A right triangle has acute angles P and R. The sine of angle P is defined by \frac{a}{13} … — SAT

A right triangle has acute angles $P$ and $R$ . The sine of angle $P$ is defined by $\frac{a}{13}$ and the cosine of angle $R$ is $\frac{5}{13}$ , where $a$ is a positive constant. What is the value of $\sin(R)$ ?

A
$\frac{5}{13}$
$\frac{12}{13}$ Answer
C
$\frac{12}{5}$
D
$\frac{\sqrt{194}}{13}$

Answer

\frac{12}{13}

The correct answer is the value

\frac{12}{13}

. Since

P

and

R

are the acute angles of a right triangle, they are complementary, which means

\sin(P) = \cos(R) = \frac{5}{13}

. By using the Pythagorean identity

\sin^2(R) + \cos^2(R) = 1

, we can substitute the value of

\cos(R)

to get

\sin^2(R) + \left(\frac{5}{13}\right)^2 = 1

. Solving for

\sin(R)

yields

\sin^2(R) = 1 - \frac{25}{169} = \frac{144}{169}

. Since

R

is an acute angle, taking the positive square root yields

\sin(R) = \frac{12}{13}

Step-by-Step Solution

Apply the complementary angle identity to determine the value of the constant

a

Since

P

and

R

are the acute angles of a right triangle, they are complementary, meaning

P + R = 90^\circ

. The co-function identity states that

\sin(P) = \cos(R)

. Given

\cos(R) = \frac{5}{13}

, we have

\sin(P) = \frac{a}{13} = \frac{5}{13}

, which means

a = 5

This establishes the relationship between the trigonometric ratios of the two acute angles.

Use the Pythagorean identity to find

\sin(R)

Using the Pythagorean identity

\sin^2(R) + \cos^2(R) = 1

, substitute

\cos(R) = \frac{5}{13}

into the equation:

\sin^2(R) + \left(\frac{5}{13}\right)^2 = 1 \implies \sin^2(R) + \frac{25}{169} = 1 \implies \sin^2(R) = 1 - \frac{25}{169} = \frac{144}{169}

. Taking the positive square root because

R

is an acute angle gives

\sin(R) = \frac{12}{13}

This determines the sine of the angle from its known cosine value.

Key Concept

Trigonometric ratios of complementary angles and the Pythagorean identity in right triangles.

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