Angle of elevation and depression, bearing and navigation, special triangles, Law of Sines and Cosines, Heron's formula, physics applications, and 3D problems — all with fully worked examples.
Every right triangle application starts here. In a right triangle with acute angle A, hypotenuse c, opposite side a, and adjacent side b:
SOH
Sine
sin A = opposite / hypotenuse
sin A = a / c
CAH
Cosine
cos A = adjacent / hypotenuse
cos A = b / c
TOA
Tangent
tan A = opposite / adjacent
tan A = a / b
Reciprocal Functions
csc A = 1 / sin A = c / a
sec A = 1 / cos A = c / b
cot A = 1 / tan A = b / a
Example 1: Find all sides and angles given A = 32° and hypotenuse c = 15
Step 1: Find angle B.
B = 90° − 32° = 58°
Step 2: Find side a (opposite to A).
sin 32° = a / 15 → a = 15 × sin 32° = 15 × 0.5299 ≈ 7.95
Step 3: Find side b (adjacent to A).
cos 32° = b / 15 → b = 15 × cos 32° = 15 × 0.8480 ≈ 12.72
Check: a² + b² = 63.2 + 161.8 = 225 = 15² ✓
Special right triangles have exact side ratios derived from equilateral triangles and squares. Memorize these — they give exact answers on every exam without a calculator.
Short leg (opp 30°): x
Long leg (opp 60°): x√3
Hypotenuse (opp 90°): 2x
Ratio: 1 : √3 : 2
Derived by cutting an equilateral triangle (all sides equal, all angles 60°) in half. The cut creates a 30-60-90 triangle with hypotenuse = original side.
Example: hypotenuse = 10
Short leg = 10/2 = 5
Long leg = 5√3 ≈ 8.66
Leg 1 (opp 45°): x
Leg 2 (opp 45°): x
Hypotenuse (opp 90°): x√2
Ratio: 1 : 1 : √2
Derived from a square with side x. The diagonal cuts the square into two 45-45-90 triangles. The diagonal has length x√2 by the Pythagorean theorem.
Example: leg = 7
Other leg = 7
Hypotenuse = 7√2 ≈ 9.90
Key Exam Trick
These triangles encode all the "nice" trig values: sin 30° = 1/2, cos 30° = √3/2, tan 30° = 1/√3 = √3/3; sin 45° = cos 45° = √2/2, tan 45° = 1; sin 60° = √3/2, cos 60° = 1/2, tan 60° = √3. These same values appear on the unit circle — the triangles and the unit circle are the same thing in different dress.
"Solve the triangle" means find every unknown side and angle. Right triangles have a 90° angle built in, so you need only two more pieces of information.
Given: One acute angle + hypotenuse
Given: One acute angle + one leg
Given: Two legs
Given: Hypotenuse + one leg
Example 2: Solve the right triangle with legs a = 5 and b = 12
Step 1: Find hypotenuse c.
c = √(5² + 12²) = √(25 + 144) = √169 = 13
Step 2: Find angle A (opposite leg = 5).
A = arctan(5/12) = arctan(0.4167) ≈ 22.6°
Step 3: Find angle B.
B = 90° − 22.6° ≈ 67.4°
Verify: 5-12-13 is a Pythagorean triple ✓
Measured upward from horizontal to the line of sight toward an object above you. Observer is at ground level; object is above. The horizontal ground, the vertical height, and the line of sight form the right triangle.
tan(elevation) = height / horizontal distance
sin(elevation) = height / line-of-sight
Measured downward from horizontal to the line of sight toward an object below you. Observer is elevated; object is below. By alternate interior angles, the depression angle at the top equals the elevation angle at the bottom.
tan(depression) = drop / horizontal distance
elevation angle = depression angle (alternate interior angles)
Example 3 (Elevation): From a point 80 ft from the base of a tower, the angle of elevation to the top is 52°. Find the tower's height.
Draw: horizontal base = 80 ft, right angle at tower base, elevation angle = 52° at observer.
