A Point is needed to define a circle
Hope this helps!!!
![\large\begin{array}{l} \textsf{From the picture, we get}\\\\ \mathsf{tan\,\theta=\dfrac{2}{3}}\\\\ \mathsf{\dfrac{sin\,\theta}{cos\,\theta}=\dfrac{2}{3}}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\qquad\mathsf{(i)} \end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctextsf%7BFrom%20the%20picture%2C%20we%20get%7D%5C%5C%5C%5C%20%5Cmathsf%7Btan%5C%2C%5Ctheta%3D%5Cdfrac%7B2%7D%7B3%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Cdfrac%7Bsin%5C%2C%5Ctheta%7D%7Bcos%5C%2C%5Ctheta%7D%3D%5Cdfrac%7B2%7D%7B3%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7B3%5C%2Csin%5C%2C%5Ctheta%3D2%5C%2Ccos%5C%2C%5Ctheta%7D%5Cqquad%5Cmathsf%7B%28i%29%7D%20%5Cend%7Barray%7D)
![\large\begin{array}{l} \textsf{Square both sides of \mathsf{(i)} above:}\\\\ \mathsf{(3\,sin\,\theta)^2=(2\,cos\,\theta)^2}\\\\ \mathsf{9\,sin^2\,\theta=4\,cos^2\,\theta}\qquad\quad\textsf{(but }\mathsf{cos^2\theta=1-sin^2\,\theta}\textsf{)}\\\\ \mathsf{9\,sin^2\,\theta=4\cdot (1-sin^2\,\theta)}\\\\ \mathsf{9\,sin^2\,\theta=4-4\,sin^2\,\theta}\\\\ \mathsf{9\,sin^2\,\theta+4\,sin^2\,\theta=4} \end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctextsf%7BSquare%20both%20sides%20of%20%5Cmathsf%7B%28i%29%7D%20above%3A%7D%5C%5C%5C%5C%20%5Cmathsf%7B%283%5C%2Csin%5C%2C%5Ctheta%29%5E2%3D%282%5C%2Ccos%5C%2C%5Ctheta%29%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7B9%5C%2Csin%5E2%5C%2C%5Ctheta%3D4%5C%2Ccos%5E2%5C%2C%5Ctheta%7D%5Cqquad%5Cquad%5Ctextsf%7B%28but%20%7D%5Cmathsf%7Bcos%5E2%5Ctheta%3D1-sin%5E2%5C%2C%5Ctheta%7D%5Ctextsf%7B%29%7D%5C%5C%5C%5C%20%5Cmathsf%7B9%5C%2Csin%5E2%5C%2C%5Ctheta%3D4%5Ccdot%20%281-sin%5E2%5C%2C%5Ctheta%29%7D%5C%5C%5C%5C%20%5Cmathsf%7B9%5C%2Csin%5E2%5C%2C%5Ctheta%3D4-4%5C%2Csin%5E2%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%5Cmathsf%7B9%5C%2Csin%5E2%5C%2C%5Ctheta%2B4%5C%2Csin%5E2%5C%2C%5Ctheta%3D4%7D%20%5Cend%7Barray%7D)
![\large\begin{array}{l} \mathsf{13\,sin^2\,\theta=4}\\\\ \mathsf{sin^2\,\theta=\dfrac{4}{13}}\\\\ \mathsf{sin\,\theta=\sqrt{\dfrac{4}{13}}}\\\\ \textsf{(we must take the positive square root, because }\theta \textsf{ is an}\\\textsf{acute angle, so its sine is positive)}\\\\ \mathsf{sin\,\theta=\dfrac{2}{\sqrt{13}}} \end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7Bl%7D%20%5Cmathsf%7B13%5C%2Csin%5E2%5C%2C%5Ctheta%3D4%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5E2%5C%2C%5Ctheta%3D%5Cdfrac%7B4%7D%7B13%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C%5Ctheta%3D%5Csqrt%7B%5Cdfrac%7B4%7D%7B13%7D%7D%7D%5C%5C%5C%5C%20%5Ctextsf%7B%28we%20must%20take%20the%20positive%20square%20root%2C%20because%20%7D%5Ctheta%20%5Ctextsf%7B%20is%20an%7D%5C%5C%5Ctextsf%7Bacute%20angle%2C%20so%20its%20sine%20is%20positive%29%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C%5Ctheta%3D%5Cdfrac%7B2%7D%7B%5Csqrt%7B13%7D%7D%7D%20%5Cend%7Barray%7D)
