Question

find the sine, cosine, and tangent of $\angle j$.

(image of a right triangle $ijk$ with right angle at $i$, $ij = 2$, hypotenuse $jk = 4$)

write your answer in simplified, rationalized form. do not round.
$\sin(j) = \square$
$\cos(j) = \square$
$\tan(j) = \square$

Explanation:

Step1: Identify sides relative to ∠J

In right triangle \( \triangle IJK \) with right angle at \( I \), for \( \angle J \):

• Opposite side (to \( \angle J \)): \( IK \)
• Adjacent side (to \( \angle J \)): \( IJ = 2 \)
• Hypotenuse: \( JK = 4 \)

First, find \( IK \) using Pythagorean theorem: \( IK = \sqrt{JK^2 - IJ^2} = \sqrt{4^2 - 2^2} = \sqrt{16 - 4} = \sqrt{12} = 2\sqrt{3} \)

Step2: Calculate \( \sin(J) \)

\( \sin(\theta) = \frac{\text{opposite}}{\text{hypotenuse}} \), so \( \sin(J) = \frac{IK}{JK} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \)

Step3: Calculate \( \cos(J) \)

\( \cos(\theta) = \frac{\text{adjacent}}{\text{hypotenuse}} \), so \( \cos(J) = \frac{IJ}{JK} = \frac{2}{4} = \frac{1}{2} \)

Step4: Calculate \( \tan(J) \)

\( \tan(\theta) = \frac{\text{opposite}}{\text{adjacent}} \), so \( \tan(J) = \frac{IK}{IJ} = \frac{2\sqrt{3}}{2} = \sqrt{3} \)

Answer:

\( \sin(J) = \frac{\sqrt{3}}{2} \)
\( \cos(J) = \frac{1}{2} \)
\( \tan(J) = \sqrt{3} \)