SAT · Geometry and Trigonometry · TOPIC LESSON
Use right-triangle structure, special ratios, and angle-based relationships to find exact or approximate measurements.
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BEFORE YOU START
Original educational lesson. Work through the examples before revealing the practice answers. This lesson supports SAT Math and Reading and Writing skill study; it does not reproduce official questions or guarantee a score.
LEARN THROUGH ORIGINAL PRACTICE
The teaching passages, worked examples and exercises in this lesson are original. They are not official College Board questions or past-exam reproductions. Blueprint and observed-practice analytics describe separate factual metadata; those source questions are not displayed here.
01 / UNDERSTAND & APPLY
Right-triangle methods require a 90° angle. The hypotenuse lies opposite that right angle and is always the longest side. The other two sides are the legs. Rotating or reflecting the drawing does not change these roles.
For trigonometry, choose one of the acute angles as the reference angle. Of the two legs, the opposite leg lies across from that angle and the adjacent leg touches it. Although the hypotenuse also touches the acute angle, the word adjacent in a trig ratio means the adjacent leg, never the hypotenuse.
Changing the reference from one acute angle to the other swaps the opposite and adjacent legs. The hypotenuse stays the same. Label these three roles before writing a ratio; this is more reliable than memorizing which side happens to be on the left or bottom.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL PRACTICE · 1
Hypotenuse AB; opposite leg BC
The side across from the right angle C is AB. The side across from A is BC. Side AC is the adjacent leg relative to A.
02 / UNDERSTAND & APPLY
For a right triangle with legs a and b and hypotenuse c, a² + b² = c². If the hypotenuse is unknown, add the leg squares and take a positive square root. If a leg is unknown, subtract the known leg square from the hypotenuse square before taking the root.
The converse tests whether three lengths form a right triangle: square the longest candidate side and compare it with the sum of the other two squares. The theorem cannot be applied to an arbitrary triangle as though one angle were right.
Pythagorean triples such as 3–4–5, 5–12–13, and 8–15–17 can speed recognition. Any common positive multiple is also a right-triangle triple. Recognition is a check, not a reason to replace a stated length with a nearby convenient one.
a² + b² = c²; c = √(a² + b²); a = √(c² − b²)
c denotes the hypotenuse, not an arbitrary side.
ORIGINAL WORKED EXAMPLE
A right triangle has hypotenuse 29 cm and one leg 20 cm. Find the other leg.
Answer 21 cm
ORIGINAL PRACTICE · 1
Yes
The longest side is 26. Since 10² + 24² = 100 + 576 = 676 = 26², the converse of the Pythagorean theorem establishes a right angle opposite the 26-unit side.
ORIGINAL PRACTICE · 2
√149
The hypotenuse squared is 7² + 10² = 149. Its positive length is √149. Adding the lengths to get 17 or subtracting their squares would be incorrect.
03 / UNDERSTAND & APPLY
A coordinate displacement can be split into a horizontal change and a vertical change, which are perpendicular. This gives the distance formula. Squaring removes the signs of the coordinate differences, so either subtraction order works if used consistently.
A rectangle’s diagonal is the hypotenuse of a right triangle whose legs are the side lengths. A diagonal across a rectangular box can be found in two steps: find a face diagonal, then use it with the third perpendicular dimension. Many diagrams hide a right triangle inside a larger shape.
An altitude can create right triangles, but it may add an unknown segment rather than immediately solve the problem. Only use equal halves if symmetry, congruence, or explicit midpoint information establishes them.
d = √((x₂ − x₁)² + (y₂ − y₁)²)
Horizontal and vertical coordinate changes are the legs of a right triangle.
ORIGINAL WORKED EXAMPLE
Find the distance between P(−3, 2) and Q(9, 7).
Answer 13 units
ORIGINAL PRACTICE · 1
12
The width and height are perpendicular legs; the diagonal is the hypotenuse. Thus h² = 20² − 16² = 144 and h = 12.
04 / UNDERSTAND & APPLY
A right triangle with equal acute angles has equal legs. If each leg is s, the Pythagorean theorem gives hypotenuse √(s² + s²) = s√2. The side ratio is therefore 1:1:√2, with the √2 entry opposite 90°.
When the hypotenuse is given, divide by √2 to find each leg. Multiplying by √2 goes in the other direction. A square’s diagonal creates two such triangles, so this same relationship connects a square’s side and diagonal.
Exact radicals are often more useful than decimals. For example, 12/√2 = 6√2 after rationalizing the denominator. Either expression has the same value; do not round √2 before completing further calculations.
