Question

consider right triangle △jkl below. which expressions represent the length of side $overline{kl}$?

Explanation:

Step1: Identify triangle type and given info

We have a right triangle \( \triangle JKL \) with \( \angle J = 90^\circ \), \( \angle L = 60^\circ \), \( JL = 3 \), \( JK = 5.2 \)? Wait, no, wait. Wait, in right triangle, right angle at \( J \), so sides: \( JL = 3 \) (adjacent to \( 60^\circ \)), \( JK \) (opposite to \( 60^\circ \)), and \( KL \) is hypotenuse. Wait, or maybe using trigonometry or Pythagoras. Wait, first, let's recall: in a right triangle, hypotenuse can be found by Pythagoras: \( KL^2 = JK^2 + JL^2 \), but also, using trigonometric ratios. Since \( \angle L = 60^\circ \), \( \cos(60^\circ) = \frac{JL}{KL} \), so \( KL = \frac{JL}{\cos(60^\circ)} \), or \( \sin(60^\circ) = \frac{JK}{KL} \), so \( KL = \frac{JK}{\sin(60^\circ)} \). Also, Pythagoras: \( KL = \sqrt{JK^2 + JL^2} \). Wait, let's check the given sides. Wait, the diagram: \( JL = 3 \), \( JK = 5.2 \)? Wait, no, maybe I misread. Wait, the right angle is at \( J \), so \( J \) is right angle, so \( JL \) and \( JK \) are legs, \( KL \) is hypotenuse. So \( JL = 3 \), \( \angle L = 60^\circ \). So using cosine: \( \cos(60^\circ) = \frac{JL}{KL} \), so \( KL = \frac{JL}{\cos(60^\circ)} \). Since \( \cos(60^\circ) = 0.5 \), so \( KL = \frac{3}{0.5} = 6 \), but also, using Pythagoras: if \( JK \) is the other leg, \( \tan(60^\circ) = \frac{JK}{JL} \), so \( JK = 3\tan(60^\circ) = 3\sqrt{3} \approx 5.196 \approx 5.2 \), which matches the given \( JK = 5.2 \). So then, hypotenuse \( KL \) can be found by Pythagoras: \( KL = \sqrt{JK^2 + JL^2} = \sqrt{(3\sqrt{3})^2 + 3^2} = \sqrt{27 + 9} = \sqrt{36} = 6 \), or by trigonometry: \( KL = \frac{JL}{\cos(60^\circ)} = \frac{3}{0.5} = 6 \), or \( KL = \frac{JK}{\sin(60^\circ)} = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}} = 6 \). So the expressions for \( KL \) could be \( \sqrt{3^2 + (3\sqrt{3})^2} \), \( \frac{3}{\cos(60^\circ)} \), \( \frac{3\sqrt{3}}{\sin(60^\circ)} \), etc. But since the problem is about which expressions represent \( KL \), let's confirm.

Step2: Analyze possible expressions

1. Using Pythagorean theorem: \( KL = \sqrt{JL^2 + JK^2} \). Since \( JL = 3 \) and \( JK = 5.2 \) (or \( 3\sqrt{3} \)), so \( KL = \sqrt{3^2 + 5.2^2} \) (but since \( 5.2 \approx 3\sqrt{3} \), exact value is \( \sqrt{3^2 + (3\sqrt{3})^2} = 6 \)).
2. Using trigonometric ratio: \( \cos(60^\circ) = \frac{JL}{KL} \implies KL = \frac{JL}{\cos(60^\circ)} \). Since \( \cos(60^\circ) = \frac{1}{2} \), this becomes \( KL = \frac{3}{\frac{1}{2}} = 6 \).
3. Using sine: \( \sin(60^\circ) = \frac{JK}{KL} \implies KL = \frac{JK}{\sin(60^\circ)} \). Since \( JK = 3\sqrt{3} \) and \( \sin(60^\circ) = \frac{\sqrt{3}}{2} \), this becomes \( KL = \frac{3\sqrt{3}}{\frac{\sqrt{3}}{2}} = 6 \).

So possible expressions are \( \sqrt{3^2 + 5.2^2} \) (or \( \sqrt{3^2 + (3\sqrt{3})^2} \)), \( \frac{3}{\cos 60^\circ} \), \( \frac{5.2}{\sin 60^\circ} \), etc.

Answer:

Possible expressions include \( \boldsymbol{\sqrt{3^2 + 5.2^2}} \), \( \boldsymbol{\frac{3}{\cos 60^\circ}} \), \( \boldsymbol{\frac{5.2}{\sin 60^\circ}} \) (or exact forms using \( 3\sqrt{3} \) instead of \( 5.2 \)). If we consider the exact value, since \( 5.2 \approx 3\sqrt{3} \), the exact expression is \( \sqrt{3^2 + (3\sqrt{3})^2} = 6 \), or trigonometric expressions as above.