SAT Math: Right triangles and trigonometry

Right triangles and trigonometry.

Geometry and Trigonometry · Right triangles and trigonometry
Both modules
14 practice questions
Calculator allowed (Desmos)

Practise right triangles and trigonometry →Adaptive SAT module

What the test covers

Pythagoras, special right triangles, SOHCAHTOA, sin x = cos(90 - x)
Radians and degrees

Key ideas

Pythagoras: \(a^2+b^2=c^2\). Common triples: 3-4-5, 5-12-13, 8-15-17, 7-24-25.
SOH CAH TOA: \(\sin=\tfrac{\text{opp}}{\text{hyp}}\), \(\cos=\tfrac{\text{adj}}{\text{hyp}}\), \(\tan=\tfrac{\text{opp}}{\text{adj}}\).
\(\sin x^\circ=\cos(90-x)^\circ\). Special triangles: \(30\)-\(60\)-\(90\) sides \(1:\sqrt3:2\); \(45\)-\(45\)-\(90\) sides \(1:1:\sqrt2\).
Radians: multiply degrees by \(\tfrac{\pi}{180}\).

Common mistakes

Check whether Desmos is in degree or radian mode.
'Opposite' depends on which angle is named.

Do it in Desmos

Switch Desmos to degrees (the spanner icon) before evaluating trig values.

IB link: IB SL 3.2-3.4.

Worked example

A right triangle has legs of length 9 and 12. What is the sine of the angle opposite the side of length 9?

\(\frac{3}{5}\)
\(\frac{3}{4}\)
\(\frac{4}{5}\)
\(\frac{4}{3}\)

Answer: A: \(\frac{3}{5}\)

The hypotenuse is \(\sqrt{81+144}=15\). \(\sin=\tfrac{\text{opposite}}{\text{hypotenuse}}=\tfrac9{15}=\tfrac35\).

Practice questions

Try each question before opening the solution. Desmos or your own calculator is allowed on every SAT Math question. Every answer here was re-checked independently by computer before it was published.

Question 1

In the equation \(\sin(x^\circ)=\cos(40^\circ)\), \(0

Student-produced response: type your answer.

Show the answer and solution

Answer: 50

\(\sin(x^\circ)=\cos((90-x)^\circ)\), so \(90-x=40\) and \(x=50\).

Question 2

In a right triangle with angles \(30^\circ\), \(60^\circ\) and \(90^\circ\), the hypotenuse has length 10. What is the length of the longer leg?

\(5\sqrt{3}\)
\(5\sqrt{2}\)
\(10\sqrt{3}\)
\(5\)

Show the answer and solution

Answer: A: \(5\sqrt{3}\)

The shorter leg (opposite \(30^\circ\)) is half the hypotenuse, 5. The longer leg is \(5\sqrt3\) (or \(10\sin60^\circ\)).

Question 3

What is \(150^\circ\) in radians?

\(\frac{3\pi}{4}\)
\(\frac{5\pi}{3}\)
\(\frac{5\pi}{6}\)
\(\frac{2\pi}{3}\)

Show the answer and solution

Answer: C: \(\frac{5\pi}{6}\)

\(150\times\tfrac{\pi}{180}=\tfrac{5\pi}{6}\).

Question 4

A right triangle has legs of length 6 and 8. What is the cosine of the angle opposite the side of length 6?

\(\frac{3}{5}\)
\(\frac{3}{4}\)
\(\frac{4}{5}\)
\(\frac{4}{3}\)

Show the answer and solution

Answer: C: \(\frac{4}{5}\)

The hypotenuse is \(\sqrt{6^2+8^2}=10\). Opposite \(6\), adjacent \(8\), hypotenuse \(10\): the cosine is \(\frac{4}{5}\).

Question 5

A right triangle has legs of length 60 and 63. What is the sine of the angle opposite the side of length 60?

\(\frac{21}{20}\)
\(\frac{20}{29}\)
\(\frac{21}{29}\)
\(\frac{20}{21}\)

Show the answer and solution

Answer: B: \(\frac{20}{29}\)

The hypotenuse is \(\sqrt{60^2+63^2}=87\). Opposite \(60\), adjacent \(63\), hypotenuse \(87\): the sine is \(\frac{20}{29}\).

Question 6

A right triangle has legs of length 20 and 21. What is the cosine of the angle opposite the side of length 20?

\(\frac{20}{21}\)
\(\frac{21}{20}\)
\(\frac{21}{29}\)
\(\frac{20}{29}\)

Show the answer and solution

Answer: C: \(\frac{21}{29}\)

The hypotenuse is \(\sqrt{20^2+21^2}=29\). Opposite \(20\), adjacent \(21\), hypotenuse \(29\): the cosine is \(\frac{21}{29}\).

Question 7

A right triangle has legs of length 3 and 4. What is the tangent of the angle opposite the side of length 3?

\(\frac{3}{4}\)
\(\frac{3}{5}\)
\(\frac{4}{3}\)
\(\frac{4}{5}\)

Show the answer and solution

Answer: A: \(\frac{3}{4}\)

The hypotenuse is \(\sqrt{3^2+4^2}=5\). Opposite \(3\), adjacent \(4\), hypotenuse \(5\): the tangent is \(\frac{3}{4}\).

Question 8

A right triangle has legs of length 6 and 8. What is the sine of the angle opposite the side of length 6?

\(\frac{3}{4}\)
\(\frac{4}{5}\)
\(\frac{3}{5}\)
\(\frac{4}{3}\)

Show the answer and solution

Answer: C: \(\frac{3}{5}\)

The hypotenuse is \(\sqrt{6^2+8^2}=10\). Opposite \(6\), adjacent \(8\), hypotenuse \(10\): the sine is \(\frac{3}{5}\).

Question 9

In the equation \(\sin(x^\circ)=\cos(20^\circ)\), \(0

Student-produced response: type your answer.

Show the answer and solution

Answer: 70

\(\sin(x^\circ)=\cos((90-x)^\circ)\), so \(90-x=20\) and \(x=70\).

Question 10

In the equation \(\cos(x^\circ)=\sin(70^\circ)\), \(0

Student-produced response: type your answer.

Show the answer and solution

Answer: 20

\(\cos(x^\circ)=\sin((90-x)^\circ)\), so \(90-x=70\) and \(x=20\).

Question 11

In the equation \(\cos(x^\circ)=\sin(20^\circ)\), \(0

Student-produced response: type your answer.

Show the answer and solution

Answer: 70

\(\cos(x^\circ)=\sin((90-x)^\circ)\), so \(90-x=20\) and \(x=70\).

Question 12

In the equation \(\cos(x^\circ)=\sin(55^\circ)\), \(0

Student-produced response: type your answer.

Show the answer and solution

Answer: 35

\(\cos(x^\circ)=\sin((90-x)^\circ)\), so \(90-x=55\) and \(x=35\).

Question 13

In the equation \(\cos(x^\circ)=\sin(25^\circ)\), \(0

Student-produced response: type your answer.

Show the answer and solution

Answer: 65

\(\cos(x^\circ)=\sin((90-x)^\circ)\), so \(90-x=25\) and \(x=65\).

Keep going

Previous topic: Lines, angles, and triangles
Next topic: Circles
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