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![\large\begin{array}{l} \textsf{From (i), we find the value of }\mathsf{cos\,\theta:}\\\\ \mathsf{3\,sin\,\theta=2\,cos\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{2}\,sin\,\theta}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\diagup\!\!\!\! 2}\cdot \dfrac{\diagup\!\!\!\! 2}{\sqrt{13}}}\\\\ \mathsf{cos\,\theta=\dfrac{3}{\sqrt{13}}}\\\\ \end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctextsf%7BFrom%20%28i%29%2C%20we%20find%20the%20value%20of%20%7D%5Cmathsf%7Bcos%5C%2C%5Ctheta%3A%7D%5C%5C%5C%5C%20%5Cmathsf%7B3%5C%2Csin%5C%2C%5Ctheta%3D2%5C%2Ccos%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Cdfrac%7B3%7D%7B2%7D%5C%2Csin%5C%2C%5Ctheta%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Cdfrac%7B3%7D%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%202%7D%5Ccdot%20%5Cdfrac%7B%5Cdiagup%5C%21%5C%21%5C%21%5C%21%202%7D%7B%5Csqrt%7B13%7D%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcos%5C%2C%5Ctheta%3D%5Cdfrac%7B3%7D%7B%5Csqrt%7B13%7D%7D%7D%5C%5C%5C%5C%20%5Cend%7Barray%7D)
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![\large\begin{array}{l} \textsf{Since sine and cosecant functions are reciprocal, we have}\\\\ \mathsf{sin\,2\theta\cdot csc\,2\theta=1}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{sin\,2\theta}\qquad\quad\textsf{(but }}\mathsf{sin\,2\theta=2\,sin\,\theta\,cos\,\theta}\textsf{)}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\,sin\,\theta\,cos\,\theta}}\\\\ \mathsf{csc\,2\theta=\dfrac{1}{2\cdot \frac{2}{\sqrt{13}}\cdot \frac{3}{\sqrt{13}}}} \end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7Bl%7D%20%5Ctextsf%7BSince%20sine%20and%20cosecant%20functions%20are%20reciprocal%2C%20we%20have%7D%5C%5C%5C%5C%20%5Cmathsf%7Bsin%5C%2C2%5Ctheta%5Ccdot%20csc%5C%2C2%5Ctheta%3D1%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcsc%5C%2C2%5Ctheta%3D%5Cdfrac%7B1%7D%7Bsin%5C%2C2%5Ctheta%7D%5Cqquad%5Cquad%5Ctextsf%7B%28but%20%7D%7D%5Cmathsf%7Bsin%5C%2C2%5Ctheta%3D2%5C%2Csin%5C%2C%5Ctheta%5C%2Ccos%5C%2C%5Ctheta%7D%5Ctextsf%7B%29%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcsc%5C%2C2%5Ctheta%3D%5Cdfrac%7B1%7D%7B2%5C%2Csin%5C%2C%5Ctheta%5C%2Ccos%5C%2C%5Ctheta%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcsc%5C%2C2%5Ctheta%3D%5Cdfrac%7B1%7D%7B2%5Ccdot%20%5Cfrac%7B2%7D%7B%5Csqrt%7B13%7D%7D%5Ccdot%20%5Cfrac%7B3%7D%7B%5Csqrt%7B13%7D%7D%7D%7D%20%5Cend%7Barray%7D)
![\large\begin{array}{l} \mathsf{csc\,2\theta=\dfrac{~~~~1~~~~}{\frac{2\cdot 2\cdot 3}{(\sqrt{13})^2}}}\\\\ \mathsf{csc\,2\theta=\dfrac{~~1~~}{\frac{12}{13}}}\\\\ \boxed{\begin{array}{c}\mathsf{csc\,2\theta=\dfrac{13}{12}} \end{array}}\qquad\checkmark \end{array}](https://tex.z-dn.net/?f=%5Clarge%5Cbegin%7Barray%7D%7Bl%7D%20%5Cmathsf%7Bcsc%5C%2C2%5Ctheta%3D%5Cdfrac%7B~~~~1~~~~%7D%7B%5Cfrac%7B2%5Ccdot%202%5Ccdot%203%7D%7B%28%5Csqrt%7B13%7D%29%5E2%7D%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bcsc%5C%2C2%5Ctheta%3D%5Cdfrac%7B~~1~~%7D%7B%5Cfrac%7B12%7D%7B13%7D%7D%7D%5C%5C%5C%5C%20%5Cboxed%7B%5Cbegin%7Barray%7D%7Bc%7D%5Cmathsf%7Bcsc%5C%2C2%5Ctheta%3D%5Cdfrac%7B13%7D%7B12%7D%7D%20%5Cend%7Barray%7D%7D%5Cqquad%5Ccheckmark%20%5Cend%7Barray%7D)
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Tags: <em>trigonometry trig function cosecant csc double angle identity geometry</em>
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Answer:
It is 100. You round the 6 to the 9 which makes the 9 round the 99 which makes it 100.
Answer:
x=2, y=1
Step-by-step explanation:
4x+8y=16
4x-8y=0
Add the two equations together:
8x=16
Divide both sides by 8:
x=2
Plug this back into one of the original equations to find y:
4(2)-8y=0
8-8y=0
y=1
Hope this helps!
Answer:
B
Step-by-step explanation:
I'm pretty sure that the shaded space represents all possible solutions, meaning that the X has to be greater than or equal to 2.