Question

find the tangent of $\angle i$.

triangle with right angle at k, side ik labeled $\sqrt{41}$, side ij labeled $\sqrt{57}$

write your answer in simplified, rationalized form. do not round.

$\tan(i) = \boxed{\dfrac{\square}{\square}}$ (with square root button option)

Explanation:

Step1: Recall tangent in right triangle

In a right triangle, \(\tan(\theta)=\frac{\text{opposite}}{\text{adjacent}}\) for an acute angle \(\theta\). For \(\angle I\), we need to find the lengths of the opposite and adjacent sides.

Step2: Find length of \(JK\) (opposite to \(\angle I\))

Using Pythagorean theorem in \(\triangle IJK\) (right - angled at \(K\)): \(IJ^{2}=IK^{2}+JK^{2}\). We know \(IJ = \sqrt{57}\), \(IK=\sqrt{41}\). Let \(JK = x\). Then \((\sqrt{57})^{2}=(\sqrt{41})^{2}+x^{2}\). So \(57 = 41+x^{2}\), \(x^{2}=57 - 41=16\), \(x = 4\) (since length is positive).

Step3: Identify opposite and adjacent for \(\angle I\)

For \(\angle I\), the opposite side is \(JK = 4\) and the adjacent side is \(IK=\sqrt{41}\).

Step4: Calculate \(\tan(I)\)

\(\tan(I)=\frac{\text{opposite}}{\text{adjacent}}=\frac{JK}{IK}=\frac{4}{\sqrt{41}}\). Rationalize the denominator: \(\frac{4}{\sqrt{41}}\times\frac{\sqrt{41}}{\sqrt{41}}=\frac{4\sqrt{41}}{41}\)

Answer:

\(\frac{4\sqrt{41}}{41}\)