Question

in the right triangle shown, $m\angle k = 60^{\circ}$ and $kl = 2$.
how long is $jl$
choose 1 answer:
a $2\sqrt{2}$
b $2\sqrt{3}$
c $4$
d $4\sqrt{2}$
e $4\sqrt{3}$

Explanation:

Step1: Identify triangle type and angles

This is a right triangle at \( L \), so \( \angle L = 90^\circ \). Given \( \angle K = 60^\circ \), then \( \angle J = 30^\circ \) (since triangle angles sum to \( 180^\circ \)).

Step2: Recall 30-60-90 triangle ratios

In a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \), where the side opposite \( 30^\circ \) is the shortest (let's call it \( a \)), opposite \( 60^\circ \) is \( a\sqrt{3} \), and hypotenuse is \( 2a \).

Step3: Determine which side is which

Here, \( KL = 2 \) is adjacent to \( 60^\circ \) (or opposite \( 30^\circ \), since \( \angle J = 30^\circ \)). Wait, actually, \( KL \) is adjacent to \( \angle K = 60^\circ \), and \( JL \) is opposite \( \angle K = 60^\circ \). Wait, no: \( \angle K = 60^\circ \), so the side opposite \( \angle K \) is \( JL \), and the side adjacent (or the shorter leg, since \( \angle J = 30^\circ \), so the side opposite \( 30^\circ \) is \( KL \)? Wait, no: \( \angle J = 30^\circ \), so the side opposite \( 30^\circ \) is \( KL \) (length 2). Then the side opposite \( 60^\circ \) (which is \( JL \)) should be \( KL \times \sqrt{3} \)? Wait, no, wait: in 30-60-90, the sides are: opposite 30°: \( a \), opposite 60°: \( a\sqrt{3} \), hypotenuse: \( 2a \). So if \( KL \) is opposite \( 30^\circ \) (since \( \angle J = 30^\circ \)), then \( KL = a = 2 \). Then the side opposite \( 60^\circ \) (which is \( JL \)) is \( a\sqrt{3} = 2\sqrt{3} \)? Wait, no, wait the diagram: \( L \) is right angle, \( K \) has 60°, so \( KL \) is one leg, \( JL \) is the other leg, \( KJ \) is hypotenuse. So \( \angle K = 60^\circ \), so \( \tan(60^\circ) = \frac{JL}{KL} \). Since \( \tan(60^\circ) = \sqrt{3} \) and \( KL = 2 \), then \( JL = KL \times \tan(60^\circ) = 2 \times \sqrt{3} = 2\sqrt{3} \). Wait, but wait, maybe I mixed up. Wait, \( \angle K = 60^\circ \), so \( \tan(\angle K) = \frac{opposite}{adjacent} = \frac{JL}{KL} \). So \( \tan(60^\circ) = \frac{JL}{2} \), so \( JL = 2 \times \tan(60^\circ) = 2\sqrt{3} \). Wait, but the options have B as \( 2\sqrt{3} \), but the selected answer was C (4). Wait, maybe I made a mistake. Wait, no, wait: maybe \( KL \) is the shorter leg (opposite 30°), so the hypotenuse would be \( 2 \times 2 = 4 \), but \( JL \) is the longer leg (opposite 60°), which is \( 2\sqrt{3} \). Wait, but the selected answer in the image is C (4), which is wrong. Wait, no, maybe I misidentified the angles. Wait, the right angle is at \( L \), so \( \angle L = 90^\circ \), \( \angle K = 60^\circ \), so \( \angle J = 30^\circ \). So the side opposite \( 30^\circ \) is \( KL \) (length 2), so the hypotenuse \( KJ \) is \( 2 \times 2 = 4 \), and the side opposite \( 60^\circ \) (which is \( JL \)) is \( 2 \times \sqrt{3} = 2\sqrt{3} \). Wait, but the options: A is \( 2\sqrt{2} \) (no, that's 45-45-90), B is \( 2\sqrt{3} \), C is 4, D is \( 4\sqrt{2} \), E is \( 4\sqrt{3} \). Wait, maybe the user made a mistake in selecting, but according to the math, \( JL \) should be \( 2\sqrt{3} \) (option B). Wait, but let's recheck: \( \tan(60^\circ) = \frac{JL}{KL} \), \( KL = 2 \), so \( JL = 2 \times \tan(60^\circ) = 2\sqrt{3} \). So the correct answer is B. Wait, but the image shows C selected, but that's incorrect. So the correct answer is B.

Answer:

B. \( 2\sqrt{3} \)