+0  
 
0
68
4
avatar+183 

I don't think I am understanding the question at all here!! Can you please advise ...thanks .

 

The answer is 931 meters. (I am getting 2 different readings for each point [1.42 and .438] - If I add

both answers and average then my answer would be 930 metres ?/  

 Mar 3, 2020

Best Answer 

 #1
avatar+24388 
+3

Question 23

 

\(\begin{array}{|rcll|} \hline \text{cos-rule}: \\ \hline 5^2 &=& a^2+b^2-2ab\cos(64^\circ6'-30^\circ 45') \\ 5^2 &=& a^2+b^2-2ab\cos(63^\circ66'-30^\circ 45') \\ 5^2 &=& a^2+b^2-2ab\cos(33^\circ 21') \\ 5^2 &=& a^2+b^2-2ab\cos(33.35^\circ) \\ && \begin{array}{rcll} \mathbf{\tan(10^\circ 12')} &=& \mathbf{\dfrac{h}{a}} \\ a &=& \dfrac{h}{\tan(10^\circ 12')} \\ \mathbf{a} &=& \mathbf{\dfrac{h}{\tan(10.2^\circ)}} \\\\ \end{array} \\ && \begin{array}{rcll} \mathbf{\tan(6^\circ 18')} &=& \mathbf{\dfrac{h}{b}} \\ b &=& \dfrac{h}{\tan(6^\circ 18')} \\ \mathbf{b} &=& \mathbf{\dfrac{h}{\tan(6.3^\circ)}} \\\\ \end{array} \\ 5^2 &=& \dfrac{h^2}{\tan^2(10.2^\circ)}+\dfrac{h^2}{\tan^2(6.3^\circ)} - \dfrac{2h^2\cos(33.35^\circ)}{\tan(10.2^\circ)\tan(6.3^\circ)} \\\\ 25 &=& h^2\left( \dfrac{1}{\tan^2(10.2^\circ)}+\dfrac{1}{\tan^2(6.3^\circ)} - \dfrac{2\cos(33.35^\circ)}{\tan(10.2^\circ)\tan(6.3^\circ)} \right) \\\\ 25 &=& h^2( 30.8887666206+82.0453112207 - 84.1035186173 ) \\ 25 &=& 28.8305592241h^2 \\ h^2 &=& \dfrac{25}{28.8305592241} \\ h &=& \dfrac{5}{\sqrt{28.8305592241}} \\ h &=& \dfrac{5}{5.36940957872} \\ h &=& 0.93120108025\ \text{km} \\ \mathbf{h} &=& \mathbf{931.20108025\ \text{m} } \\ \hline \end{array}\)

 

The height of the beacon above A is \(\mathbf{931.2\ \text{meters}}\)

 

laugh

 Mar 3, 2020
edited by heureka  Mar 3, 2020
 #1
avatar+24388 
+3
Best Answer

Question 23

 

\(\begin{array}{|rcll|} \hline \text{cos-rule}: \\ \hline 5^2 &=& a^2+b^2-2ab\cos(64^\circ6'-30^\circ 45') \\ 5^2 &=& a^2+b^2-2ab\cos(63^\circ66'-30^\circ 45') \\ 5^2 &=& a^2+b^2-2ab\cos(33^\circ 21') \\ 5^2 &=& a^2+b^2-2ab\cos(33.35^\circ) \\ && \begin{array}{rcll} \mathbf{\tan(10^\circ 12')} &=& \mathbf{\dfrac{h}{a}} \\ a &=& \dfrac{h}{\tan(10^\circ 12')} \\ \mathbf{a} &=& \mathbf{\dfrac{h}{\tan(10.2^\circ)}} \\\\ \end{array} \\ && \begin{array}{rcll} \mathbf{\tan(6^\circ 18')} &=& \mathbf{\dfrac{h}{b}} \\ b &=& \dfrac{h}{\tan(6^\circ 18')} \\ \mathbf{b} &=& \mathbf{\dfrac{h}{\tan(6.3^\circ)}} \\\\ \end{array} \\ 5^2 &=& \dfrac{h^2}{\tan^2(10.2^\circ)}+\dfrac{h^2}{\tan^2(6.3^\circ)} - \dfrac{2h^2\cos(33.35^\circ)}{\tan(10.2^\circ)\tan(6.3^\circ)} \\\\ 25 &=& h^2\left( \dfrac{1}{\tan^2(10.2^\circ)}+\dfrac{1}{\tan^2(6.3^\circ)} - \dfrac{2\cos(33.35^\circ)}{\tan(10.2^\circ)\tan(6.3^\circ)} \right) \\\\ 25 &=& h^2( 30.8887666206+82.0453112207 - 84.1035186173 ) \\ 25 &=& 28.8305592241h^2 \\ h^2 &=& \dfrac{25}{28.8305592241} \\ h &=& \dfrac{5}{\sqrt{28.8305592241}} \\ h &=& \dfrac{5}{5.36940957872} \\ h &=& 0.93120108025\ \text{km} \\ \mathbf{h} &=& \mathbf{931.20108025\ \text{m} } \\ \hline \end{array}\)

 

The height of the beacon above A is \(\mathbf{931.2\ \text{meters}}\)

 

laugh

heureka Mar 3, 2020
edited by heureka  Mar 3, 2020
 #3
avatar+183 
+2

Thanks for that...what a great job Heureka!!

 

I tried to solve this problem using sine rule. Therefore I got values of a and b independent of height of beacon! A bit puzzling.. 

 

Regards and thanks -  have a great day! 

OldTimer  Mar 4, 2020
 #4
avatar+24388 
+3

Thank you, OldTimer !

 

laugh

heureka  Mar 4, 2020
 #2
avatar+155 
0

i dont even know what this is ()_()

 Mar 3, 2020

34 Online Users

avatar
avatar
avatar
avatar