Question

solve the right triangle. write your answers in simplified, rationalized form. do not round. ij = hi = m∠i = °

Explanation:

Step1: Find \(m\angle I\)

In a right - triangle, the sum of the two non - right angles is \(90^{\circ}\).
Since \(\angle H = 90^{\circ}\) and \(\angle J=45^{\circ}\), then \(m\angle I=90^{\circ}-m\angle J\).
\(m\angle I = 90^{\circ}-45^{\circ}=45^{\circ}\)

Step2: Find \(IJ\) (hypotenuse)

We know that \(\cos J=\frac{HJ}{IJ}\), and \(J = 45^{\circ}\), \(HJ = 8\sqrt{30}\).
Since \(\cos45^{\circ}=\frac{\sqrt{2}}{2}\), then \(IJ=\frac{HJ}{\cos45^{\circ}}\).
Substitute \(HJ = 8\sqrt{30}\) into the formula: \(IJ=\frac{8\sqrt{30}}{\frac{\sqrt{2}}{2}}=8\sqrt{30}\times\frac{2}{\sqrt{2}}=8\sqrt{15}\times\sqrt{2}\times\frac{2}{\sqrt{2}} = 16\sqrt{15}\)

Step3: Find \(HI\)

Since \(\tan J=\frac{HI}{HJ}\), and \(\tan45^{\circ}=1\), \(HJ = 8\sqrt{30}\)
\(HI=HJ\times\tan45^{\circ}\)
\(HI = 8\sqrt{30}\times1=8\sqrt{30}\)

Answer:

\(IJ = 16\sqrt{15}\)
\(HI=8\sqrt{30}\)
\(m\angle I = 45^{\circ}\)