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.

Result

h =  69.73 m
d1 =  165 m
d2 =  116 m

Solution:

a=57 m A=25 rad=25 π180 =25 3.1415926180 =0.43633=5π/36 B=37 rad=37 π180 =37 3.1415926180 =0.64577  tanA=ha+b tanB=hb b=htanB  atanA+btanA=h atanA+htanBtanA=h  h=a tan(A)1tan(A)/tan(B)=57 tan(0.4363)1tan(0.4363)/tan(0.6458)69.72869.73 ma=57 \ \text{m} \ \\ A=25 ^\circ \rightarrow\ \text{rad}=25 ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ =25 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ =0.43633=5π/36 \ \\ B=37 ^\circ \rightarrow\ \text{rad}=37 ^\circ \cdot \ \dfrac{ \pi }{ 180 } \ =37 ^\circ \cdot \ \dfrac{ 3.1415926 }{ 180 } \ =0.64577 \ \\ \ \\ \tan A=\dfrac{ h }{ a+b } \ \\ \tan B=\dfrac{ h }{ b } \ \\ b=\dfrac{ h }{ \tan B } \ \\ \ \\ a \tan A + b \tan A=h \ \\ a \tan A + \dfrac{ h }{ \tan B } \tan A=h \ \\ \ \\ h=\dfrac{ a \cdot \ \tan(A) }{ 1 - \tan(A)/\tan(B) }=\dfrac{ 57 \cdot \ \tan(0.4363) }{ 1 - \tan(0.4363)/\tan(0.6458) } \doteq 69.728 \doteq 69.73 \ \text{m}
b=htan(B)=69.728tan(0.6458)92.5348 m  d1=h2+(a+b)2=69.7282+(57+92.5348)2164.9938165 mb=\dfrac{ h }{ \tan(B) }=\dfrac{ 69.728 }{ \tan(0.6458) } \doteq 92.5348 \ \text{m} \ \\ \ \\ d_{1}=\sqrt{ h^2 + (a+b)^2 }=\sqrt{ 69.728^2 + (57+92.5348)^2 } \doteq 164.9938 \doteq 165 \ \text{m}
d2=h2+b2=69.7282+92.53482115.8662116 md_{2}=\sqrt{ h^2 + b^2 }=\sqrt{ 69.728^2 + 92.5348^2 } \doteq 115.8662 \doteq 116 \ \text{m}



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