We have to find the length of IH.
We can draw this as:
Both segments EF and EI are radius of the circle.
Also, EH is the projection of EF into the horizontal axis, and can be written as:
[tex]EH=EF\cdot\cos (\alpha)[/tex]
We can also relate FH, that has a length that is half of FG, as the projection of EF in the vertical axis. This can be written as:
[tex]\begin{gathered} FH=\frac{FG}{2}=EF\cdot\sin (\alpha) \\ \frac{6.5}{2}=4.1\cdot\sin (\alpha) \\ \sin (\alpha)=\frac{6.5}{2}\cdot\frac{1}{4.1}\approx0.79 \\ \alpha=\arcsin (0.79) \\ \alpha\approx52.44\degree \end{gathered}[/tex]
We can use this result to solve for IH as:
[tex]\begin{gathered} IH=EI-EH \\ EI=EF \\ EH=EF\cdot\cos (\alpha) \\ \Rightarrow IH=EF-EF\cdot\cos (\alpha) \\ IH=EF(1-\cos (\alpha)) \\ IH=4.1\cdot(1-\cos (52.44\degree)) \\ IH\approx4.1(1-0.61) \\ IH\approx4.1\cdot0.39 \\ IH\approx1.6 \end{gathered}[/tex]
Answer: IH = 1.6 meters
