Question

If sinθ =2/3 , which of the following are possible?

A.secθ=-3/2 and tanθ=2/√5
B.cosθ=√5/3 and tanθ= 2/√5
C.secθ= 3/√5 and tanθ= 2/√5
D.cosθ=-√5/3 and tanθ=2/√5

Answer 1

2,3 sqrt 5 triangle.....

Answer 2

C.secθ= 3/√5 and tanθ= 2/√5

Answer 3

and B.cosθ=√5/3 and tanθ= 2/√5

Answer 4

thanks

Answer 5

\[\sin(\theta)=\frac{2}{3}\]



\[\sin^2(\theta)=(\frac{2}{3})^2\]



\[\sin^2(\theta)=\frac{4}{9}\]


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\[\sin^2(\theta)+\cos^2(\theta)=1\]



\[\frac{4}{9}+\cos^2(\theta)=1\]



\[\cos^2(\theta)=1-\frac{4}{9}\]



\[\cos^2(\theta)=\frac{5}{9}\]



\[\cos(\theta)=\sqrt{\frac{5}{9}}\] or \[\cos(\theta)=-\sqrt{\frac{5}{9}}\]



\[\cos(\theta)=\frac{\sqrt{5}}{3}\] or \[\cos(\theta)=-\frac{\sqrt{5}}{3}\]



\[\sec(\theta)=\frac{3}{\sqrt{5}}\] or \[\sec(\theta)=-\frac{3}{\sqrt{5}}\]



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\[\tan(\theta)=\frac{\sin(\theta)}{\cos(\theta)}\]



\[\tan(\theta)=\frac{\frac{2}{3}}{\frac{\sqrt{5}}{3}}\]



\[\tan(\theta)=\frac{2}{\sqrt{5}}\]



So the answer is choice C

Answer 6

tsk, tsk, gotta watch those pesky quadrants....