Pré-Vestibular ⇒ Trigonometria no Triangulo Retângulo Tópico resolvido

[gab1234]

Em um triângulo retângulo ABC, reto em A, tem-se que tg B + tg C =25/12
O valor de sen B + sen C é:
a) 25/12
b) 12/25
c) 7/5
d) 5/7

Resposta

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c

[petras]

gab1234,
Sendo a = hipotenusa
b cateto oposto a B e c cateto oposto a C?

[tex3]\mathsf{tgB + tgC = \frac{25}{12}\rightarrow \frac{b}{c}+\frac{c}{b} = \frac{25}{12}→ \\
\frac{\underbrace{b^2 + c^2}}{b.c} = \frac{25}{12} →\frac{a^2}{b.c} = \frac{25}{12}\rightarrow \boxed{\frac{bc}{a^2}=\frac{12}{25}}\\
\therefore senB + senC = \frac{b}{a} + \frac{c}{a}→(senB + senC)^2 = (\frac{b}{a} + \frac{c}{a})^2 =\\
(\frac{b}{a})^2 + (\frac{c}{a})^2 + \underbrace{2. \frac{b}{a}. \frac{c}{a}} = \frac{\cancel{b^2+c^2}}{\cancel{a^2}} + 2. \frac{12}{25} =1+\frac{24}{25}= \frac{49}{25}\\
\therefore \boxed{\color{red}senB + senC = \frac{7}{5}}}[/tex3]

Código: Selecionar tudoSHIFT+Click na eq = ZOOM

[tex3]\mathsf{tgB + tgC = \frac{25}{12}\rightarrow \frac{b}{c}+\frac{c}{b} = \frac{25}{12}→ \\
\frac{\underbrace{b^2 + c^2}}{b.c} = \frac{25}{12} →\frac{a^2}{b.c} = \frac{25}{12}\rightarrow \boxed{\frac{bc}{a^2}=\frac{12}{25}}\\
\therefore senB + senC = \frac{b}{a} + \frac{c}{a}→(senB + senC)^2 = (\frac{b}{a} + \frac{c}{a})^2 =\\
(\frac{b}{a})^2 + (\frac{c}{a})^2 + \underbrace{2. \frac{b}{a}. \frac{c}{a}} = \frac{\cancel{b^2+c^2}}{\cancel{a^2}} + 2. \frac{12}{25} =1+\frac{24}{25}= \frac{49}{25}\\
\therefore \boxed{\color{red}senB + senC = \frac{7}{5}}}[/tex3]