Question: Find $m\angle I$. [asy] pair I,J,K; I=(0,0); J=(0,2sqrt(6)); K=(4sqrt(2),0); draw(I—J—K—I);…

Find $m\angle I$.

[asy] pair I,J,K; I=(0,0); J=(0,2sqrt(6)); K=(4sqrt(2),0); draw(I—J—K—I); draw(rightanglemark(K,I,J,18)); label(“$I$“,I,SW); label(“$J$“,J,NW); label(“$K$“,K,SE); label(“$2\sqrt{6}$”,(J+I)/2,W); label(“$4\sqrt{2}$”,(I+K)/2,S); [/asy]

Solution

To find \( m \angle I \) in the right triangle \( \triangle JIK \), we can use trigonometry. First, identify the sides: - \( JI = 2\sqrt{6} \) (opposite) - \( KI = 4\sqrt{2} \) (adjacent) To find \( \angle I \), use the tangent function: \[ \tan \angle I = \frac{\text{opposite}}{\text{adjacent}} = \frac{JI}{KI} \] Substitute the known values: \[ \tan \angle I = \frac{2\sqrt{6}}{4\sqrt{2}} \] Simplify the fraction: \[ \tan \angle I = \frac{\sqrt{6}}{2\sqrt{2}} \] Rationalize the denominator: \[ \tan \angle I = \frac{\sqrt{6}}{2\sqrt{2}} \times \frac{\sqrt{2}}{\sqrt{2}} = \frac{\sqrt{12}}{4} = \frac{2\sqrt{3}}{4} = \frac{\sqrt{3}}{2} \] Therefore: \[ \tan \angle I = \frac{\sqrt{3}}{2} \] Now, find \(\angle I\) using a calculator: \[ \angle I = \tan^{-1} \left( \frac{\sqrt{3}}{2} \right) \] This will give the measure of \(\angle I\).