Question

Dado o triângulo ABC, é traçada a altura AH. a partir das informações da figura, determine:

Answer 1

a)

${6}^{2}={5}^{2}+{7}^{2}-2.5.7.\cos(\alpha)\\36=25+49-70\cos(\alpha)\\70\cos(\alpha)=74-36$

$70\cos(\alpha)=38\\\cos(\alpha)=\dfrac{38}{70}$

$\boxed{\boxed{\mathsf{\cos(\alpha)=\dfrac{19}{35}}}}$

b)O triângulo ABH é retângulo daí

$\overline{BH}=5\cos(\alpha)\\\overline{BH}=5.\dfrac{19}{35}$

$\boxed{\boxed{\mathsf{\overline{BH}=\dfrac{19}{7}}}}$

c) usando o teorema de Pitágoras no triângulo ABH temos

${5}^{2}={(\dfrac{19}{7})}^{2}+{\overline{AH}}^{2}\\{\overline{AH}}^{2}=25-\dfrac{361}{49}$

${\overline{Ah}}^{2}=\dfrac{1225-361}{49}\\{\overline{AH}}^{2}=\dfrac{864}{49}$

$\overline{AH}=\sqrt{\dfrac{864}{49}}$

$\boxed{\boxed{\mathsf{\overline{AH}=\dfrac{12\sqrt{6}}{7}}}}$