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 with \( \angle L = 90^\circ \), \( \angle K = 60^\circ \), so \( \angle J = 30^\circ \). In a 30-60-90 triangle, the sides are in the ratio \( 1 : \sqrt{3} : 2 \) (opposite 30°, 60°, 90° respectively).

Step2: Determine sides relative to angles

\( KL = 2 \) is adjacent to \( 60^\circ \) (or opposite \( 30^\circ \)), and \( JL \) is opposite \( 60^\circ \). Let the side opposite \( 30^\circ \) be \( x_1 \), opposite \( 60^\circ \) be \( x_2 \), hypotenuse \( h \). Here, \( KL = 2 \) is \( x_1 \) (opposite \( 30^\circ \))? Wait, no: \( \angle K = 60^\circ \), so \( \angle J = 30^\circ \). So side opposite \( \angle J = 30^\circ \) is \( KL = 2 \), side opposite \( \angle K = 60^\circ \) is \( JL = x \). In 30-60-90, side opposite 30° is \( s \), opposite 60° is \( s\sqrt{3} \), hypotenuse \( 2s \). Here, \( s = KL = 2 \), so \( JL = s\sqrt{3} = 2\sqrt{3} \)? Wait, no: Wait, \( \angle L = 90^\circ \), so \( KL \) and \( JL \) are legs, \( KJ \) is hypotenuse. \( \angle K = 60^\circ \), so \( \tan(60^\circ) = \frac{JL}{KL} \). \( \tan(60^\circ) = \sqrt{3} \), \( KL = 2 \), so \( JL = KL \times \tan(60^\circ) = 2 \times \sqrt{3} = 2\sqrt{3} \).

Answer:

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