Réponse:
[tex] \tan(16) = \frac{ lo}{pl} = \frac{lo}{2} \: donc \: lo = 2 \times \tan(16) = 0.57km[/tex]
[tex] {ot}^{2} = {lp}^{2} + {lo \:}^{2} \: donc \: po = \sqrt{ {2}^{2} + {0.57}^{2} } =2.08km[/tex]
[tex] \tan(70) = \frac{ot}{po} \: \: donc \: \: ot = po \times \tan(70) = 2.08 \times 2.74 = 5.72 \\ ot = 5.72km[/tex]
[tex]a) \: \: \: \: ot = lt - lo[/tex]
