Question

Two buildings on a plane ground are 20 m apart. From the top of the smaller building one can see the base of the taller building at an angle of depression of ${50}^{\circ }$ and its top at an angle of elevation of ${25}^{\circ }.$ Find the height of the bigger building. $[Tan{50}^{\circ }=1.2Tan{25}^{\circ }=0.46]$

Answer 1

In the given figure, let CD be the smaller building and AB be the taller building

In $\Delta ADE$

$Tan{50}^{\circ }=\frac{AE}{DE}=\frac{AE}{20}$ (as AC = DE = 20 m)

$AE=20.Tan{50}^{\circ }=20\times 1.2=24m$

In $\Delta DBE$

$Tan{25}^{\circ }=\frac{BE}{DE}=\frac{BE}{20}$ (as AC = DE = 20 m)

$BE=20.Tan{25}^{\circ }=20\times 0.46=9.3m$

Height of the taller building $=AB=AE+EB=24+9.3=33.3m$