Sagot :
☺ Salut ☺
[tex]\rule{6cm}{1mm}[/tex]
[tex]On\;a\;:[/tex]
[tex]BO = 15\;m[/tex]
[tex]\widehat{AOB} = {24}^{o}[/tex]
[tex]\widehat{BOC} = {53}^{o}[/tex]
[tex]ABO\;et\;BCO\;sont\; deux\; triangles\; \\rectangle\;en\;B.[/tex]
[tex]On\; sait\;que\;:[/tex]
[tex]tan = \dfrac{C.opp}{C.adj}[/tex]
[tex]\rule{6cm}{1mm}[/tex]
[tex] Calculons\;AB\; et\;BC\;[/tex]
[tex]tan\hat{O} = \dfrac{AB}{BO}[/tex]
[tex]tan{24}^{o}= \dfrac{AB}{15 }[/tex]
[tex]AB = 15 \times tan{24}^{o}[/tex]
[tex]\blue{AB = 6,6\;m}[/tex]
[tex]\\[/tex]
[tex]tan\hat{O} = \dfrac{BC}{BO}[/tex]
[tex]tan{53}^{o} = \dfrac{AB}{15 }[/tex]
[tex]BC = 15 \times tan{53}^{o}[/tex]
[tex]\blue{BC = 19,9\;m}[/tex]
[tex]\\[/tex]
[tex]Calculons\;la\; hauteur(h_{M})\;du\;mat\;:[/tex]
[tex]h_{M} = AB + BC[/tex]
[tex]h_{M} = 6,6\;m\;+\;19,9\;m[/tex]
[tex]\boxed{\boxed{\green{h_{M} = 26,5\;m}}}[/tex]
[tex]\rule{6cm}{1mm}[/tex]