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Triangular Trigonometry

Authored by Jennifer Laskowitz

9th - 12th Grade

10 Questions

Used 4+ times

Triangular Trigonometry
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1.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Given that the sinθ=45,  what is the secθ?\sin\theta=\frac{4}{5},\ \ what\ is\ the\ \sec\theta?  

 secθ=54\sec\theta=\frac{5}{4}  

 secθ=34\sec\theta=\frac{3}{4}  

 secθ=53\sec\theta=\frac{5}{3}  

 secθ=35\sec\theta=\frac{3}{5}  

2.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Given that the  cosθ=38, what is the tanθ?\cos\theta=\frac{3}{8},\ what\ is\ the\ \tan\theta?  

 tanθ=553\tan\theta=\frac{\sqrt{55}}{3}  

 tanθ=558\tan\theta=\frac{\sqrt{55}}{8}  

 tanθ=83\tan\theta=\frac{8}{3}  

 tanθ=35555\tan\theta=\frac{3\sqrt{55}}{55}  

3.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Given that the

 secθ=109, what is the cotθ?\sec\theta=\frac{10}{9},\ what\ is\ the\ \cot\theta?  

 cotθ=91919\cot\theta=\frac{9\sqrt{19}}{19}  

 cotθ=919\cot\theta=9\sqrt{19}  

 cotθ=199\cot\theta=\frac{\sqrt{19}}{9}  

 cotθ=910\cot\theta=\frac{9}{10}  

4.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Find the  sinθ\sin\theta given the point (-1,5) which is located on the terminal side of an angle  θ\theta  .

 sinθ=52626\sin\theta=\frac{5\sqrt{26}}{26}  

 sinθ=52626\sin\theta=-\frac{5\sqrt{26}}{26}  

 sinθ=265\sin\theta=\frac{\sqrt{26}}{5}  

 sinθ=26\sin\theta=-\sqrt{26}  

5.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Find the  secθ\secθ   given the point (4,-3) which is located on the terminal side of an angle  θ\theta  .


secθ=45\sec\theta=\frac{4}{5}

secθ=45\sec\theta=-\frac{4}{5}

secθ=35\sec\theta=\frac{3}{5}

secθ=53\sec\theta=-\frac{5}{3}

secθ=54\sec\theta=\frac{5}{4}

6.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Find the  tanθ\tanθ   given the point (-5,8) which is located on the terminal side of an angle  θ\theta  .

 tanθ=85\tan\theta=\frac{8}{5}  

 tanθ=85\tan\theta=-\frac{8}{5}  

 tanθ=895\tan\theta=\frac{\sqrt{89}}{5}  

 tanθ=58989\tan\theta=-\frac{5\sqrt{89}}{89}  

 tanθ=898\tan\theta=-\frac{\sqrt{89}}{8}  

7.

MULTIPLE CHOICE QUESTION

5 mins • 1 pt

Find the  cosθ\cosθ   given the point (-6,-4) which is located on the terminal side of an angle  θ\theta  .

 cosθ=31313\cos\theta=\frac{3\sqrt{13}}{13}  

 cosθ=31313\cos\theta=-\frac{3\sqrt{13}}{13}  

 cosθ=133\cos\theta=\frac{\sqrt{13}}{3} 

 cosθ=133\cos\theta=-\frac{\sqrt{13}}{3}  

 cosθ=2132\cos\theta=-\frac{2\sqrt{13}}{2}  

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