The problem asks us to find the value of $\sin L$ rounded to the nearest hundredth. We are given a right triangle $LKJ$ with a right angle at vertex $K$. The length of side $KJ$ is 2 and the length of side $LJ$ is 7.

1. Problem Description

The problem asks us to find the value of $\sin L$ rounded to the nearest hundredth. We are given a right triangle $LKJ$ with a right angle at vertex $K$. The length of side $KJ$ is 2 and the length of side $LJ$ is

7.

2. Solution Steps

First, we need to find the length of side $LK$. We can use the Pythagorean theorem:

$a^2 + b^2 = c^2$

where $a$ and $b$ are the lengths of the legs of the right triangle and $c$ is the length of the hypotenuse.

In this case, $LK^2 + KJ^2 = LJ^2$, so $LK^2 + 2^2 = 7^2$.

$LK^2 + 4 = 49$

$LK^2 = 49 - 4$

$LK^2 = 45$

$LK = \sqrt{45} = 3\sqrt{5}$

Now we can find $\sin L$. The sine of an angle in a right triangle is the ratio of the length of the opposite side to the length of the hypotenuse.

$\sin L = \frac{\text{opposite}}{\text{hypotenuse}} = \frac{KJ}{LJ} = \frac{2}{7}$

To find the value of $\sin L$ rounded to the nearest hundredth, we divide 2 by 7:

$\frac{2}{7} \approx 0.2857$

Rounding to the nearest hundredth, we get 0.

2

9.

3. Final Answer

0.29