Set up: tan(52°) = height / 80
height = 80 × tan(52°) = 80 × 1.2799 ≈ 102.4 ft
Example 4 (Depression): A lifeguard in a tower 12 m high spots a swimmer. The angle of depression to the swimmer is 18°. How far is the swimmer from the base of the tower?
The depression angle at the top = elevation angle at the swimmer = 18°.
tan(18°) = 12 / d → d = 12 / tan(18°)
d = 12 / 0.3249 ≈ 36.9 m
Example 5 (Two Angles): A building's height is observed from two points on the ground, both on the same side. From point A (100 m away), the elevation is 38°. From point B (closer), the elevation is 55°. How far is B from the building?
Let d = distance from B to building, h = height.
From A: tan(38°) = h / 100 → h = 100 × tan(38°) ≈ 78.13 m
From B: tan(55°) = h / d → d = h / tan(55°)
d = 78.13 / 1.4281 ≈ 54.7 m
Bearing is measured clockwise from due North. A bearing of 0° (or 360°) is due North; 90° is due East; 180° is due South; 270° is due West. You may also see compass notation like "N 45° E" (45 degrees east of north, equivalent to bearing 045°) or "S 30° W" (30 degrees west of south, equivalent to bearing 210°).
Setup Protocol for Bearing Problems
Example 6: A ship travels 40 miles on a bearing of N 32° E. How far north and how far east has it traveled?
The angle from north is 32°. The hypotenuse (path) = 40 miles.
North component (adjacent to 32° from north):
North = 40 × cos(32°) = 40 × 0.8480 ≈ 33.9 miles north
East component (opposite to 32° from north):
East = 40 × sin(32°) = 40 × 0.5299 ≈ 21.2 miles east
Example 7: A plane flies 120 km due East, then turns and flies 90 km due North. What is the bearing from the starting point to the final position?
The path forms a right triangle with legs East = 120 km and North = 90 km.
Step 1: Find the straight-line distance.
d = √(120² + 90²) = √(14400 + 8100) = √22500 = 150 km
Step 2: Find the angle from North.
tan(θ) = East/North = 120/90 = 1.333 → θ = arctan(1.333) ≈ 53.1°
Step 3: State the bearing.
Bearing = N 53.1° E, or 053.1° ≈ 053°
Example 8 (Two-Leg Voyage): A boat travels 25 miles on bearing 040°, then 35 miles on bearing 130°. Find the distance and bearing back to start.
Leg 1 (bearing 040°): N 40° E
East₁ = 25 sin(40°) ≈ 16.07 km; North₁ = 25 cos(40°) ≈ 19.15 km
Leg 2 (bearing 130°): S 50° E (angle from south = 180° − 130° = 50°)
East₂ = 35 sin(50°) ≈ 26.81 km; South₂ = 35 cos(50°) ≈ 22.50 km (i.e., North₂ = −22.50)
Net displacement:
Total East = 16.07 + 26.81 = 42.88 km
Total North = 19.15 − 22.50 = −3.35 km (i.e., 3.35 km South)
Return distance:
d = √(42.88² + 3.35²) = √(1838.7 + 11.2) ≈ 42.9 km
Return bearing (from current position back to start):
Angle from South toward West: arctan(42.88/3.35) ≈ 85.5° west of south → bearing ≈ 270° − (90° − 85.5°) = 265.5°, or approximately S 85.5° W
Also written as: sin A / a = sin B / b = sin C / c
AAS
Two angles + non-included side
ASA
Two angles + included side
SSA
Two sides + non-included angle (ambiguous)
Example 9 (AAS — Surveying): Two rangers observe a fire from stations A and B, which are 6 km apart. Ranger A measures the angle to the fire as 47° from the AB line. Ranger B measures 64°. How far is the fire from each station?
Label: A = 47°, B = 64°, side AB = c = 6 km. Find sides a and b.
Step 1: Find angle C (angle at the fire).
