Question

usando a lei dos cosseno qual é o cos A no triângulo dado​

Answer 1

Resposta:

a²=b²+c²-2*b*c * cos(β)     .....β ângulo entre b  e c

4²=3²+3²- 2*3*3 * cos(α)

16=9+9-18* cos(α)

cos(α) = (16-9-9)/(-18)

cos(α)  = -2/(-18)

cos(α)  =1/9

Answer 2

Resposta:

$\cos{\alpha} = \dfrac{2}{18}$

Explicação passo a passo:

$\overline{AB} = 3\\\overline{AC} = 3\\\overline{BC} = 4\\\overline{BC} = \sqrt{\overline{AB}^2 + \overline{AC}^2 - (2 \times \overline{AB}\times \overline{AC}) \cos{\alpha}}\\4 = \sqrt{3^2 + 3^2 - (2\times 3 \times 3) \cos{\alpha}}\\4 = \sqrt{18 - 18\cos{\alpha}}\\16 = 18 - 18\cos{\alpha}\\18\cos{\alpha} = 2\\\cos{\alpha} = \dfrac{2}{18}\\\alpha = \arccos{\dfrac{2}{18}}\\\alpha = 83,62^{\circ}$