Digital SAT · Math · Unit 4 · Geometry & Trigonometry

Lesson 4.3 · Trigonometry

On the test: 1–2 questions a test — almost always a right triangle and one ratio; now and then a radian angle on the unit circle

Big idea. A trig ratio is just a fraction of two sides, and it belongs to the ANGLE, not to the size of the triangle. Name the sides from the angle you are asked about (opposite, adjacent, hypotenuse), find any missing side with Pythagoras, and read the fraction. The two acute angles of a right triangle swap sine and cosine. On the unit circle, cos is the x and sin is the y.

🎛 Move it · SOH · CAH · TOA

Pick a ratio and watch which two sides light up. Grow the triangle — the ratio holds. Turn the angle — it changes.

angle A37°size (scale)1×

sin A = opposite / hypotenuse = 6.02 / 10 = 0.6 (SOH · CAH · TOA).

Drag the size: every side grows, the ratio does NOT move — it belongs to the ANGLE. That is why similar triangles have equal sin, cos, tan. Drag the angle: now it changes.

Given a ratio like cos A = 3/5? Build the sides as 3 parts and 5 parts (and 4 by Pythagoras), then find what one part is worth.

🧑‍🏫 Rule 1 · Name the sides from the anglestep 1 of 8

1
the three names
$sin = \frac{opp}{hyp}, cos = \frac{adj}{hyp}, tan = \frac{opp}{adj}$
The hypotenuse faces the right angle. Opposite faces YOUR angle. Adjacent touches your angle (and is not the hypotenuse).

🧑‍🏫 Rule 2 · The missing side first (Pythagoras)step 1 of 6

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📋 The question · SAT · August US 2024 · Math 800 Book Q11
$figure from the test$
In the triangle shown $R S = 157$ . What is the value of $sin R$ ?
(type the answer)
what sin R needs
$sin R = \frac{T S}{T R}$
August US 2024: the side opposite $R$ is $T S$ — not given.

🧑‍🏫 Rule 3 · A ratio builds the sides in partsstep 1 of 5

1
📋 The question · SAT · May US 2024 · Math 800 Book Q6
In $△ A B C$ , $∠ B$ is a right angle and the length of $\overline{B C}$ is 204 millimeters. If $cos A = \frac{3}{5}$ , what is the length, in millimeters, of $\overline{A B}$ ?
1. A)255
2. B)204
3. C)153
4. D)51
build
$cos A = \frac{3}{5} : A B = 3 p, A C = 5 p, B C = 4 p$
May US 2024: a 3-4-5 triangle scaled by some $p$ .

🎛 Move it · where the values come from

Tap a value: it is read straight off one of the two special triangles.

	sin	cos	tan
30°
45°
60°

Tap any value to see which special triangle it comes from. Notice the pattern: sin 30 = cos 60, sin 60 = cos 30 (complementary angles), and tan = sin ÷ cos.

🎛 Move it · the special triangles

45-45-90 and 30-60-90 — the same shapes as in Lesson 4.2.

leg x4

45-45-90 = half a square. Legs equal (x), hypotenuse x√2 — the square’s diagonal.

Perimeter = 2x + x√2 = 13.66. Seeing “18 + 18√2”? Factor: x(2 + √2) = x√2(√2 + 1), so the HYPOTENUSE x√2 = 18. Going back: leg = hypotenuse ÷ √2.

🧑‍🏫 Rule 4 · The values to know by heartstep 1 of 3

1
📋 The question · Gen Z Study · special values
What is the value of $sin 30° + cos 60° + tan 45°$ ?
(type the answer)
add them
$\frac{1}{2} + \frac{1}{2} + 1 = 2$
$sin 30 = cos 60 = \frac{1}{2}$ , $tan 45 = 1$ .

