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relações trigonometricas 9ºC

Authored by Luís Valverde

Mathematics

9th Grade

Used 23+ times

relações trigonometricas 9ºC
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9 questions

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1.

FILL IN THE BLANKS QUESTION

20 sec • 1 pt

Qual a função trigonométrica que me dá a relação entre o cateto adjacente e a hipotenusa?

(a)  

2.

MULTIPLE CHOICE QUESTION

20 sec • 1 pt

Media Image

Qual o valor do  sen αsen\ \alpha  no triângulo da imagem?

 53\frac{5}{3}  

 35\frac{3}{5}  

 34\frac{3}{4}  

 43\frac{4}{3}  

3.

MULTIPLE SELECT QUESTION

5 mins • 1 pt

Seja  α\alpha  a amplitude de um ângulo, e 
 sen α=45sen\ \alpha=\frac{4}{5} .
Assinala as afirmações corretas.

 tg α=43tg\ \alpha=\frac{4}{3}  

 tg α=34tg\ \alpha=\frac{3}{4}  

 cos⁡ α=35\cos\ \alpha=\frac{3}{5}   

 tg2α−cos⁡ α=5345tg^2\alpha-\cos\ \alpha=\frac{53}{45}  

4.

MULTIPLE CHOICE QUESTION

2 mins • 1 pt

 Considera um ângulo β\beta tal que  tg β=34tg\ \beta=\frac{3}{4} .

Qual o valor de  cos⁡ β\cos\ \beta ? 

 12\frac{1}{2}  

 45\frac{4}{5}  

 35\frac{3}{5}  

4

5.

MULTIPLE SELECT QUESTION

30 sec • 1 pt

Considera θ\theta  um ângulo agudo.
Seleciona as afirmações corretas.

 (sen θ−cos⁡ θ)(sen θ+cos⁡ θ)+2 cos⁡2θ=1\left(sen\ \theta-\cos\ \theta\right)\left(sen\ \theta+\cos\ \theta\right)+2\ \cos^2\theta=1  

 (cos⁡ θ+sen θ)2=1\left(\cos\ \theta+sen\ \theta\right)^2=1  

 tg θ×sen θ+cos⁡ θ=1cos⁡ θtg\ \theta\times sen\ \theta+\cos\ \theta=\frac{1}{\cos\ \theta}  

 tg2θ=1cos⁡2θtg^2\theta=\frac{1}{\cos^2\theta}  

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Se o valor do sen δ=74sen\ \delta=\frac{\sqrt{7}}{4} .

Qual o valor de  tg δ×sen δ−cos⁡ δtg\ \delta\times sen\ \delta-\cos\ \delta ? 

 7−34\frac{\sqrt{7}-3}{4}  

 3−74\frac{3-\sqrt{7}}{4}  

 −16-\frac{1}{6}  

 16\frac{1}{6}  

7.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Qual a expressão que me permite obter a tangente dado o coseno de um ângulo agudo  α\alpha ?

 tg α=cos⁡2 α+sen2 αtg\ \alpha=\sqrt{\cos^2\ \alpha+sen^2\ \alpha}  

 tg α=1−cos⁡2 αtg\ \alpha=\sqrt{1-\cos^2\ \alpha}  

 tg α=1−cos⁡2αcos⁡αtg\ \alpha=\frac{\sqrt{1-\cos^2\alpha}}{\cos\alpha}  

 tg α=1−cos⁡2αcos⁡2 αtg\ \alpha=\frac{\sqrt{1-\cos^2\alpha}}{\cos^2\ \alpha}  

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