C = 180° − 47° − 64° = 69°
Step 2: Set up ratio using known pair (c, C).
c / sin C = 6 / sin 69° = 6 / 0.9336 ≈ 6.426
Step 3: Find distances.
a = 6.426 × sin 47° = 6.426 × 0.7314 ≈ 4.70 km (from B to fire)
b = 6.426 × sin 64° = 6.426 × 0.8988 ≈ 5.78 km (from A to fire)
In SSA, knowing two sides and a non-included angle does not uniquely determine the triangle. Given angle A (acute) and sides a and b:
| Condition | Triangles | Reason |
|---|---|---|
| a < b sin A | 0 | Side a too short to reach the base |
| a = b sin A | 1 (right triangle) | Side a exactly reaches, forms 90° |
| b sin A < a < b | 2 | Two different triangles possible |
| a ≥ b | 1 | Only one orientation possible |
If angle A is obtuse: 0 triangles when a ≤ b; 1 triangle when a > b.
Example 10 (SSA Ambiguous — Two Triangles): A = 38°, a = 20, b = 28. Find all possible triangles.
Step 1: Compute b sin A to classify.
b sin A = 28 × sin 38° = 28 × 0.6157 ≈ 17.24
Since 17.24 < 20 (a) < 28 (b) → two triangles exist.
Step 2: Find sin B.
sin B / 28 = sin 38° / 20 → sin B = 28 × 0.6157 / 20 ≈ 0.8620
B₁ = arcsin(0.8620) ≈ 59.5° (acute)
B₂ = 180° − 59.5° = 120.5° (obtuse)
Step 3: Check validity (A + B < 180°).
Triangle 1: 38° + 59.5° = 97.5° < 180° ✓ → C₁ = 82.5°
Triangle 2: 38° + 120.5° = 158.5° < 180° ✓ → C₂ = 21.5°
Step 4: Find c for each.
c₁ = 20 × sin 82.5° / sin 38° = 20 × 0.9914 / 0.6157 ≈ 32.2
c₂ = 20 × sin 21.5° / sin 38° = 20 × 0.3665 / 0.6157 ≈ 11.9
a² = b² + c² − 2bc cos A
b² = a² + c² − 2ac cos B
c² = a² + b² − 2ab cos C
cos A = (b² + c² − a²) / (2bc)
cos B = (a² + c² − b²) / (2ac)
cos C = (a² + b² − c²) / (2ab)
SAS
Two sides + included angle → find third side, then use Law of Sines for angles
SSS
Three sides → find angles using angle form; find largest angle first
Example 11 (SAS — Navigation): Two boats leave a dock. Boat A travels 15 km on bearing 055° and Boat B travels 22 km on bearing 145°. How far apart are the boats?
The angle between the two paths: 145° − 55° = 90°. So the included angle C = 90°.
Wait — cos 90° = 0, so Law of Cosines simplifies to the Pythagorean theorem!
c² = 15² + 22² = 225 + 484 = 709
c = √709 ≈ 26.6 km
In general (non-perpendicular example): a = 18, b = 25, C = 110°
c² = 18² + 25² − 2(18)(25) cos 110°
c² = 324 + 625 − 900 × (−0.3420)
c² = 949 + 307.8 = 1256.8 → c ≈ 35.5
Example 12 (SSS — Architecture): A triangular roof truss has sides 24 ft, 18 ft, and 30 ft. Find all interior angles.
a = 24, b = 18, c = 30 (longest side, find its angle first).
Step 1: Find angle C (opposite c = 30).
cos C = (24² + 18² − 30²) / (2 × 24 × 18)
cos C = (576 + 324 − 900) / 864 = 0 / 864 = 0
C = 90° ← This is a 3-4-5 right triangle scaled by 6! (18-24-30 = 6×3-4-5)
Step 2: Find angle A.
sin A / 24 = sin 90° / 30 → sin A = 24/30 = 0.8 → A ≈ 53.1°
Step 3: Find angle B.
B = 180° − 90° − 53.1° ≈ 36.9°
Connection to Pythagorean Theorem
When C = 90°, cos C = 0, so the Law of Cosines becomes c² = a² + b², which is exactly the Pythagorean theorem. The Law of Cosines is the generalization of the Pythagorean theorem to all triangles.
s = (a + b + c) / 2
(semi-perimeter)
K = √[s(s−a)(s−b)(s−c)]
Use when all three sides are known and no angle has been found. Perfect for SSS problems.
