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An engineer estimates the angle of elevation to the top of the building to be 50°. After moving 1.5 meters further away, the angle of elevation was 40°. How high is the top of the building?

Guest May 5, 2017
 #1
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An engineer estimates the angle of elevation to the top of the building to be 50°.
After moving 1.5 meters further away, the angle of elevation was 40°.
How high is the top of the building?

 

\(\begin{array}{|lrcll|} \hline (1) & \tan(50^{\circ}) &=& \frac{h}{x} \\ & x &=& \frac{h}{\tan(50^{\circ})} \\\\ (2) & \tan(40^{\circ}) &=& \frac{h}{1.5+x} \\ & \tan(40^{\circ}) &=& \frac{h}{1.5+\frac{h}{\tan(50^{\circ})}} \\ & \tan(40^{\circ})\cdot \left( 1.5+\frac{h}{\tan(50^{\circ})}\right) &=& h \\ & 1.5\cdot \tan(40^{\circ}) + h\cdot \frac{ \tan(40^{\circ}) } { \tan(50^{\circ}) } &=& h \\ & h-h\cdot \frac{ \tan(40^{\circ}) } { \tan(50^{\circ}) } &=& 1.5\cdot \tan(40^{\circ}) \\ & h\cdot \left(1- \frac{ \tan(40^{\circ} ) } { \tan(50^{\circ}) } \right) &=& 1.5\cdot \tan(40^{\circ}) \\ & h\cdot \left( \frac{ \tan(50^{\circ})-\tan(40^{\circ} ) } { \tan(50^{\circ}) } \right) &=& 1.5\cdot \tan(40^{\circ}) \\ & h &=& 1.5\cdot \left( \frac{ \tan(50^{\circ})\cdot\tan(40^{\circ}) }{ \tan(50^{\circ})-\tan(40^{\circ} ) } \right) \\ & h &=& 1.5\cdot \left( \frac{ 1.19175359259\cdot 0.83909963118}{ 1.19175359259-0.83909963118 ) } \right) \\ & h &=& 1.5\cdot \left( \frac{ 1 }{ 0.35265396142 } \right) \\ & h &=& 1.5\cdot 2.83564090981 \\ & h &=& 4.25346136471\ m \\ \hline \end{array}\)

 

The top of the building is 4.25 m high.

 

laugh

heureka  May 5, 2017
edited by heureka  May 5, 2017

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