🎛 Move it · two angles of a right triangle

The red side is opposite A and next to B.

angle A35°

The red side is opposite A and next to B. So sin A (opposite/hypotenuse) and cos B (adjacent/hypotenuse) are the SAME fraction. That is why sin A = cos B exactly when A + B = 90° — the two acute angles of one right triangle. (“co-sine” = the sine of the COmplement.)

🧑‍🏫 Rule 5 · The two acute angles swap sin and cosstep 1 of 8

1
📋 The question · SAT · May 2024 · Math 800 Book Q8
In right triangle $R S T$ , the sum of the measures of angle $R$ and angle $S$ is 90 degrees. The value of $sin (R)$ is $\frac{3 17}{13}$ . What is the value of $cos (S)$ ?
1. A) $\frac{4 17}{51}$
2. B) $\frac{3 17}{13}$
3. C) $\frac{13 17}{51}$
4. D) $\frac{3 17}{4}$
R + S = 90
$cos S = sin R = \frac{3 17}{13}$
May 2024: B — no computing.

🧑‍🏫 Rule 6 · cos(90 − x) = sin x, and sin² + cos² = 1step 1 of 6

1
the identity
$cos (90 - x) = sin x, sin (90 - x) = cos x$
The same fact as Rule 5, written with one letter.

🎛 Move it · the ramp

Change the height and the run. tan θ is the slope; the law caps it at 1/12.

height22.3run x300

tan θ = rise ÷ run = 22.3 ÷ 300 = 0.074 — tan is the SLOPE of the ramp. The law wants tan θ ≤ 1/12 ≈ 0.0833: legal.

Turn the limit around: h/x ≤ 1/12 ⇔ x ≥ 12h = 267.6. A smaller tan needs a LONGER run — the inequality flips because x is on the bottom.

🧑‍🏫 Rule 7 · tan θ = rise ÷ runstep 1 of 3

1
📋 The question · SAT · March 2024 · Math 800 Book Q3
$figure from the test$
According to a US law, ramps for use by the general public must form an angle with the ground such that $tan θ \leq \frac{1}{12}$ . If the ramp in the figure conforms to this law and the height of the ramp is 28.2 inches, what is the least possible value of $x$ , in inches?
(type the answer)
the law
$\frac{28.2}{x} \leq \frac{1}{12}$
March 2024: height over run.

🧑‍🏫 Rule 8 · A ratio inside an areastep 1 of 2

1
📋 The question · SAT · May 2024 · Math 800 Book Q7
In triangle $X Y Z$ , angle $Y$ is a right angle, the measure of angle $Z$ is $39°$ , and the length of $\overline{Y Z}$ is 22 units. If the area, in square units, of triangle $X Y Z$ can be represented by the expression $k tan 39°$ , where $k$ is a constant, what is the value of $k$ ?
(type the answer)
the other leg
$tan 39° = \frac{X Y}{Y Z} \Rightarrow X Y = 22 tan 39°$
May 2024: $X Y$ faces $Z$ , $Y Z$ touches it.

🎛 Move it · degrees ↔ radians

Turn the angle: degrees and radians side by side.

angle120°radius r3

180° = π radians (half a turn), 360° = 2π. So: degrees → radians: × π/180 · radians → degrees: × 180/π.

A radian is the angle whose arc is exactly one radius long — so the arc is always θ (in radians) × r. Whole turns (2π) can be added to any angle and land in the same place.

🎛 Move it · the unit circle

Turn θ: the point is (cos θ, sin θ). Watch the signs change by quadrant, then add whole turns — the point doesn’t move.

angle θ225°extra full turns0

On a circle of radius 1, the point at angle θ IS (cos θ, sin θ): cos = how far across, sin = how far up. The blue and red legs and the radius make a right triangle with hypotenuse 1 — so sin² + cos² = 1 (Pythagoras).

Quadrant 3: Tan positive (sin, cos negative). The size comes from the reference angle 45° (the angle to the nearest x-axis); the quadrant only decides the signs.

