tree): \(\text{Adj} = 65\text{ ft}, \theta = 54^\circ\). Setup: \(\tan(54^\circ) = \frac{h}{65}\). Calculation: \(h = 65 \cdot \tan(54^\circ) \approx \mathbf{89.5\text{ ft}}\).
Problem 2 (Lifeguard Rescue): \(\text{Opp} = 28\text{ ft}, \theta = 18^\circ\). Setup with \(\cot\): \(d = 28 \cdot \cot(18^\circ) = \frac{28}{\tan(18^\circ)} \approx \mathbf{86.2\text{ ft}}\).
Problem 3 (Architect / Eye Height): Step 1: \(h_{\text{eye}} = 90 \cdot \tan(28^\circ) \approx 47.85\text{ ft}\). Step 2: \(H = 47.85 + 5.5 = \mathbf{53.4\text{ ft}}\).
Exit Ticket: Complete Key & 10-Point Scoring Rubric
Question 1 (4 pts):
Opp=7, Adj=24, Hyp=25
\(\sin\theta = \mathbf{7/25}\) (1 pt)
\(\csc\theta = \mathbf{25/7}\) (1 pt)
\(\cos\theta = \mathbf{24/25}\) (1 pt)
\(\sec\theta = \mathbf{25/24}\) (1 pt)
Question 2 (3 pts):
Calculator reciprocal operations
A: \(\sec(50^\circ) = \frac{1}{\cos(50^\circ)} \approx \mathbf{1.5557}\) (1 pt)
B: \(\cot(25^\circ) = \frac{1}{\tan(25^\circ)} \approx \mathbf{2.1445}\) (1 pt)
C: \(x = 16\cos(37^\circ) \approx \mathbf{12.8}\) (1 pt)
Question 3 (3 pts):
Kite flight modeling
Triangle diagram (1 pt)
Equation: \(\sin(52^\circ) = \frac{h}{160}\) (1 pt)
Height: \(h = 160\sin(52^\circ) \approx \mathbf{126.1\text{ ft}}\) (1 pt)
The Six Trig Ratios — Educator Guide Page 3 of 3 Classroom Ready
Right Triangle Guided Notes Interactive Lecture Companion The 6 Trigonometric Ratios
Right Triangle Trigonometry: Guided Notes
Name:
Date: Period:
Essential Question: How do primary and reciprocal trigonometric ratios model geometric and real-world relationships?
1 Right Triangle Anatomy (Reference Angle \(\theta\))
Fill in key terms
θ [ A ] [ B ] [ C ]
[ A ] Hypotenuse: Always the side; directly opposite the ° angle.
[ B ] Opposite Leg: Across the triangle from reference angle .
[ C ] Adjacent Leg: Next to angle \(\theta\), forming one of its rays.
2 The Six Trigonometric Ratios (Primary & Reciprocal)
One “co-” per pair
SINE & COSECANT
\(\sin\theta =\) /
\(\csc\theta =\) /
\(\csc\theta = \frac{1}{\sin\theta}\)
COSINE & SECANT
\(\cos\theta =\) /
\(\sec\theta =\) /
\(\sec\theta = \frac{1}{\cos\theta}\)
TANGENT & COTANGENT
\(\tan\theta =\) /
\(\cot\theta =\) /
\(\cot\theta = \frac{1}{\tan\theta}\)
CFU #1
Check for Understanding: Evaluate All 6 Ratios for Angle \(\alpha\)
2 min
α 20 15 25
\(\sin\alpha =\)
\(\cos\alpha =\)
\(\tan\alpha =\)
\(\csc\alpha =\)
\(\sec\alpha =\)
\(\cot\alpha =\)
3 Calculator Protocol: Reciprocals & Mode Check
No “SEC” Key!
How to Evaluate Reciprocal Ratios:
Use division: \(\sec(\theta) = \frac{1}{\cos(\theta)}\) OR \([\cos(\theta)]^{-1}\).
NEVER use \(\sin^{-1}, \cos^{-1}, \tan^{-1}\) (those find angles!).
Practice Evaluations (4 dec.):
\(\sec(40^\circ) = \frac{1}{\cos(40^\circ)} \approx\)
\(\csc(55^\circ) = \frac{1}{\sin(55^\circ)} \approx\)
CFU #2 Check for Understanding: Spot the Calculator Error
Which student is wrong?
Taylor enters: \(1 / \cos(60^\circ) = 2.0\)
Correct method?
Morgan enters: \(\cos^{-1}(60^\circ) =\) ERROR
Why did Morgan get an error?
Right Triangle Guided Notes Page 1 of 2 Turn over for 3-Step Solving & Modeling →
Part 2: Solving & Applied Modeling
Finding Unknown Sides & Real-Life Scenarios
Method: Label → Select Ratio → Solve
4 The 3-Step Solving Strategy & Reciprocal Shortcut
Primary Setup (\(\cos\))
\(\cos(34^\circ) = \frac{15}{x}\)
Variable in denominator \(\rightarrow\) Swap and divide:
\(x = \frac{15}{\cos(34^\circ)} \approx\)
Reciprocal Shortcut (\(\sec\))
\(\sec(34^\circ) = \frac{x}{15}\)
Variable in numerator \(\rightarrow\) Simply multiply:
\(x = 15 \cdot \sec(34^\circ) \approx\)
CFU #3 Check for Understanding: Ratio Selection & Solve
2 min
48° h 30
1. Primary ratio: | Reciprocal ratio:
2. Equation setup:
3. Solved height: \(h \approx\)
5 Real-World Modeling: Elevation vs. Depression
1 2
Angle 1 = Angle of : Look from horizontal.
