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Triangle ABC has a right angle at C. sinA sinB and sinC forms a geometric sequence. Please determine the value of sinA. 

 Mar 25, 2019
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Triangle ABC has a right angle at C. sinA sinB and sinC forms a geometric sequence.

Please determine the value of sinA. 

 

\(\text{Let $r=$ ratio } \)

 

\(\begin{array}{lcll} C=90^\circ \\ \sin(C) = \sin(90^\circ) = 1 \\ \end{array} \)

 

\(\begin{array}{lcll} B=90^\circ-A \\ \sin(B) = \sin(90^\circ-A) = \cos(A) \\ \end{array} \)

 

Geometric sequence:

\(\begin{array}{|rcll|} \hline a_1 &=& a \\ &=& \sin(A) \\\\ a_2 &=&a\cdot r \quad &| \quad a=\sin(A) \\ &=& \sin(A)\cdot r \quad &| \quad a_2 = \sin(B)=\cos(A)\\ \mathbf{\cos(A)} & \mathbf{=}& \mathbf{\sin(A)\cdot r} \qquad (1) \\\\ a_3 &=& a\cdot r^2 \quad &| \quad a=\sin(A) \\ &=& \sin(A)\cdot r^2 \quad &| \quad a_3 =\sin(C)=1 \\ \mathbf{1} &\mathbf{=}& \mathbf{\sin(A)\cdot r^2} \qquad (2) \\ \hline \end{array}\)

 

\(\begin{array}{|rcll|} \hline \cos(A) &=&\sin(A)\cdot r & (1) \\ r &=& \dfrac{\cos(A)}{\sin(A)} \\ \hline 1 &=& \sin(A)\cdot r^2 & (2) \quad | \quad r = \dfrac{\cos(A)}{\sin(A)} \\ 1 &=& \sin(A)\cdot \left(\dfrac{\cos(A)}{\sin(A)}\right)^2 \\ 1 &=& \sin(A)\cdot \dfrac{\cos^2(A)}{\sin^2(A)} \\ 1 &=& \dfrac{\cos^2(A)}{\sin(A)} \\ \sin(A) &=& \cos^2(A) & \quad | \quad \cos^2(A)=1-\sin^2(A) \\ \sin(A) &=& 1-\sin^2(A) \\ \sin^2(A)+\sin(A) -1 &=& 0 \\\\ \sin(A) &=& \dfrac{-1\pm \sqrt{1-4\cdot(-1)} } {2} \\ \sin(A) &=& \dfrac{-1\pm \sqrt{5} } {2} \\ \sin(A) &=& \dfrac{-1{\color{red}+}\sqrt{5} } {2} \\ \mathbf{\sin(A) } &\mathbf{=}& \mathbf{\dfrac{-1 + \sqrt{5} } {2} } \\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline r^2 &=& \dfrac{1}{\sin(A)} \\ r &=& \dfrac{1}{\sqrt{\sin(A)}} \\ \mathbf{r} &\mathbf{=}& \mathbf{ \dfrac{1}{\sqrt{\dfrac{-1 + \sqrt{5} } {2}}} }\\ \hline \end{array} \)

 

\(\begin{array}{|rcll|} \hline a_1 = \sin(A) &=& \dfrac{-1 + \sqrt{5} } {2} \\ a_2 = \sin(B) &=& \sqrt{\dfrac{-1 + \sqrt{5} } {2}} \\ a_3 = \sin(C) &=& 1 \\ \hline \end{array} \)

 

The value of \(\sin(A)\) is  \(\mathbf{\dfrac{-1 + \sqrt{5} } {2} } \) 

 

laugh

 Mar 25, 2019

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