Question

Question 16
The angles of elevation of the top of a tower from two points at a distance of 4 m and 9 m from the base of the tower and in the same straight line with it are complementary. Prove that the height of the tower is 6 m.

Answer 1

Let AB be the height of tower and x be the one of the complementry angle.
So, (90-x) be the another complementary angle.
C and D be the two points with distance 4 m and 9 m from the base respectively.

As per question,
In right $\Delta ABC,$
tan $x=\frac{AB}{BC}$
$\Rightarrow tanx=\frac{AB}{4}$
$\Rightarrow AB=4tanx...\left(i\right)$
Also,
In right $\Delta ABD,$
tan $({90}^{\circ }-x)=\frac{AB}{BD}$
$\Rightarrow cotx=\frac{AB}{9}$
$\Rightarrow AB=9cotx...\left(ii\right)$
Multiplying equation (i) and (ii)
$A{B}^2=9cotx\times 4tanx$
$\Rightarrow A{B}^2=36$
$\Rightarrow AB=\pm 6$
Height cannot be negative.

Therefore, the height of the tower is 6m.