Question

Qual o valor aproximado do sen a?

Answer 1

Área de uma triângulo

$\displaystyle \sf S = \sqrt{p\cdot (p-a)\cdot (p-b)\cdot (p-c)} \\\\ S = \frac{a\cdot b \cdot sen(\alpha) }{2} \\\\\ onde : \\\\\ p = \text{semi per{\'i}metro } \\\\ a,b \ e \ c = lados \ do \ triangulo \\\\ \alpha = angulo \ entre\ os \ lados \ adjacentes$

1º vamos achar o semiperímtro :

$\displaystyle \sf p = \frac{4+5+6}{2} \to p = \frac{15}{2} \\\\\\ (p-a) = \frac{15}{2}-4 \to (p-a) = \frac{7}{2} \\\\\\ (p-b) =\frac{15}{2}-5 \to (p-b)= \frac{5}{2} \\\\\\ (p-c) = \frac{15}{2}-6 \to (p-c) = \frac{3}{2} \\\\ Da{\'i} \ a \ {\'A}rea } \ do \ triangulo\ vale : \\\\ S = \sqrt{\frac{15}{2}\cdot \frac{7}{2}\cdot \frac{5}{2}\cdot \frac{3}{2}} \\\\\\ S = \sqrt{\frac{15^2}{2^4} \cdot 7} \to S = \frac{15}{2^2}\cdot \sqrt{7} \to S =\frac{15}{4}\sqrt{7}$

Porém podemos calcular a área da ouitra forma :

$\displaystyle \sf S = \frac{4\cdot 5\cdot sen(\alpha)}{2} =10\cdot sen(\alpha) \\\\ Da{\'i} \ \text{igualando as areas, temos }}: \\\\ 10\cdot sen(\alpha) =\frac{15}{4} \cdot \sqrt{7} \\\\\\ sen(\alpha) = \frac{15}{4\cdot 10}\sqrt{7} \\\\\ sen(\alpha) =\frac{3}{8}\cdot \sqrt{7}=0,375\cdot \sqrt{7} \\\\ \sqrt{7} \approx 2,645 \\\\\ sen(\alpha) \approx 0,375\cdot 2,645 \\\\\ \huge\boxed{\sf sen(\alpha) \approx 0,99 }\checkmark$

letra a