Question

find the tangent of ∠i.
h i
2√43
√129
j (right angle at j)
write your answer in simplified, rationalized form. do not round.
tan(i) =
(buttons for fraction and square root)
submit

Explanation:

Step1: Identify the right triangle sides

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

• Opposite side to \( \angle I \) is \( HJ=\sqrt{129} \)
• Adjacent side to \( \angle I \) is \( IJ \). First, find \( IJ \) using Pythagorean theorem: \( HI^2 = HJ^2+IJ^2 \). Given \( HI = 2\sqrt{43} \), \( HJ=\sqrt{129} \). So, \( (2\sqrt{43})^2=(\sqrt{129})^2 + IJ^2 \).
• Calculate \( (2\sqrt{43})^2 = 4\times43 = 172 \), \( (\sqrt{129})^2 = 129 \). Then \( 172=129 + IJ^2 \), so \( IJ^2=172 - 129 = 43 \), thus \( IJ=\sqrt{43} \).

Step2: Recall tangent definition

Tangent of an angle in right triangle is \( \tan(\theta)=\frac{\text{opposite}}{\text{adjacent}} \). For \( \angle I \), \( \tan(I)=\frac{HJ}{IJ} \).

Step3: Substitute values

Substitute \( HJ = \sqrt{129} \) and \( IJ=\sqrt{43} \): \( \tan(I)=\frac{\sqrt{129}}{\sqrt{43}} \). Rationalize the denominator by multiplying numerator and denominator by \( \sqrt{43} \): \( \frac{\sqrt{129}\times\sqrt{43}}{\sqrt{43}\times\sqrt{43}}=\frac{\sqrt{129\times43}}{43} \). But \( 129 = 3\times43 \), so \( \sqrt{129\times43}=\sqrt{3\times43\times43}=43\sqrt{3} \). Thus, \( \frac{43\sqrt{3}}{43}=\sqrt{3} \). Wait, wait, no: Wait, \( 129 = 3\times43 \), so \( \sqrt{129}=\sqrt{3\times43}=\sqrt{3}\times\sqrt{43} \). So \( \frac{\sqrt{129}}{\sqrt{43}}=\frac{\sqrt{3}\times\sqrt{43}}{\sqrt{43}}=\sqrt{3} \)? Wait, no, wait my mistake earlier: Wait, \( HJ=\sqrt{129} \), \( IJ=\sqrt{43} \). Wait, \( \tan(I)=\frac{\text{opposite}}{\text{adjacent}}=\frac{HJ}{IJ}=\frac{\sqrt{129}}{\sqrt{43}} \). Simplify: \( \frac{\sqrt{129}}{\sqrt{43}}=\sqrt{\frac{129}{43}}=\sqrt{3} \) (since \( 129\div43 = 3 \)). Wait, that's simpler. Because \( 129 = 3\times43 \), so \( \frac{129}{43}=3 \), so \( \sqrt{\frac{129}{43}}=\sqrt{3} \). Wait, no, wait: Wait, \( \frac{\sqrt{129}}{\sqrt{43}}=\sqrt{\frac{129}{43}}=\sqrt{3} \) (because \( 129\div43 = 3 \)). Alternatively, \( \frac{\sqrt{129}}{\sqrt{43}}=\frac{\sqrt{3\times43}}{\sqrt{43}}=\sqrt{3} \) (canceling \( \sqrt{43} \) from numerator and denominator).

Wait, let's recheck Pythagorean step: \( HI = 2\sqrt{43} \), so \( HI^2 = 4\times43 = 172 \). \( HJ^2 = 129 \). Then \( IJ^2 = 172 - 129 = 43 \), so \( IJ = \sqrt{43} \). Then opposite side to \( \angle I \) is \( HJ=\sqrt{129} \), adjacent is \( IJ=\sqrt{43} \). So \( \tan(I)=\frac{\sqrt{129}}{\sqrt{43}} \). Simplify: \( \frac{\sqrt{129}}{\sqrt{43}}=\sqrt{\frac{129}{43}}=\sqrt{3} \) (since \( 129 = 3\times43 \), so \( \frac{129}{43}=3 \)). So \( \tan(I)=\sqrt{3} \)? Wait, no, wait: Wait, \( \sqrt{129} = \sqrt{3\times43} \), \( \sqrt{43} \) is in denominator. So \( \frac{\sqrt{3\times43}}{\sqrt{43}}=\sqrt{3} \). Yes, because \( \frac{\sqrt{a\times b}}{\sqrt{b}}=\sqrt{a} \) when \( b>0 \).

Answer:

\( \sqrt{3} \)