Réponse :
On a
[tex]sin60 = \frac{opp}{hyp} \\opp = sin60 \times hyp\\opp = \frac{1}{2} \times 15\\opp = 7,5 m[/tex]
et on a h = opp + la mesure de Paul
h = 7,5m + 1,62 m
h = 9,12 m
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Réponse :
On a
[tex]sin60 = \frac{opp}{hyp} \\opp = sin60 \times hyp\\opp = \frac{1}{2} \times 15\\opp = 7,5 m[/tex]
et on a h = opp + la mesure de Paul
h = 7,5m + 1,62 m
h = 9,12 m