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March 25th, 2017, 01:37 AM  #1 
Newbie Joined: Mar 2017 From: Manila Posts: 3 Thanks: 0  Trigonometry Angle of elevation, depression
A man is looking out of his apartment window at the office building across the street. From the man’s point of view, the angle of elevation to the top of the office building is 57.5 degrees , while the angle of depression to the base of the office building is 41.2 degrees . Find: a. the altitude of the man’s eyes AND the height of the office building if the apartment and the office building are 14.6 meters apart b. the altitude of the man’s eyes AND the distance between the office building and the apartment if the height of the office building is 41.8 meters How would I solve this question?? I don't understand this question at all... Pls help! Thanks! 
March 25th, 2017, 05:57 AM  #2 
Senior Member Joined: Jun 2015 From: England Posts: 566 Thanks: 146 
Have you drawn a diagram (two actually one for each part of the question)? Angles of elevation and depression are measured from the horizontal. So this gives you two right angled triangles in each case with the angle of elevation or depression providing another angle and that part of the question one of the sides so you can calculate the other side by trigonometry. What else did you not understand? 
March 25th, 2017, 06:03 AM  #3  
Math Team Joined: Jul 2011 From: Texas Posts: 2,507 Thanks: 1234  Quote:
$\color{red}{y}$ = height of the man's eye above the ground $\color{red}{Hy}$ = height of the building $\color{red}{x}$ = horizontal distance between the buildings two equations ... $\tan(57.5) = \dfrac{Hy}{x}$ $\tan(41.2) = \dfrac{y}{x}$ for part (a), you are given $x = 14.6 \, m$, solve for $H \text{ and } y$ for part (b), you are given $H = 41.8 \, m$, solve for $x \text{ and } y$  

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angle, depression, elevation, trigonometry 
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