Angles of elevation

From points A and B on level ground, the angles of elevation of the top of a building are 25° and 37° respectively. If |AB| = 57m, calculate, to the nearest meter, the distances of the top of the building from A and B if they are both on the same side of the building.

Correct result:

h =  69.73 m
d₁ =  165 m
d₂ =  116 m

Solution:

$$\begin{gathered} a=57\text{ m } \\ A=25^{\circ }\to \text{ rad}=25^{\circ }\cdot {\displaystyle \frac{\pi }{180}}=25^{\circ }\cdot {\displaystyle \frac{3.1415926}{180}}=0.43633=5\pi \mathrm{/}36 \\ B=37^{\circ }\to \text{ rad}=37^{\circ }\cdot {\displaystyle \frac{\pi }{180}}=37^{\circ }\cdot {\displaystyle \frac{3.1415926}{180}}=0.64577 \\ \tan A={\displaystyle \frac{h}{a+b}} \\ \tan B={\displaystyle \frac{h}{b}} \\ b={\displaystyle \frac{h}{\tan B}} \\ a\tan A+b\tan A=h \\ a\tan A+{\displaystyle \frac{h}{\tan B}}\tan A=h \\ h={\displaystyle \frac{a\cdot \tan (A)}{1-\tan (A)\mathrm{/}\tan (B)}}={\displaystyle \frac{57\cdot \tan (0.4363)}{1-\tan (0.4363)\mathrm{/}\tan (0.6458)}}=69.73\text{ m} \end{gathered}$$

$$\begin{gathered} b={\displaystyle \frac{h}{\tan (B)}}={\displaystyle \frac{69.728}{\tan (0.6458)}}\doteq 92.5322\text{ m } \\ {d}_1=\sqrt{h^2+(a+b{)}^2}=\sqrt{69.728^2+(57+92.5322{)}^2}=165\text{ m} \end{gathered}$$

$d_2=\sqrt{h^2+b^2}=\sqrt{69.728^2+92.5322^2}=116\text{ m}$

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