K = ½ ab sin C
K = ½ ac sin B
K = ½ bc sin A
Use when two sides and the included angle are known (SAS). This formula appears in surveying, land area calculations, and physics.
Example 13 (Heron's — Surveying): A triangular field has sides 120 m, 85 m, and 95 m. Find its area.
Step 1: Compute s.
s = (120 + 85 + 95) / 2 = 300 / 2 = 150
Step 2: Compute each factor.
s − a = 150 − 120 = 30
s − b = 150 − 85 = 65
s − c = 150 − 95 = 55
Step 3: Compute area.
K = √[150 × 30 × 65 × 55]
K = √[16,087,500]
K ≈ 4011 m²
Example 14 (Trig Area — Architecture): A triangular skylight has two sides of 4 m and 6 m with an included angle of 72°. Find its area.
K = ½ × 4 × 6 × sin 72°
K = 12 × 0.9511
K ≈ 11.41 m²
Land surveyors measure a baseline and two angles at each endpoint to locate a distant point without walking to it. This is AAS: known baseline + two angles → Law of Sines gives all distances. Modern GPS replaced optical triangulation but the mathematics is identical.
A roof with a 7:12 pitch (rises 7 inches for every 12 inches of horizontal run) creates a right triangle. The pitch angle = arctan(7/12) ≈ 30.3°. Rafter length = horizontal run / cos(30.3°). Architects use SOH-CAH-TOA and the Pythagorean theorem daily.
Bridge trusses are built from triangles because triangles are rigid. Each triangle has known side lengths (SSS) so the Law of Cosines gives all angles, which then determine how forces distribute through each member. Area formulas give material cross-sections for load calculations.
Example 15 (Architecture — Rafter): A house has a horizontal span of 28 ft and a roof pitch of 9:12. Find the rafter length and the roof angle.
Pitch 9:12 means the roof rises 9 inches for every 12 inches of run.
Roof angle: tan(θ) = 9/12 = 0.75 → θ = arctan(0.75) ≈ 36.87°
Half-span (horizontal run for one side) = 28/2 = 14 ft.
Rafter length = 14 / cos(36.87°) = 14 / 0.8 = 17.5 ft
Alternatively: rise = 14 × (9/12) = 10.5 ft, then rafter = √(14² + 10.5²) = √(196 + 110.25) = √306.25 ≈ 17.5 ft ✓
When an object rests on a ramp (inclined plane) tilted at angle θ, gravity acts straight down. Trigonometry decomposes the weight vector W into two perpendicular components using the right triangle formed by the ramp direction and the vertical.
Weight vector: W = mg (straight down)
Component along ramp (parallel): W‖ = W sin θ = mg sin θ
Component perpendicular to ramp: W⊥ = W cos θ = mg cos θ
W⊥ equals the normal force (no vertical acceleration perpendicular to the ramp). W‖ is the force pulling the object down the slope (causing sliding).
Why sin θ for the slope component?
The angle θ of the ramp from horizontal is the same as the angle between the weight vector and the perpendicular to the ramp (alternate interior angles with parallel horizontal lines). So the component along the ramp is opposite to θ in the right triangle, giving W sin θ. The perpendicular component is adjacent to θ, giving W cos θ.
Example 16 (Physics): A 50 kg crate sits on a ramp inclined at 25°. What force does the crate exert along the ramp? What is the normal force? (g = 9.8 m/s²)
Weight: W = 50 × 9.8 = 490 N
Force along ramp (parallel component):
W‖ = 490 × sin(25°) = 490 × 0.4226 ≈ 207.1 N
Normal force (perpendicular component):
N = 490 × cos(25°) = 490 × 0.9063 ≈ 444.1 N
Check: √(207.1² + 444.1²) = √(42890 + 197225) = √240115 ≈ 490 N ✓
Example 17 (Force Resultant): Two forces of 30 N and 40 N act at a right angle to each other. Find the magnitude and direction of the resultant.
The two forces form legs of a right triangle; the resultant is the hypotenuse.
