We use cookies to personalise content and advertisements and to analyse access to our website. Furthermore, our partners for online advertising receive pseudonymised information about your use of our website. cookie policy and privacy policy.
 
+0  
 
-1
118
1
avatar+283 

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
 #1
avatar+22896 
+3

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

7 Online Users