Question

Two towers are 40m apart. The angle of elevation of the top of the taller tower from the base and top of the shorter tower is ${70}^{\circ }$ and ${40}^{\circ }$ respectively. Find the height of the shorter tower. $[Tan{40}^{\circ }=0.83,Tan{70}^{\circ }=2.74]$

Answer 1

Let AB and CD be the shorter and the taller tower respectively.

$InRight-angled△ADEtan{40}^{\circ }=\frac{DE}{AE}=\frac{DE}{40}(AE=BC=40m)DE=40.tan{40}^{\circ }=40\times 0.83=33.56m\text{In Right -angled}△BDCtan{70}^{\circ }=\frac{CD}{BC}=\frac{CD}{40}CD=40.tan{70}^{\circ }=40\times 2.74=109.89mCE=CD-DECE=109.89-33.56=76.33m\text{Hence, height of the shorter tower}=AB=CE=76.33m$