Question

b)determine a medida aproximada de cada ângulo
2)Usando as relações trigonométricas,calcule a medida aproximada dos ângulos indicados
3)usando a calculadora determine a medida do lado indicado por x

Answer 1

Explicação passo-a-passo:

b)

a)

$\sf cos~\hat{A}=0,857$

$\sf \hat{A}=arccos~0,857$

$\sf \red{\hat{A}=31^{\circ}}$

b)

$\sf sen~\hat{B}=0,453$

$\sf \hat{B}=arcsen~0,453$

$\sf \red{\hat{B}=27^{\circ}}$

c)

$\sf tg~\hat{C}=1,145$

$\sf \hat{C}=arctg~1,145$

$\sf \red{\hat{C}=49^{\circ}}$

d)

$\sf cos~\hat{D}=0,975$

$\sf \hat{D}=arccos~0,975$

$\sf \red{\hat{D}=12,84^{\circ}}$

e)

$\sf tg~\hat{E}=1,62$

$\sf \hat{E}=arctg~1,62$

$\sf \red{\hat{E}=58,31^{\circ}}$

2)

a)

$\sf sen~x=\dfrac{cateto~oposto}{hipotenusa}$

$\sf sen~x=\dfrac{2}{5}$

$\sf sen~x=0,4$

$\sf x=arcsen~0,4$

$\sf \red{x=23,58^{\circ}}$

b)

$\sf cos~x=\dfrac{cateto~adjacente}{hipotenusa}$

$\sf cos~x=\dfrac{3}{4}$

$\sf cos~x=0,75$

$\sf x=arccos~0,75$

$\sf \red{x=41,41^{\circ}}$

c)

$\sf cos~x=\dfrac{cateto~adjacente}{hipotenusa}$

$\sf cos~x=\dfrac{3,5}{4}$

$\sf cos~x=0,875$

$\sf x=arccos~0,875$

$\sf \red{x=29^{\circ}}$

d)

$\sf tg~x=\dfrac{cateto~oposto}{cateto~adjacente}$

$\sf tg~x=\dfrac{4,2}{1,7}$

$\sf tg~x=2,47$

$\sf x=arctg~2,47$

$\sf \red{x=68^{\circ}}$

3)

a)

$\sf tg~30^{\circ}=\dfrac{cateto~oposto}{cateto~adjacente}$

$\sf \dfrac{\sqrt{3}}{3}=\dfrac{2\sqrt{3}}{x}$

$\sf x\sqrt{3}=3\cdot2\sqrt{3}$

$\sf x\sqrt{3}=6\sqrt{3}$

$\sf x=\dfrac{6\sqrt{3}}{\sqrt{3}}$

$\sf \red{x=6~cm}$

b)

$\sf cos~60^{\circ}=\dfrac{cateto~adjacente}{hipotenusa}$

$\sf \dfrac{1}{2}=\dfrac{x}{8,6}$

$\sf 2x=8,6\cdot1$

$\sf 2x=8,6$

$\sf x=\dfrac{8,6}{2}$

$\sf \red{x=4,3~cm}$

c)

$\sf sen~45^{\circ}=\dfrac{cateto~oposto}{cateto~adjacente}$

$\sf \dfrac{\sqrt{2}}{2}=\dfrac{x}{5\sqrt{2}}$

$\sf 2x=5\sqrt{2}\cdot\sqrt{2}$

$\sf 2x=5\cdot2$

$\sf 2x=10$

$\sf x=\dfrac{10}{2}$

$\sf \red{x=5~cm}$