a)4√2
_________________________________
ca: 4
ca: 4co: ?
ca: 4co: ?h: ?
[tex] \tan \alpha =\frac{co}{ca} \\ despejar \: el \: co: \\ \tan \alpha \times ca = co \\ \tan(45) \times 4 = co \\ 1 \times 4 = co \\ 4 = co[/tex]
Una vez tenido todos los catetos (lados), hallar la hipotenusa
[tex]h = \sqrt{ {ca}^{2} + {co}^{2} } \\ h = \sqrt{ {4}^{2} + {4}^{2} } \\ h = \sqrt{16 + 16 } \\ h = \sqrt{32 } \\ h =5.6568 \: cm[/tex]
5.6568 = 4√2
5.6568 = 4 × 1.4142
5.6568 = 5.6568
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Coronita porfa xd
¿Cuál es la medida de la hipotenusa?
a)4√2
_________________________________
ca: 4
ca: 4co: ?
ca: 4co: ?h: ?
[tex] \tan \alpha =\frac{co}{ca} \\ despejar \: el \: co: \\ \tan \alpha \times ca = co \\ \tan(45) \times 4 = co \\ 1 \times 4 = co \\ 4 = co[/tex]
Una vez tenido todos los catetos (lados), hallar la hipotenusa
[tex]h = \sqrt{ {ca}^{2} + {co}^{2} } \\ h = \sqrt{ {4}^{2} + {4}^{2} } \\ h = \sqrt{16 + 16 } \\ h = \sqrt{32 } \\ h =5.6568 \: cm[/tex]
5.6568 = 4√2
5.6568 = 4 × 1.4142
5.6568 = 5.6568