Question

consider right triangle △jkl below. which expressions are equivalent to sin(∠k)? choose 2 answers:

Explanation:

To solve for the expressions equivalent to \(\sin(\angle K)\) in right triangle \(\triangle JKL\) (with the right angle at \(J\)):

Step 1: Recall the definition of sine in a right triangle

In a right triangle, the sine of an acute angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.

For \(\angle K\):

• The side opposite \(\angle K\) is \(JL\) (since it does not form \(\angle K\)).
• The side adjacent to \(\angle K\) is \(JK\).
• The hypotenuse (longest side, opposite the right angle) is \(KL\).

Step 2: Apply the sine definition to \(\angle K\)

By definition:
\[
\sin(\angle K) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JL}{KL}
\]

Step 3: Use complementary angles (optional, if needed)

In a right triangle, the acute angles are complementary (their sum is \(90^\circ\)). Thus, \(\angle K + \angle L = 90^\circ\), so \(\angle K = 90^\circ - \angle L\).

For \(\angle L\):

• The side opposite \(\angle L\) is \(JK\).
• The hypotenuse is still \(KL\).

By definition:
\[
\sin(\angle L) = \frac{JK}{KL}, \quad \cos(\angle L) = \frac{JL}{KL}
\]

But since \(\angle K = 90^\circ - \angle L\), we use the co-function identity:
\[
\sin(90^\circ - \theta) = \cos(\theta)
\]
Thus:
\[
\sin(\angle K) = \sin(90^\circ - \angle L) = \cos(\angle L)
\]

Key Expressions for \(\sin(\angle K)\)

From the above:

1. \(\boldsymbol{\frac{JL}{KL}}\) (direct application of sine definition for \(\angle K\)).
2. \(\boldsymbol{\cos(\angle L)}\) (using the co-function identity, since \(\angle K\) and \(\angle L\) are complementary).

(Note: The specific answer depends on the provided options, but the reasoning follows the sine definition and co-function identity for complementary angles in a right triangle.)

Answer:

To solve for the expressions equivalent to \(\sin(\angle K)\) in right triangle \(\triangle JKL\) (with the right angle at \(J\)):

Step 1: Recall the definition of sine in a right triangle

In a right triangle, the sine of an acute angle is defined as the ratio of the length of the opposite side to the length of the hypotenuse.

For \(\angle K\):

• The side opposite \(\angle K\) is \(JL\) (since it does not form \(\angle K\)).
• The side adjacent to \(\angle K\) is \(JK\).
• The hypotenuse (longest side, opposite the right angle) is \(KL\).

Step 2: Apply the sine definition to \(\angle K\)

By definition:
\[
\sin(\angle K) = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{JL}{KL}
\]

Step 3: Use complementary angles (optional, if needed)

In a right triangle, the acute angles are complementary (their sum is \(90^\circ\)). Thus, \(\angle K + \angle L = 90^\circ\), so \(\angle K = 90^\circ - \angle L\).

For \(\angle L\):

• The side opposite \(\angle L\) is \(JK\).
• The hypotenuse is still \(KL\).

By definition:
\[
\sin(\angle L) = \frac{JK}{KL}, \quad \cos(\angle L) = \frac{JL}{KL}
\]

But since \(\angle K = 90^\circ - \angle L\), we use the co-function identity:
\[
\sin(90^\circ - \theta) = \cos(\theta)
\]
Thus:
\[
\sin(\angle K) = \sin(90^\circ - \angle L) = \cos(\angle L)
\]

Key Expressions for \(\sin(\angle K)\)

From the above:

1. \(\boldsymbol{\frac{JL}{KL}}\) (direct application of sine definition for \(\angle K\)).
2. \(\boldsymbol{\cos(\angle L)}\) (using the co-function identity, since \(\angle K\) and \(\angle L\) are complementary).

(Note: The specific answer depends on the provided options, but the reasoning follows the sine definition and co-function identity for complementary angles in a right triangle.)