Add full turns (2π each) — the point does not move: 5π/4 lands where 5π/4 does. To simplify a huge angle, subtract 2π until it is between 0 and 2π.

🧑‍🏫 Rule 9 · The point is (cos θ, sin θ)step 1 of 6

1
📋 The question · Gen Z Study · unit circle point
On the unit circle, the point at angle $\frac{3 π}{2}$ from the positive $x$ -axis is:
1. A) $(1, 0)$
2. B) $(0, 1)$
3. C) $(- 1, 0)$
4. D) $(0, - 1)$
axis angles
$\frac{3 π}{2} = 270° \to (0, - 1)$
Three quarter-turns: straight down. D.

🎛 Move it · ASTC and the reference angle

Pick a reference angle: its four relatives light up, one per quadrant. Then pick a function and a sign — only the quadrants that fit stay lit.

reference angle θ:function:sign:

ASTC — “All Students Take Calculus”, counter-clockwise from quadrant 1: All three positive in Q1 · only Sin in Q2 · only Tan in Q3 · only Cos in Q4. (Why: cos is x, sin is y, tan is y/x — read the signs of x and y in each quadrant.)

Reference angle = the acute angle to the nearest x-axis. From an angle to its reference: Q1: ref = angle · Q2: ref = 180° − angle (π − angle) · Q3: ref = angle − 180° (angle − π) · Q4: ref = 360° − angle (2π − angle).

Backwards — the corresponding angles: one reference angle θ gives one angle in each quadrant: θ, 180° − θ, 180° + θ, 360° − θ (radians: θ, π − θ, π + θ, 2π − θ).

🧑‍🏫 Rule 10 · ASTC — which functions are positive wherestep 1 of 7

1
the words
$All ∣ Students ∣ Take ∣ Calculus$
Quadrants I, II, III, IV, counter-clockwise from the top right.

🧑‍🏫 Rule 11 · The reference angle — and the corresponding anglesstep 1 of 10

1
what it is
$the acute angle to the nearest x -axis$
It gives the SIZE of sin, cos, tan; ASTC gives the SIGN.

🧑‍🏫 Rule 12 · Whole turns change nothingstep 1 of 4

1
📋 The question · Gen Z Study · coterminal
Which angle between $0$ and $2 π$ ends in the same place as $\frac{17 π}{6}$ ?
1. A) $\frac{π}{6}$
2. B) $\frac{5 π}{6}$
3. C) $\frac{7 π}{6}$
4. D) $\frac{11 π}{6}$
subtract 2π
$\frac{17 π}{6} - 2 π = \frac{5 π}{6}$
B.

Real question 1SAT · August US 2024 · Math 800 Book Q11type the answer

$figure from the test$

In the triangle shown, $R S = 157$ . What is the value of $sin R$ ?

Take with you

▸SOH-CAH-TOA, named from YOUR angle: opposite faces it, adjacent touches it, the hypotenuse faces the right angle. Sin and cos are never above 1.
▸Missing side? Pythagoras first. A ratio like 3/5 builds the sides as 3p, 4p, 5p — find p.
▸30°: 1/2, √3/2, √3/3 · 45°: √2/2, √2/2, 1 · 60°: √3/2, 1/2, √3. tan = sin ÷ cos.
▸A + B = 90 ⇔ sin A = cos B; cos(90 − x) = sin x; sin² + cos² = 1.
▸tan is a slope (rise ÷ run); with x in the denominator, “tan ≤ c” becomes “x ≥ height/c”.
▸π = 180°. On the unit circle the point is (cos θ, sin θ); whole turns (2π) change nothing.
▸ASTC (All Students Take Calculus): Q1 all +, Q2 sin +, Q3 tan +, Q4 cos +. Reference angle: 180 − θ, θ − 180, 360 − θ; back again: θ, 180 − θ, 180 + θ, 360 − θ — the reference gives the size, ASTC the sign.

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