Legs s, s; hypotenuse s√2
Applies only to 45°–45°–90° triangles.
ORIGINAL WORKED EXAMPLE
A square has diagonal 12 cm. Find its side length and area.
Answer Side 6√2 cm; area 72 cm²
ORIGINAL PRACTICE · 1
Hypotenuse 9√2; area 81/2
The other leg is also 9. The hypotenuse is 9√2, and the two legs are perpendicular, so area is (1/2)(9)(9) = 81/2.
05 / UNDERSTAND & APPLY
An altitude splits an equilateral triangle into two congruent right triangles. If the original side is 2s, the altitude creates a short leg s and a hypotenuse 2s. The remaining leg is √((2s)² − s²) = s√3.
The side opposite 30° is the short leg s, the side opposite 60° is the long leg s√3, and the side opposite 90° is the hypotenuse 2s. Associate each length with its opposite angle; the orientation of the diagram is irrelevant.
If a long leg is given, divide it by √3 to recover s, then double s for the hypotenuse. If the hypotenuse is given, halve it first. These are different starting operations, so identify the known side before using the ratio.
Opposite 30° : opposite 60° : opposite 90° = 1 : √3 : 2
Multiply all three entries by the same positive factor.
Full-size scientific diagram. Swipe or scroll horizontally; keyboard users can focus the diagram and use arrow keys.
ORIGINAL WORKED EXAMPLE
A 30°–60°–90° triangle has long leg 12√3 m. Find the other two sides and its area.
Answer Short leg 12 m; hypotenuse 24 m; area 72√3 m²
ORIGINAL PRACTICE · 1
5√3
An altitude bisects the base into two 5-unit segments and forms a 30°–60°–90° triangle with hypotenuse 10. The altitude is the long leg, so it is 5√3.
ORIGINAL PRACTICE · 2
9√3
The short leg is half the hypotenuse, or 9. The side opposite 60° is the long leg, 9√3. The side 9 lies opposite 30°, not 60°.
06 / UNDERSTAND & APPLY
For an acute angle θ in a right triangle, sine is opposite/hypotenuse, cosine is adjacent/hypotenuse, and tangent is opposite/adjacent. The mnemonic SOH–CAH–TOA can help recall these definitions, but the side labels must be tied to θ.
The ratios depend on the angle rather than the triangle’s size. Any two right triangles with the same acute angle are similar by AA. Scaling multiplies both numerator and denominator by the same factor, leaving the ratio unchanged.
Sine and cosine of an acute angle lie strictly between 0 and 1 because a leg is shorter than the hypotenuse. Tangent can exceed 1 because the opposite leg can exceed the adjacent leg. All three ratios are positive for an acute angle.
sin θ = opposite/hypotenuse; cos θ = adjacent/hypotenuse; tan θ = opposite/adjacent
Ratios are dimensionless when all side lengths use the same unit.
ORIGINAL WORKED EXAMPLE
Relative to acute angle θ, a right triangle has opposite leg 9 and adjacent leg 12. Find sin θ, cos θ, and tan θ.
Answer sin θ = 3/5; cos θ = 4/5; tan θ = 3/4
ORIGINAL PRACTICE · 1
8/17
The hypotenuse is 17. Relative to A, the opposite leg is 15 and the adjacent leg is 8, so cosine is adjacent/hypotenuse = 8/17.
ORIGINAL PRACTICE · 2
5/13
The length scale factor is 3, so the corresponding opposite side becomes 15 and the hypotenuse 39. Their ratio 15/39 simplifies to 5/13. Trig values do not scale with side lengths.
07 / UNDERSTAND & APPLY
Circle the known side and the requested side relative to the given acute angle. Use the trig ratio containing those two roles: opposite with hypotenuse suggests sine; adjacent with hypotenuse suggests cosine; opposite with adjacent suggests tangent.
Write the equation before solving. If sin θ = x/h, then x = h sin θ. If sin θ = a/x, then x = a/sin θ. Multiplication versus division follows from the unknown’s position; it is not determined by which trig function you picked.
When the angle is 30°, 45°, or 60°, special-triangle ratios give exact values: sin 30° = 1/2, cos 30° = √3/2, sin 45° = cos 45° = √2/2, and tan 60° = √3. Use exact values when the question requests an exact answer.
ORIGINAL WORKED EXAMPLE
A right triangle has a 30° angle, and the opposite leg has length 7. Find the hypotenuse h.
