In right triangle ABC, the right angle is at vertex B, AB = 12, and BC = 5. A line segment… — ACT

In right triangle $ABC$ , the right angle is at vertex $B$ , $AB = 12$ , and $BC = 5$ . A line segment $BD$ is drawn perpendicular to the hypotenuse $AC$ such that $D$ lies on $AC$ . From point $D$ , a perpendicular line segment $DE$ is drawn to side $AB$ , where $E$ lies on $AB$ . What is the length of segment $DE$ ?

A
$\frac{300}{169}$
B
$\frac{60}{13}$
$\frac{720}{169}$ Answer
D
$\frac{144}{13}$
E
$\frac{1728}{169}$

Answer

\frac{720}{169}

The correct answer is

\frac{720}{169}

. First, find the hypotenuse of the right triangle

ABC

using the Pythagorean theorem:

AC = \sqrt{AB^2 + BC^2} = \sqrt{12^2 + 5^2} = 13

. In right triangle

ABC

, the sine of angle

A

\sin(\angle A) = \frac{BC}{AC} = \frac{5}{13}

. Next, in right triangle

ABD

(which has a right angle at

D

), the length of the altitude

BD

can be found using the sine of angle

A

BD = AB \sin(\angle A) = 12 \cdot \frac{5}{13} = \frac{60}{13}

. In right triangle

BDE

(which has a right angle at

E

), the angle

\angle BDE

is equal to

\angle A

because both are complementary to

\angle ABD

(or

\angle EBD

). Therefore, the adjacent side

DE

is found using the cosine of angle

\angle BDE

DE = BD \cos(\angle BDE) = BD \cos(\angle A) = \frac{60}{13} \cdot \frac{12}{13} = \frac{720}{169}

Step-by-Step Solution

Find the hypotenuse

AC

of the right triangle

ABC

using the Pythagorean theorem.

AC = 13

The hypotenuse is needed to find the trigonometric ratios of angle

A

Determine

\sin(\angle A)

and

\cos(\angle A)

from right triangle

ABC

\sin(\angle A) = \frac{5}{13}

and

\cos(\angle A) = \frac{12}{13}

These trigonometric ratios are needed for the calculations in the nested right triangles.

Find the length of altitude

BD

in right triangle

ABD

using

\sin(\angle A)

BD = \frac{60}{13}

BD

serves as the hypotenuse for the next right triangle

BDE

Identify that

\angle BDE = \angle A

and calculate the length of

DE

in right triangle

BDE

DE = \frac{720}{169}

Since

\angle BDE = \angle A

DE = BD \cos(\angle BDE) = BD \cos(\angle A) = \frac{60}{13} \cdot \frac{12}{13} = \frac{720}{169}

Key Concept

Applying SOHCAHTOA (sine and cosine definitions) sequentially across nested right triangles by identifying equal angles.

Estimated Time:2m 0s

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