Given:-
From a window 100ft above the ground in building A, the top and bottom of building B are sighted so that the angles are 70 degrees and 30 degrees respectively.
To find:-
The height of building B.
So now, the image of the given data is,
So now we find the value of PS. so we get,
[tex]\begin{gathered} \tan \text{ 30=}\frac{100}{PS} \\ \frac{1}{\sqrt[]{3}}=\frac{100}{PS} \\ PS=100\sqrt[]{3} \end{gathered}[/tex]So now we find the height of QS,
[tex]\begin{gathered} \tan \text{ 70=}\frac{QS}{PS} \\ 2.7474=\frac{QS}{100\sqrt[]{3}} \\ QS=100\sqrt[]{3}\times2.7474 \\ QS=475.84 \end{gathered}[/tex]So the total height is,
[tex]100+475.84=575.84[/tex]So the height of building B is 575.84
