Question

A $12m$ tower and $8m$ pole are $22m$ apart. As shown in the figure, supporting wires are tied from the top of both the tower and pole to a hook fixed at the ground between them.


The length of the supporting wire is $10m$, which is attached to the pole, and $20m,$ which is attached to the tower. The hook is $6m$ away from the pole. Also, the acute angle made by the wire joined to the pole and the ground is approximately ${53}^o$. Find out the acute angle made by the wire hooked to the top of the tower with the ground.

• A. Approx ${37}^o$
• B. Approx ${53}^o$
• C. ${56}^o$
• D. ${35}^o$

Answer 1

The correct option is A Approx ${37}^o$

The schematic is as follows:


Now, in $△ABO$ and $△ODC$,
$\frac{AB}{OD}=\frac{8}{16}=\frac{1}{2}\frac{OB}{CD}=\frac{6}{12}=\frac{1}{2}\frac{AO}{OC}=\frac{10}{20}=\frac{1}{2}$

As all the corresponding sides are in same ratio so $△ABO~△ODC\left(SSSsimilarity\right)$
Here, vertex $A$ corresponds to vertex $O.$

So, $\angle AOB=\angle OCD$, $\angle B=\angle D$ and $\angle OAB=\angle COD$ (corresponding angles of similar triangles are equal).

$\angle AOB=53°\left(given\right)$

So, for $△ABO$,

$\begin{array}{ccc}\angle AOB+\angle OAB+\angle ABO& =& 180°\\ 53°+90°+\angle OAB& =& 180°\\ \angle OAB& =& 180°-143°\\ \angle OAB& =& 37°\end{array}$

We know that angle made by the supporting wire with ground to the top of tower is $\angle COD$.

From similar triangles, $\angle COD=\angle OAB$.

So, $\angle COD=37°$

Hence, option $\left(a\right)$ is correct.