Resultant = √(30² + 40²) = √(900 + 1600) = √2500 = 50 N
Direction (angle from the 40 N force):
θ = arctan(30/40) = arctan(0.75) ≈ 36.87°
This is a 3-4-5 triple scaled by 10!
Three-dimensional problems use two or more right triangles in perpendicular planes. The strategy is always: identify the horizontal right triangle on the ground plane, then identify the vertical right triangle that contains the height, and solve them in sequence.
Example 18 (Flagpole): A flagpole casts a shadow 18 m long when the sun's angle of elevation is 40°. How tall is the flagpole?
Vertical right triangle: base = 18 m (shadow), elevation angle = 40°.
tan(40°) = height / 18
height = 18 × tan(40°) = 18 × 0.8391 ≈ 15.1 m
Example 19 (Building Height from Two Points): Observer at point A (due south of the building) measures the elevation angle to the top as 42°. Observer at point B (due east of the building) measures 35°. Points A and B are 60 m apart and on the same level as the building base. Find the building's height.
Let h = height, d_A = horizontal distance from A to building, d_B = horizontal distance from B to building.
From A: tan(42°) = h / d_A → d_A = h / tan(42°) = h / 0.9004 ≈ 1.1106h
From B: tan(35°) = h / d_B → d_B = h / tan(35°) = h / 0.7002 ≈ 1.4281h
Since A is due south and B is due east of the building, A, B, and the building base form a right triangle on the ground.
d_A² + d_B² = 60² (Pythagorean theorem on the ground plane)
(1.1106h)² + (1.4281h)² = 3600
1.2334h² + 2.0395h² = 3600
3.2729h² = 3600
h² = 3600 / 3.2729 ≈ 1100.0
h ≈ 33.2 m
Example 20 (Mountain Peak): From a point on level ground, the angle of elevation to a mountain peak is 15°. After walking 1000 m directly toward the mountain, the angle of elevation is 22°. Find the height of the mountain above the observation level.
Let h = height, d = horizontal distance from the second point to the base of the mountain.
From second point: tan(22°) = h / d → d = h / tan(22°) ≈ 2.4751h
From first point: tan(15°) = h / (d + 1000) → d + 1000 = h / tan(15°) ≈ 3.7321h
Subtract the two expressions for d:
(d + 1000) − d = 3.7321h − 2.4751h
1000 = 1.2570h
h = 1000 / 1.2570 ≈ 795 m
Confusing opposite and adjacent sides
Always identify which angle you are working with first. Opposite is the side directly across from that angle. Adjacent is the leg that forms the angle along with the hypotenuse. Labeling the sides before writing any ratio saves many errors.
Using elevation angle as if it were the angle in the right triangle
The elevation angle IS the angle in the right triangle formed by horizontal ground, vertical height, and the line of sight. The right angle is at the base of the object. The elevation angle is at the observer. The third angle completes 180°. Draw the picture every time.
Measuring bearing from south or east instead of north
Bearing is ALWAYS measured clockwise from due North. North = 0°, East = 90°, South = 180°, West = 270°. Compass notation N 30° E means rotate 30° east from north, giving bearing 030°.
Forgetting the ambiguous case (SSA)
Whenever you have SSA, immediately compute b sin A. If b sin A < a < b, you have two triangles and must find both. Report both solutions. Ignoring this is one of the most common lost points on exams.
Applying SOH-CAH-TOA to non-right triangles
SOH-CAH-TOA works ONLY in right triangles. For any triangle without a 90° angle, use the Law of Sines or Law of Cosines. The right angle must be explicitly given or proven — never assume it.
Forgetting to check that angles sum to 180°
Always add all angles at the end. The sum must be 180° (within rounding tolerance). If it is not, you made an arithmetic or setup error. This check takes five seconds and catches most mistakes.
The angle of elevation is the angle measured upward from a horizontal line to the line of sight toward an object above you. If you stand on the ground and look up at the top of a building, the angle between the ground and your line of sight is the angle of elevation. It is always measured from the horizontal, never from the vertical. In right triangle problems, this angle is one of the acute angles of the triangle formed by the horizontal ground, the vertical height of the object, and the slanted line of sight.
