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 )=0.436332312999 B=(37∘→rad)=(37⋅ π180 )=0.645771823238  tan⁡A=ha+b tan⁡B=hb b=htan⁡B  atan⁡A+btan⁡A=h atan⁡A+htan⁡Btan⁡A=h  h=a⋅ tan⁡(A)1−tan⁡(A)/tan⁡(B)=57⋅ tan⁡(0.4363)1−tan⁡(0.4363)/tan⁡(0.6458)≐69.728=69.73  m a = 57 \ m \ \\ A = (25^\circ \rightarrow rad) = (25 \cdot \ \dfrac{ \pi }{ 180 } \ ) = 0.436332312999 \ \\ B = (37^\circ \rightarrow rad) = (37 \cdot \ \dfrac{ \pi }{ 180 } \ ) = 0.645771823238 \ \\ \ \\ \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 = 69.73 \ \text { m }
b=htan⁡(B)=69.728tan⁡(0.6458)≐92.5348 m  d1=h2+(a+b)2=69.7282+(57+92.5348)2≐164.9938=165  m b = \dfrac{ h }{ \tan(B) } = \dfrac{ 69.728 }{ \tan(0.6458) } \doteq 92.5348 \ m \ \\ \ \\ d_{ 1 } = \sqrt{ h^2 + (a+b)^2 } = \sqrt{ 69.728^2 + (57+92.5348)^2 } \doteq 164.9938 = 165 \ \text { m }
d2=h2+b2=69.7282+92.53482≐115.8662=116  m d_{ 2 } = \sqrt{ h^2 + b^2 } = \sqrt{ 69.728^2 + 92.5348^2 } \doteq 115.8662 = 116 \ \text { m }

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