In right triangle ABC, the right angle is located at vertex C. Point M is the midpoint of … — ACT

In right triangle $ABC$ , the right angle is located at vertex $C$ . Point $M$ is the midpoint of leg $BC$ . The length of leg $AC$ is $12$ units, and $\tan(\angle MAC) = \frac{1}{3}$ . What is the value of $\sin(\angle BAC)$ ?

$\frac{2\sqrt{13}}{13}$ Answer
B
$\frac{3\sqrt{13}}{13}$
C
$\frac{2}{3}$
D
$\frac{\sqrt{10}}{10}$
E
$\frac{2\sqrt{5}}{5}$

Answer

The value of

\sin(\angle BAC)

\frac{2\sqrt{13}}{13}

In right triangle

ACM

\tan(\angle MAC) = \frac{MC}{AC} = \frac{1}{3}

. Since

AC = 12

, we find

MC = 4

. Because

M

is the midpoint of side

BC

BC = 2 \times 4 = 8

. In right triangle

ABC

, the hypotenuse is

AB = \sqrt{12^2 + 8^2} = \sqrt{208} = 4\sqrt{13}

. The sine of angle

BAC

is defined as

\frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{8}{4\sqrt{13}} = \frac{2\sqrt{13}}{13}

Step-by-Step Solution

Find the length of segment

MC

using right triangle

ACM

MC = 4

In right triangle

ACM

with right angle at

C

\tan(\angle MAC) = \frac{\text{opposite}}{\text{adjacent}} = \frac{MC}{AC}

. Given

\tan(\angle MAC) = \frac{1}{3}

and

AC = 12

\frac{MC}{12} = \frac{1}{3} \implies MC = 4

Determine the length of side

BC

BC = 8

Since

M

is the midpoint of leg

BC

BC = 2 \times MC = 2 \times 4 = 8

Calculate hypotenuse

AB

of right triangle

ABC

AB = 4\sqrt{13}

By the Pythagorean theorem in

\triangle ABC

AB = \sqrt{AC^2 + BC^2} = \sqrt{12^2 + 8^2} = \sqrt{144 + 64} = \sqrt{208} = 4\sqrt{13}

Calculate

\sin(\angle BAC)

\sin(\angle BAC) = \frac{2\sqrt{13}}{13}

In right triangle

ABC

\sin(\angle BAC) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{BC}{AB} = \frac{8}{4\sqrt{13}} = \frac{2}{\sqrt{13}} = \frac{2\sqrt{13}}{13}

Key Concept

Applying SOHCAHTOA definitions and the Pythagorean theorem in multi-step right triangle geometry.

Estimated Time:2m 0s

Question

Answer

Step-by-Step Solution

Key Concept

Common Mistakes

Topics

More questions from this topic