Angle 2 = Angle of : Look from horizontal.
Key Fact: \(\angle 1 = \angle 2\) because horizontal lines are parallel (alternate interior angles).
CFU #4 Check for Understanding: Sketch & Setup Check
Field Check (2 min)
A security camera is mounted 40 feet up on a warehouse wall. It spots a suspicious car at an angle of depression of \(22^\circ\) .
Where is the \(22^\circ\) angle located?
Between camera sightline & horizontal Between camera sightline & wall
Distance using \(\cot\) or \(\tan\):
\(d = 40 \cdot \cot(22^\circ) = \frac{40}{\tan(22^\circ)} \approx\) ft
The 6 Trigonometric Golden Rules:
1. All 6 ratios flip in reciprocal pairs (\(\csc = 1/\sin, \sec = 1/\cos, \cot = 1/\tan\)) • 2. No sec key: use \(1 / \cos\) • 3. Angles of elevation/depression always start from a horizontal sightline !
Right Triangle Guided Notes Page 2 of 2 Classroom Ready
Right Triangle Bellringer Daily Warm-Up 5–8 Minutes • Elevation, Ratios & Depression
Right Triangle Bellringer
Name:
Date: Period:
Check Mode Ensure your calculator is in DEGREE mode before computing. Show setups for all problems.
Teacher Stamp:
1
Real-World Modeling: Tower Height
Angle of Elevation
An engineer stands 75 feet from the base of a communications tower on level ground. Using an inclinometer, she measures the angle of elevation of \(48^\circ\) to the top transmitter.
Step 1: Sketch & Label Diagram Include ground distance (75 ft), angle of elevation (\(48^\circ\)), and tower height \(h\).
Step 2: Trigonometric Equation & Calculation Set up an equation with \(\tan(48^\circ)\) or \(\cot(48^\circ)\) and solve for \(h\).
Tower Height \(h\) =
feet (nearest tenth)
2
Exact Reciprocal Ratios
Simplified fraction form
θ 15 8 17
\(\cos\theta = \frac{\text{adj}}{\text{hyp}}\)
\(\sec\theta = \frac{\text{hyp}}{\text{adj}}\)
\(\cot\theta = \frac{\text{adj}}{\text{opp}}\)
3
Real-World Modeling: Drone Survey
Angle of Depression
A search drone hovers at an altitude of 280 feet directly above a ground base. The operator sights a marker on the ground at an angle of depression of \(32^\circ\) .
Step 1: Sketch & Label Diagram Include drone altitude (280 ft), horizontal line of sight, angle (\(32^\circ\)), and ground distance \(d\).
Step 2: Trigonometric Equation & Calculation Solve for the ground distance \(d\) from the base to the marker.
Ground Distance \(d\) =
feet (nearest tenth)
Confidence: 1 - Need Help 2 - Getting There 3 - Mastered
Keep this sheet ready for whole-class debrief.
Right Triangle Bellringer Key Teacher Solution Key Daily Warm-Up • Complete Key & Teaching Notes
Right Triangle Bellringer Key
Master Solutions & Rubric
Target Time: 5–8 min warmup
Quick Walk-Around Check: Verify student calculators display DEG mode. Contrast the angle of elevation in Prob 1 (measured UP from ground) with angle of depression in Prob 3 (measured DOWN from horizontal).
1
Real-World Modeling: Tower Height (Solution)
Angle of Elevation
Step 1: Diagram Key
48° adj = 75 ft opp = h
Observer at ground level; angle measured upward.
Step 2: Equation & Calculation
Standard Setup: \(\tan(48^\circ) = \frac{h}{75} \implies h = 75 \cdot \tan(48^\circ)\)
Reciprocal Setup: \(\cot(48^\circ) = \frac{75}{h} \implies h = \frac{75}{\cot(48^\circ)}\)
\(h = 75 \cdot (1.11061...) \approx 83.2959...\)
Final Answer: \(h = 83.3\text{ feet}\)
2
Exact Reciprocal Ratios (Solutions)
Triangle: 8 - 15 - 17
θ 15 (adj) 8 (opp) 17 (hyp)
\(\cos\theta = \frac{\text{adj}}{\text{hyp}}\)
\(\frac{15}{17}\)
\(\sec\theta = \frac{\text{hyp}}{\text{adj}}\)
\(\frac{17}{15}\)
\(\cot\theta = \frac{\text{adj}}{\text{opp}}\)
\(\frac{15}{8}\)
3
Real-World Modeling: Drone Survey (Solution)
Angle of Depression
Step 1: Diagram Key
32° 32° 280' distance d
Alt. interior angles: angle at marker = \(32^\circ\)
Step 2: Equation & Calculation
Standard Setup: \(\tan(32^\circ) = \frac{280}{d} \implies d = \frac{280}{\tan(32^\circ)}\)
Reciprocal Setup: \(d = 280 \cdot \cot(32^\circ)\)
\(d = \frac{280}{0.624869} \approx 448.0939...\)
Final Answer: \(d = 448.1\text{ feet}\)
3-Minute Debrief Pacing & Transition Whole-Class Discussion
1. Elevation vs. Depression: Contrast Prob 1 (looking up from ground) with Prob 3 (looking down from horizontal altitude line).
2. Variable Position: Notice how \(\tan\) puts the variable in the numerator in Prob 1 (\(h = 75\tan 48^\circ\)) vs denominator in Prob 3 (\(d = \frac{280}{\tan 32^\circ}\)).
3. Bridge to Reciprocals: Emphasize how \(\cot\) in Prob 3 allows students to multiply rather than divide (\(d = 280\cot 32^\circ\)).
