SAT Math: Right triangles and trigonometry

Right triangles and trigonometry.

Geometry and Trigonometry · Right triangles and trigonometry
Both modules
14 practice questions · 4 with free solutions
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: $\sin = \frac{opp}{hyp}$ , $\cos = \frac{adj}{hyp}$ , $\tan = \frac{opp}{adj}$ .
$\sin x^{\circ} = \cos (90 - x)^{\circ}$ . Special triangles: $30$ - $60$ - $90$ sides $1 : \sqrt{3} : 2$ ; $45$ - $45$ - $90$ sides $1 : 1 : \sqrt{2}$ .
Radians: multiply degrees by $\frac{π}{180}$ .

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 6 and 8. What is the cosine of the angle opposite the side of length 6?

$\frac{3}{5}$
$\frac{3}{4}$
$\frac{4}{5}$
$\frac{4}{3}$

Answer: C: $\frac{4}{5}$

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

Practice questions

Try each question before opening the solution. Desmos or your own calculator is allowed on every SAT Math question. Solutions to the first 3 are open to everyone; the rest need only a free account. Every answer here was checked independently before it was published.

Question 1

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 2

In the equation $\sin (x^{\circ}) = \cos (40^{\circ})$ , $0 < x < 90$ . What is the value of $x$ ?

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 3

In the equation $\cos (x^{\circ}) = \sin (20^{\circ})$ , $0 < x < 90$ . What is the value of $x$ ?

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 4

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

$\frac{3}{5}$
$\frac{3}{4}$
$\frac{4}{5}$
$\frac{4}{3}$