Question

question
find the exact value of tan j in simplest radical form.

Explanation:

Step1: Recall the definition of tangent in a right triangle

In a right triangle, the tangent of an acute angle is the ratio of the length of the opposite side to the length of the adjacent side. For angle \( J \) in right triangle \( KJL \) (right - angled at \( K \)), \( \tan J=\frac{\text{opposite side to }J}{\text{adjacent side to }J} \).

The side opposite to angle \( J \) is \( KL = 3 \), and the side adjacent to angle \( J \) is \( KJ=\sqrt{38} \)? Wait, no, wait. Wait, in right triangle \( \triangle KJL \) with right angle at \( K \), angle at \( J \): the sides: \( KJ \) and \( KL \) are the legs, \( JL \) is the hypotenuse. Wait, let's re - identify the sides.

Wait, in right - angled triangle \( \triangle KJL \), right - angled at \( K \). So, for angle \( J \):
- The opposite side to angle \( J \) is \( KL \) (because when we look at angle \( J \), the side that does not form angle \( J \) and is opposite to it is \( KL \)).
- The adjacent side to angle \( J \) is \( KJ \) (the side that forms angle \( J \) along with the hypotenuse).

Wait, but let's confirm the lengths. We know that in a right - triangle, by Pythagoras theorem, \( (KJ)^2+(KL)^2=(JL)^2 \). Let's check: \( (\sqrt{38})^2 + 3^2=38 + 9 = 47=(\sqrt{47})^2 \), which is correct. So the triangle is right - angled at \( K \).

So, \( \tan J=\frac{\text{opposite}}{\text{adjacent}}=\frac{KL}{KJ} \)

Step2: Substitute the values of the sides

We know that \( KL = 3 \) and \( KJ=\sqrt{38} \). But we need to rationalize the denominator? Wait, no, wait, maybe I made a mistake in identifying the sides. Wait, maybe angle \( J \)'s opposite and adjacent sides are different. Wait, let's label the triangle properly. The right angle is at \( K \), so vertices are \( K \) (right angle), \( J \), and \( L \). So side \( KJ \): from \( K \) to \( J \), length \( \sqrt{38} \); side \( KL \): from \( K \) to \( L \), length \( 3 \); side \( JL \): from \( J \) to \( L \), length \( \sqrt{47} \).

For angle \( J \), the sides:
- Opposite side: the side that is not part of angle \( J \) and is opposite to it. Angle \( J \) is formed by sides \( KJ \) and \( JL \). So the side opposite to \( J \) is \( KL = 3 \).
- Adjacent side: the side that is part of angle \( J \) and is not the hypotenuse. So the adjacent side is \( KJ=\sqrt{38} \)? Wait, no, that can't be. Wait, maybe I mixed up. Wait, in a right - triangle, for angle \( \theta \), \( \tan\theta=\frac{\text{opposite leg}}{\text{adjacent leg}} \).

Wait, let's take angle \( J \). The legs are \( KJ \) and \( KL \), hypotenuse is \( JL \). So angle at \( J \): the two legs are \( KJ \) (adjacent) and \( KL \) (opposite)? Wait, no, when you are at angle \( J \), the sides: one leg is \( J K \) (from \( J \) to \( K \)) and the other leg is \( J L \)? No, no, the triangle has vertices \( K \), \( J \), \( L \) with right angle at \( K \). So the sides are:
- \( KJ \): between \( K \) and \( J \)
- \( KL \): between \( K \) and \( L \)
- \( JL \): between \( J \) and \( L \)

So angle at \( J \): the sides forming angle \( J \) are \( J K \) and \( J L \). The side opposite to angle \( J \) is \( K L \), and the side adjacent to angle \( J \) is \( J K \).

So \( \tan J=\frac{\text{opposite}}{\text{adjacent}}=\frac{KL}{KJ}=\frac{3}{\sqrt{38}} \). But we need to rationalize the denominator. Multiply numerator and denominator by \( \sqrt{38} \):

\( \frac{3\times\sqrt{38}}{\sqrt{38}\times\sqrt{38}}=\frac{3\sqrt{38}}{38} \)

Wait, but that seems odd. Wait, maybe I identified the sides wrong. Wait, let's check the Pythagorean theorem again. \( KJ=\sqrt{38} \), \( KL = 3 \), \( JL=\sqrt{47} \). \( (\sqrt{38})^2+3^2 = 38 + 9=47=(\sqrt{47})^2 \), so that's correct.

Wait, maybe angle \( J \)'s adjacent and opposite sides are different. Wait, maybe I should consider angle \( J \) with respect to the other leg. Wait, no, let's re - express the tangent formula. In a right triangle, for angle \( A \), \( \tan A=\frac{\text{length of side opposite }A}{\text{length of side adjacent }A} \).

So if we have right triangle \( \triangle KJL \), right - angled at \( K \):

- For angle \( J \):
- Opposite side: \( KL \) (length 3)
- Adjacent side: \( KJ \) (length \( \sqrt{38} \))

So \( \tan J=\frac{KL}{KJ}=\frac{3}{\sqrt{38}} \). Rationalizing the denominator:

Multiply numerator and denominator by \( \sqrt{38} \):

\( \tan J=\frac{3\sqrt{38}}{38} \)

Wait, but let's check again. Maybe I made a mistake in the side labels. Let's assume that the right angle is at \( K \), so \( \angle K = 90^{\circ} \). Then, the sides:

- \( KJ \): one leg
- \( KL \): another leg
- \( JL \): hypotenuse

So angle at \( J \): the sides:

- Opposite to \( J \): \( KL \)
- Adjacent to \( J \): \( KJ \)

So the calculation is correct.

Answer:

\( \frac{3\sqrt{38}}{38} \)