College Board Digital SAT Math (style)-style practice: In a right triangle, the side opposite angle θ is 6 and the hypotenuse is 10. What is sinθ? (A) 35 (B) 45 (C) 34 (D) 53

The correct answer is (A), (3)/(5). By SOH, sinθ = oppositehypotenuse = (6)/(10) = (3)/(5). (This is a 6-8-10 triangle, a scaled 3-4-5.)

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How do you use the Pythagorean theorem, special right triangles, and the trigonometric ratios to find sides and angles?

Right triangles and trigonometry: apply the Pythagorean theorem, the special right triangles, the sine, cosine and tangent ratios (SOH-CAH-TOA), and the sine and cosine of complementary angles.

A focused answer to the Digital SAT Geometry and Trigonometry skill of right triangles and trigonometry: the Pythagorean theorem, the special right triangles, SOH-CAH-TOA, and the complementary-angle relationship between sine and cosine.

Generated by Claude Opus 4.811 min answerUpdated 2026-06-14

Reviewed by: AI editorial process; not yet individually human-reviewed

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Quick answer

The Pythagorean theorem $a^2 + b^2 = c^2$ relates the legs and hypotenuse of a right triangle. The special right triangles are $45$ - $45$ - $90$ (sides $x : x : x\sqrt{2}$ ) and $30$ - $60$ - $90$ (sides $x : x\sqrt{3} : 2x$ ), both on the reference sheet. The trig ratios are SOH-CAH-TOA: $\sin\theta = \frac{\text{opp}}{\text{hyp}}$ , $\cos\theta = \frac{\text{adj}}{\text{hyp}}$ , $\tan\theta = \frac{\text{opp}}{\text{adj}}$ (you must memorise these; they are not on the sheet). The two acute angles of a right triangle are complementary, and $\sin\theta = \cos(90^\circ - \theta)$ , so sine and cosine of complementary angles are equal. Recognise common triples like $3$ - $4$ - $5$ and $5$ - $12$ - $13$ .

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What this skill is asking
The right-triangle toolkit
A worked trig-ratio question
The special right triangles
Complementary angles and the sine-cosine link

What this skill is asking

This is the right-triangle core of the Digital SAT (Geometry and Trigonometry domain): the Pythagorean theorem, the two special right triangles, the trigonometric ratios (sine, cosine, tangent), and the relationship between the sine and cosine of complementary angles. Note that while the special-triangle side ratios are on the reference sheet, the trig ratios themselves are not, so you must know SOH-CAH-TOA.

The right-triangle toolkit

Three tools cover right-triangle problems.

A worked trig-ratio question

Label the sides relative to the angle, then apply the ratio.

Finding a side with a trig ratio

In a right triangle, angle $\theta$ has $\cos\theta = \frac{4}{5}$ , and the hypotenuse is $20$ . Find the side adjacent to $\theta$ .

step 1 Identify the ratio

$\cos\theta = \frac{\text{adjacent}}{\text{hypotenuse}}$ (CAH).

step 2 Substitute the known values

$\frac{4}{5} = \frac{\text{adjacent}}{20}$ .

step 3 Solve for the adjacent side

adjacent $= \frac{4}{5} \times 20 = 16$ .

step 4 Confirm with Pythagoras

If adjacent is $16$ and hypotenuse $20$ , the opposite is $\sqrt{20^2 - 16^2} = \sqrt{400 - 256} = \sqrt{144} = 12$ , a $12$ - $16$ - $20$ (scaled $3$ - $4$ - $5$ ) triangle. Consistent.

The special right triangles

The two special triangles let you find all sides from one, without trig. In a $45$ - $45$ - $90$ triangle (a right isosceles triangle), the legs are equal and the hypotenuse is a leg times $\sqrt{2}$ . In a $30$ - $60$ - $90$ triangle, the side opposite $30^\circ$ is the shortest ( $x$ ), the side opposite $60^\circ$ is $x\sqrt{3}$ , and the hypotenuse (opposite $90^\circ$ ) is $2x$ . Both ratios are on the reference sheet, so the work is recognising the triangle type from its angles (or from a square's diagonal, or an equilateral triangle cut in half) and reading off the sides. These appear constantly because they give exact answers quickly.

Complementary angles and the sine-cosine link

Because a right triangle's two acute angles are complementary (they sum to $90^\circ$ , since the third angle is the $90^\circ$ right angle), the side that is opposite one acute angle is adjacent to the other. That geometric fact produces the identity $\sin\theta = \cos(90^\circ - \theta)$ : the sine of an angle equals the cosine of its complement. So if $\sin(A) = 0.6$ and $A + B = 90^\circ$ , then $\cos(B) = 0.6$ too. The SAT tests this directly with "given $\sin(x^\circ) = \cos(y^\circ)$ , and $x + y = 90$ " style questions. Recognising the complementary relationship answers them without computing any angle.

Common traps

Looking for trig ratios on the reference sheet: SOH-CAH-TOA is not provided. Memorise $\sin = \frac{\text{opp}}{\text{hyp}}$ , $\cos = \frac{\text{adj}}{\text{hyp}}$ , $\tan = \frac{\text{opp}}{\text{adj}}$ .
Mislabeling opposite and adjacent: "Opposite" and "adjacent" are relative to the angle in question; the same side is opposite one acute angle and adjacent to the other.
Mixing up the special-triangle ratios: $45$ - $45$ - $90$ uses $\sqrt{2}$ ; $30$ - $60$ - $90$ uses $\sqrt{3}$ and a hypotenuse of $2x$ . Putting $\sqrt{3}$ in the wrong place is common.
Forgetting the complementary identity: $\sin\theta = \cos(90^\circ - \theta)$ . When an equation sets a sine equal to a cosine, the angles are complementary.

Exam-style practice questions

Practice questions written in the style of College Board exam questions on this dot point, with worked answer explainers. The year tag is the paper they imitate, not the source.

Digital SAT Math (style)1 marksIn a right triangle, the side opposite angle

\theta

is 6 and the hypotenuse is 10. What is

\sin\theta

? (A)

\tfrac{3}{5}

(B)

\tfrac{4}{5}

(C)

\tfrac{3}{4}

(D)

\tfrac{5}{3}

Show worked answer →

The correct answer is (A), $\frac{3}{5}$ .

By SOH, $\sin\theta = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{6}{10} = \frac{3}{5}$ . (This is a 6-8-10 triangle, a scaled 3-4-5.)

Digital SAT Math (style)1 marksIn a right triangle,

\sin(A) = \cos(B)

where

A

and

B

are the two acute angles. If

\sin(A) = 0.6

, what is

\cos(B)

? (A)

0.4

(B)

0.6

(C)

0.8

(D)

1.0

Show worked answer →

The correct answer is (B), 0.6.

In a right triangle the two acute angles are complementary ( $A + B = 90^\circ$ ), and the sine of an angle equals the cosine of its complement: $\sin(A) = \cos(90^\circ - A) = \cos(B)$ . So $\cos(B) = \sin(A) = 0.6$ .

Related dot points

Sources & how we know this

Math Specifications — College Board (2024)
What Are Content Domains? — College Board (2024)