Part A:
From the given figure, IJ represents the hypothenuse of the right triangle IJL.
By the Pythagoras theorem,
[tex]IJ^2=24^2+32^2 \\ \\ =576+1024=1600 \\ \\ \Rightarrow IJ= \sqrt{1600} =40[/tex]
Part B:
From the given figure, angle J is obtained as follows:
[tex]\tan I= \frac{32}{24} = \frac{4}{3} \\ \\ \Rightarrow I^o=\tan^{-1}\left( \frac{4}{3} \right)[/tex]
Line JM can be obtained as follows:
[tex]\tan I= \frac{JM}{40} \\ \\ \frac{4}{3}= \frac{JM}{40} \\ \\ \Rightarrow JM= \frac{40\times4}{3} = \frac{160}{3} [/tex]
Part C:
From the given triangle, LM is one of the legs of the right triangle JLM with the other leg, JL = 32 and the hypothenuse, JM = 160/3.
By the Pythagoras theorem,
[tex]LM^2=JM^2-JL^2 \\ \\ =\left( \frac{160}{3} \right)^2-32^2= \frac{25,600}{9} -1,024 \\ \\ = \frac{16,384}{9} \\ \\ \Rightarrow LM= \sqrt{\frac{16,384}{9}} = \frac{128}{3} [/tex]
Part D:
From the figure, IM = IL + LM
[tex]24+ \frac{128}{3} = \frac{200}{3} [/tex]