The angle of elevation is measured upward from horizontal — you look up to the object. The angle of depression is measured downward from horizontal — you look down to the object. By the alternate interior angles theorem, when an observer looks down at an object below, the angle of depression at the observer equals the angle of elevation at the object. This means you can freely swap them in problems where the geometry involves parallel horizontal lines.
Bearing is measured clockwise from due North, ranging from 0 to 360 degrees. To set up a bearing problem: (1) Draw a north-pointing arrow at the starting point. (2) Rotate clockwise by the given bearing angle to get the direction of travel. (3) Identify the right triangle formed — often the north-south leg and the east-west leg. (4) Use SOH-CAH-TOA or the Law of Sines/Cosines depending on what is given. Common bearings: N 30 E means 30 degrees east of north; S 45 W means 45 degrees west of south.
30-60-90 triangle: the side opposite 30 degrees (short leg) has length x, the side opposite 60 degrees (long leg) has length x times the square root of 3, and the hypotenuse has length 2x. Ratio: 1 to root-3 to 2. 45-45-90 triangle: both legs are equal (length x) and the hypotenuse has length x times the square root of 2. Ratio: 1 to 1 to root-2. Memorize these because they appear on every trig exam and allow you to find exact values without a calculator.
Solving a right triangle completely means finding all three sides and all three angles. You are always given the right angle (90 degrees). Given one acute angle A and one side: find the second acute angle B = 90 minus A, then use SOH-CAH-TOA to find the remaining sides. Given two sides: use the Pythagorean theorem to find the third side, then use arctan, arcsin, or arccos to find the acute angles. Always verify that the two acute angles sum to 90 degrees and that the hypotenuse is the longest side.
Heron's formula computes the area of any triangle when all three side lengths are known, without needing any angle. First compute the semi-perimeter: s = (a + b + c) divided by 2. Then area K = the square root of [s times (s minus a) times (s minus b) times (s minus c)]. Use it in SSS problems when you need the area directly. It also appears in surveying when land boundaries are measured as distances without angles.
On an inclined plane tilted at angle theta, gravity (weight W pointing straight down) is resolved into two components using right triangle trig. The component along the ramp (pulling the object down the slope) equals W times sin(theta). The component perpendicular to the ramp (the normal force direction) equals W times cos(theta). This decomposition comes directly from SOH-CAH-TOA applied to the right triangle formed by the weight vector, the ramp direction, and the perpendicular to the ramp.
3D right triangle problems are solved in two steps: first identify the horizontal right triangle on the ground plane, then identify the vertical right triangle that includes the height. For a flagpole problem: the shadow cast on the ground is the base of a vertical right triangle. The angle of elevation from the shadow tip to the top of the pole is one acute angle. Apply tan(elevation angle) = height divided by shadow length to find the height. For two-point problems, set up simultaneous equations using two different elevation angles observed from two different distances.
| Situation | Given | Method |
|---|---|---|
| Right triangle | Any angle + any side | SOH-CAH-TOA |
| Right triangle | Two sides | Pythagorean theorem + arctan/arcsin/arccos |
| Special triangle | One side + 30/60/90 or 45/45/90 | Ratio: 1:√3:2 or 1:1:√2 |
| Oblique triangle AAS/ASA | Two angles + any side | Law of Sines |
| Oblique triangle SSA | Two sides + non-included angle | Law of Sines (check ambiguous case) |
| Oblique triangle SAS | Two sides + included angle | Law of Cosines (side form) |
| Oblique triangle SSS | Three sides | Law of Cosines (angle form) |
| Area (SAS) | Two sides + included angle | K = ½ ab sin C |
| Area (SSS) | Three sides | Heron's formula |
| Elevation/Depression | Angle + one distance | SOH-CAH-TOA (tan for height/distance) |
| Bearing problem | Bearing + distance | Decompose into N/S and E/W components |
| Inclined plane | Angle + weight | W sin θ along ramp, W cos θ perpendicular |
SOH-CAH-TOA foundations, inverse trig, and the unit circle connection
Deep dive into oblique triangles, all cases, and the ambiguous case with 5 worked examples
Complete overview of all precalculus topics — Stewart textbook chapter by chapter
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