Geometry and Trigonometry

1Pythagorean theorem — a missing leg

In any right triangle with legs a and b and hypotenuse c, a^2 + b^2 = c^2. The hypotenuse is always the longest side — opposite the right angle — never one of the legs you are solving for.

Given two sides, isolate the square of the unknown, take the square root, and keep the positive root; lengths on the SAT are positive. Multiples of the 3-4-5 triple (6-8-10, 9-12-15, …) and 5-12-13 appear often enough that spotting them saves the squaring step entirely.

Figure. Legs a and b with hypotenuse c; the missing leg is under the square root after moving a squared.

Finding a leg

1. Label the sidesMark the right angle, name the hypotenuse c, and put the given lengths on the two sides you know.
2. Substitute into a^2 + b^2 = c^2Plug the known values; leave the missing leg as a symbol or as x^2 on one side of the equation.
3. Isolate and rootSubtract the known square, then take the square root. Discard a negative root — a side length cannot be negative.

2SOH-CAH-TOA — sine from two sides

In a right triangle, label opposite, adjacent, and hypotenuse relative to the angle \theta you care about — not relative to the right angle. Then \sin\theta = \dfrac{\text{opp}}{\text{hyp}}, \cos\theta = \dfrac{\text{adj}}{\text{hyp}}, and \tan\theta = \dfrac{\text{opp}}{\text{adj}} (SOH-CAH-TOA).

The SAT rarely asks you to memorise decimal values; it asks for a ratio in simplest form once the sides are identified. If opposite and hypotenuse are given, sine is one division — no Pythagorean step required.

Figure. Sine is opposite over hypotenuse — read the angle first, then pick the two sides.

Reading sine

1. Pick the reference angleMark \theta at a non-right vertex. The side across from \theta is opposite; the side touching \theta (not the hypotenuse) is adjacent.
2. Identify hypotenuseThe hypotenuse is the side opposite the right angle — always the longest side.
3. Apply SOHSine is opposite over hypotenuse. Reduce the fraction if a common factor appears.

3Special right triangles

Two families appear often enough to memorise. A 45-45-90 triangle has legs in ratio 1:1 and hypotenuse \ell\sqrt{2} when each leg is \ell. A 30-60-90 triangle has sides 1 : \sqrt{3} : 2 — short leg, long leg, hypotenuse — so scaling the short leg by k scales the hypotenuse by 2k and the long leg by k\sqrt{3}.

Multiples of 3-4-5 (6-8-10, 9-12-15, …) and 5-12-13 are Pythagorean triples rather than angle-ratio templates, but spotting them skips squaring on leg problems the same way 30-60-90 skips the algebra when the given side matches a ratio slot.

Figure. 30-60-90 sides stay in ratio 1 : √3 : 2 after any common scale factor.

Using 30-60-90

1. Match the given side to a slotDecide whether the known length is the short leg, the long leg, or the hypotenuse in the 1:\sqrt{3}:2 ratio.
2. Find the scale factorDivide the given side by its unit-ratio value to get k.
3. Scale the unknownMultiply the unknown slot by k (and by \sqrt{3} when the long leg is needed).

4Cofunction identity

In a right triangle the two acute angles sum to 90^\circ, so each is the complement of the other. That geometry gives \sin\theta = \cos(90^\circ - \theta): the side opposite \theta is adjacent to (90^\circ - \theta), so the two ratios swap numerator and denominator roles against the same hypotenuse.

On the SAT this identity turns a cosine-of-complement question into a sine lookup, and vice versa — useful when only one ratio is given or when the reference angle is stated in the form 90^\circ - \theta.

Figure. In one right triangle the side opposite θ is adjacent to the complement. That single side-label swap is the cofunction identity.

Swapping sine and cosine

1. Name the complementIf the angle is \theta, its complement is 90^\circ - \theta — the other acute angle in the triangle.
2. Apply the identity\sin\theta = \cos(90^\circ - \theta) and \cos\theta = \sin(90^\circ - \theta).
3. Read the given ratioWhen a problem asks for \cos(90^\circ - \theta) and gives \sin\theta, the answer is the same value — no new triangle needed.

5Circle area and circumference

For a circle of radius r, circumference C = 2\pi r and area A = \pi r^2. The Digital SAT reference sheet lists both — memorise the relationship, not every decimal approximation.

Circumference is a length (units); area is squared units. A question that gives diameter d wants r = d/2 substituted before either formula — using d directly in \pi r^2 is the common slip.

From radius to area

1. Read off rIf diameter is given, halve it before substituting.
2. Square for areaCompute r^2, then multiply by \pi. Leave answers in terms of \pi unless the choices are decimals.
3. Double for circumferenceC = 2\pi r uses r once, not r^2 — area and circumference are different powers of r.

6Inscribed angle vs central angle

An inscribed angle has its vertex on the circle; a central angle has its vertex at the centre. Both subtend the same arc, and the inscribed angle is always half the central angle on that arc: \text{inscribed} = \dfrac{1}{2} \times \text{central}.

SAT items often give the central angle and ask for an inscribed angle on the same arc, or give an inscribed angle and ask how many degrees the arc spans — multiply or divide by 2 accordingly, and keep track of which angle sits at the centre versus on the rim.

Halving the central angle

1. Identify the arcBoth angles must intercept the same arc — the chord endpoints match.
2. Locate the verticesCentral angle: vertex at the centre. Inscribed angle: vertex on the circle.
3. Apply the half ruleInscribed = \dfrac{1}{2} \times central when the central angle is given.

7Similar triangles and area scale

Two triangles are similar when their corresponding angles are equal; then corresponding sides are proportional. If the side-length ratio is k : 1, every linear measure scales by k — but area scales by k^2 because area is two-dimensional.

A SAT question that gives two similar triangles and asks for an area ratio wants the square of the side ratio, not the side ratio itself. Set up \dfrac{A_1}{A_2} = \left(\dfrac{s_1}{s_2}\right)^2 once the matching sides are identified.

Figure. Linear scale k multiplies lengths; area scales as k squared.

Area from side ratio

1. Match corresponding sidesPair sides opposite equal angles — never cross-match.
2. Form the linear ratioWrite s_1 : s_2 for one pair of corresponding sides.
3. Square for areaArea ratio = (s_1/s_2)^2.