Question

find the cosine of ∠j.

write your answer in simplified, rationalized form. do not round.
cos (j) =

Explanation:

Step1: Identify triangle type and sides

This is a right - triangle \( \triangle HIJ \) with \( \angle H = 90^{\circ} \). The adjacent side to \( \angle J \) is \( HJ=\sqrt{35} \), the opposite side is \( HI = \sqrt{23} \), and we need to find the hypotenuse \( IJ \) first using the Pythagorean theorem \( c=\sqrt{a^{2}+b^{2}} \) (where \( a=\sqrt{35} \), \( b = \sqrt{23} \)).

\( IJ=\sqrt{(\sqrt{35})^{2}+(\sqrt{23})^{2}}=\sqrt{35 + 23}=\sqrt{58} \)

Step2: Recall cosine definition

The cosine of an angle in a right - triangle is defined as \( \cos(\theta)=\frac{\text{Adjacent side to } \theta}{\text{Hypotenuse}} \). For \( \angle J \), the adjacent side is \( HJ=\sqrt{35} \) and the hypotenuse is \( IJ=\sqrt{58} \).

So, \( \cos(J)=\frac{\sqrt{35}}{\sqrt{58}} \). To rationalize the denominator, we multiply the numerator and denominator by \( \sqrt{58} \):

\( \cos(J)=\frac{\sqrt{35}\times\sqrt{58}}{\sqrt{58}\times\sqrt{58}}=\frac{\sqrt{35\times58}}{58}=\frac{\sqrt{2030}}{58} \) (But we can also leave it as \( \frac{\sqrt{35}}{\sqrt{58}}=\frac{\sqrt{35\times58}}{58}=\frac{\sqrt{2030}}{58} \), or simplify the fraction of the square - roots before rationalizing. Wait, actually, \( \sqrt{35\times58}=\sqrt{35}\times\sqrt{58} \), and another way:

We know that \( \cos(J)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{HJ}{IJ} \), and \( HJ = \sqrt{35} \), \( IJ=\sqrt{(\sqrt{35})^{2}+(\sqrt{23})^{2}}=\sqrt{35 + 23}=\sqrt{58} \). So \( \cos(J)=\frac{\sqrt{35}}{\sqrt{58}}=\frac{\sqrt{35}\times\sqrt{58}}{58}=\frac{\sqrt{2030}}{58} \), but we can also check if we made a mistake in identifying adjacent and hypotenuse. Wait, in right - triangle \( \triangle HIJ \), \( \angle H = 90^{\circ} \), so the sides: \( HJ \) is one leg, \( HI \) is the other leg, \( IJ \) is the hypotenuse. For angle \( J \), the adjacent side is \( HJ \) (because it forms angle \( J \) with the hypotenuse \( IJ \)), and the opposite side is \( HI \). So by definition of cosine, \( \cos(J)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{HJ}{IJ}=\frac{\sqrt{35}}{\sqrt{58}} \), and after rationalizing, \( \frac{\sqrt{35\times58}}{58}=\frac{\sqrt{2030}}{58} \). But we can also note that \( \sqrt{35\times58}=\sqrt{35}\times\sqrt{58} \), and \( 35\times58 = 35\times(50 + 8)=1750+280 = 2030 \). Alternatively, we can write \( \cos(J)=\frac{\sqrt{35}}{\sqrt{58}}=\frac{\sqrt{35}}{\sqrt{58}}\times\frac{\sqrt{58}}{\sqrt{58}}=\frac{\sqrt{2030}}{58} \), but we can also simplify the fraction before rationalizing. Wait, no, the key is that \( \cos(J)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{\sqrt{35}}{\sqrt{58}} \), and when we rationalize, we get \( \frac{\sqrt{35\times58}}{58}=\frac{\sqrt{2030}}{58} \), but actually, \( 35\times58 = 35\times58 = 2030 \), and we can also check if \( \sqrt{2030} \) can be simplified. The prime factorization of \( 2030=2\times5\times7\times29 \), so it cannot be simplified further. But wait, maybe we made a mistake in the adjacent side. Wait, in \( \triangle HIJ \), vertices are \( H \) (right angle), \( I \), \( J \). So the sides: \( HJ \) is from \( H \) to \( J \), length \( \sqrt{35} \), \( HI \) is from \( H \) to \( I \), length \( \sqrt{23} \), \( IJ \) is from \( I \) to \( J \), hypotenuse. So angle at \( J \): the sides forming angle \( J \) are \( HJ \) and \( IJ \), so adjacent side is \( HJ \), opposite is \( HI \). So cosine of \( J \) is \( \frac{HJ}{IJ}=\frac{\sqrt{35}}{\sqrt{58}}=\frac{\sqrt{35\times58}}{58}=\frac{\sqrt{2030}}{58} \), but we can also write it as \( \frac{\sqrt{35}}{\sqrt{58}}=\frac{\sqrt{35}}{\sqrt{58}}\approx\frac{5.916}{7.616} \), but we need to keep it in simplified, rationalized form. So the correct rationalized form is \( \frac{\sqrt{2030}}{58} \) or we can also note that \( \sqrt{35}\times\sqrt{58}=\sqrt{35\times58}=\sqrt{2030} \), so \( \cos(J)=\frac{\sqrt{2030}}{58} \), but we can also check if we can simplify the fraction. Wait, \( 2030 = 2\times5\times7\times29 \), no square factors, so this is the simplified form. Alternatively, we can write \( \cos(J)=\frac{\sqrt{35}}{\sqrt{58}}=\frac{\sqrt{35}}{\sqrt{58}}\times\frac{\sqrt{58}}{\sqrt{58}}=\frac{\sqrt{2030}}{58} \).

Wait, another way: Let's re - check the Pythagorean theorem. \( (\sqrt{35})^2=35 \), \( (\sqrt{23})^2 = 23 \), so \( 35 + 23=58 \), so hypotenuse \( IJ=\sqrt{58} \). Then \( \cos(J)=\frac{\text{Adjacent}}{\text{Hypotenuse}}=\frac{HJ}{IJ}=\frac{\sqrt{35}}{\sqrt{58}} \), and rationalizing gives \( \frac{\sqrt{35}\times\sqrt{58}}{58}=\frac{\sqrt{2030}}{58} \).

Answer:

\( \boldsymbol{\frac{\sqrt{2030}}{58}} \) (or \( \frac{\sqrt{35}}{\sqrt{58}} \) before rationalizing, but the rationalized form is \( \frac{\sqrt{2030}}{58} \))