Answer 14
ORIGINAL PRACTICE · 1
15
Write opposite/36 = 5/12. Thus opposite = 36(5/12) = 15. The ratio gives a length scale factor of 3 relative to legs 5 and 12.
ORIGINAL PRACTICE · 2
10
Cosine links adjacent leg to hypotenuse: adjacent = 20 cos 60° = 20(1/2) = 10. The longer leg 10√3 is opposite the 60° angle.
08 / UNDERSTAND & APPLY
The two acute angles in a right triangle sum to 90°. A leg that is opposite one acute angle is adjacent to the other, while the hypotenuse is shared. Therefore the sine of one angle equals the cosine of its complement.
This relationship can simplify expressions without evaluating any trig function numerically. For example, sin 18° and cos 72° have the same value because their angles add to 90°. Equal trig values here come from a specific angle relationship, not from sine and cosine being interchangeable.
On the acute-angle interval, sine increases and cosine decreases, so an equation sin A = cos B with A and B both acute implies A + B = 90°. If angles outside this range are allowed, periodicity can introduce other solutions; do not silently extend the acute-angle conclusion.
sin θ = cos(90° − θ); cos θ = sin(90° − θ)
For complementary acute angles, the opposite and adjacent roles swap.
ORIGINAL WORKED EXAMPLE
Angles (2x + 7)° and (3x − 2)° are acute, and sin(2x + 7)° = cos(3x − 2)°. Find x.
Answer x = 17
ORIGINAL PRACTICE · 1
7/25
Angles A and B are complementary. The same side is opposite A and adjacent to B, so sin A = cos B = 7/25. No missing-side calculation is necessary.
09 / UNDERSTAND & APPLY
A ratio describes a family of similar right triangles. If sin θ = 8/17 for an acute angle, you may use a representative triangle with opposite leg 8 and hypotenuse 17, even if the actual side lengths are unknown. Pythagoras then finds the remaining leg.
Dividing a² + b² = c² by c² gives sin²θ + cos²θ = 1. Here sin²θ means (sin θ)², not sin(θ²). Also, tan θ = sin θ/cos θ when cosine is nonzero. These identities summarize the same side relationships.
The positive square root is correct for acute-angle sine and cosine. For general angles on the unit circle, signs depend on the quadrant; the Circles lesson develops that extension. Do not apply an acute-triangle sign assumption to an unrestricted angle.
sin²θ + cos²θ = 1; tan θ = sin θ/cos θ
These follow from the right-triangle side definitions for acute θ.
ORIGINAL WORKED EXAMPLE
For acute θ, sin θ = 8/17. Find cos θ and tan θ.
Answer cos θ = 15/17; tan θ = 8/15
ORIGINAL PRACTICE · 1
5/13
A representative triangle has adjacent leg 12 and hypotenuse 13, so the opposite leg is √(169 − 144) = 5. Sine is 5/13, using the positive root for an acute angle.
10 / UNDERSTAND & APPLY
In a line-of-sight model, an angle of elevation is measured upward from a horizontal line. Draw the horizontal distance as one leg and the vertical height difference as the other; the line of sight is the hypotenuse. A distance described along a ramp or cable is usually slanted, not horizontal.
If an observer’s eyes are above the ground, the right triangle measures the height above eye level. Add the eye height to obtain the object’s total height. State assumptions such as level ground and a vertical object rather than treating them as universal.
Use degree mode for degree angles and radian mode for radian angles. A quick degree-mode check is sin 30° = 0.5. Inverse trig finds an angle: θ = tan⁻¹(opposite/adjacent). The notation tan⁻¹ here means inverse tangent, not 1/tan. Keep intermediate decimals unrounded and round the requested final measurement only.
ORIGINAL WORKED EXAMPLE
On level ground, an observer’s eyes are 1.6 m above the ground and 18 m horizontally from a vertical tower. The angle of elevation to its top is 38°. Find the tower’s height to the nearest tenth of a meter.
Answer 15.7 m
ORIGINAL PRACTICE · 1
1.46 m
The ramp is the hypotenuse, and the rise is opposite 12°. Thus rise = 7 sin 12° ≈ 1.45538 m, which rounds to 1.46 m. Using tangent with 7 as adjacent would treat the slanted ramp as horizontal.
ORIGINAL PRACTICE · 2
37°
tan θ = 6/8 = 0.75. Use inverse tangent in degree mode: θ ≈ 36.8699°, which rounds to 37°. Taking 1/tan(0.75) would not solve this angle equation.
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