Right Triangle Trigonometry | SAT Math Practice

The key idea

In a right triangle,

sin = \frac{opposite}{hypotenuse}

cos = \frac{adjacent}{hypotenuse}

tan = \frac{opposite}{adjacent}

. The two acute angles are complementary, so

sin A = cos B

and

tan B = \frac{1}{t a n A}

. A ratio gives side lengths up to a scale factor

k

The trap

Treating the ratio's numbers as the actual side lengths.

Can Desmos do it?

Desmos evaluates trig functions, but a sine-equals-cosine question asks you to recognize complementary angles, not to solve an equation numerically.

College Board skill: Right triangles and trigonometry · Geometry and Trigonometry

Try one at each level

One question from each level. The drill has 12 at every level, and moves you up after four right in a row.

Level 1 · Warm-up

In right triangle

A B C

, the right angle is at

C

and

tan A = \frac{12}{5}

. What is the value of

sin B

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Answer: 5/13Draw it with sides for angle

A

: opposite

12

, adjacent

5

, hypotenuse

13

(any multiple works, since only ratios matter; the missing side comes from

1 2^{2} + 5^{2} = 1 3^{2}

). For angle

B

the roles of the legs swap: the side opposite

B

5

and the side adjacent to

B

12

. So

sin B = \frac{5}{13}

Level 2 · Test level

For the acute angles

(4 x + 27) °

and

(3 x + 7) °

sin (4 x + 27) ° = cos (3 x + 7) °

. What is the value of

x

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Answer: 8The sine of an angle equals the cosine of its complement, so the angles add to

90°

(4 x + 27) + (3 x + 7) = 90

. Then

7 x = 56

and

x = 8

Level 3 · Beyond the test

In right triangle

A B C

, angle

C

is the right angle and

tan A = \frac{20}{21}

. The perimeter of the triangle is 280. What is the area of the triangle?

Show the answer

Answer: 3360The legs are

20 k

and

21 k

, and the hypotenuse is

(20 k)^{2} + (21 k)^{2} = 29 k

. The perimeter is

70 k = 280

, so

k = 4

. The area is

\frac{1}{2} (80) (84) = 3360

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