Question

$ABC$ is a triangular park with $AB=AC=100$ cm. $A$ clock tower is situated at the midpoint of $BC$. The angles of elevation of the top of the tower from $A$ and $B$ are $co{t}^{-1}3.2$ and $cose{c}^{-1}2.6$. The height of the tower is

• A. $25$ mt
• B. $50$ mt
• C. $100$ mt
• D. $50\sqrt{2}$ mt

Answer 1

Answer: (A)

$25$ mt

$y^2=A{D}^2={100}^2-B{D}^2=10000-x^2$
${\cot }^{-1}\left(3.2\right)=\alpha ,cose{c}^{-1}\left(2.6\right)=\beta )$
$\cot \alpha =3.2$ & $cosec\beta =2.6$
$\cot \alpha =\frac{y}{h}$
$\Rightarrow y=3.2\times h$
Now,
$\cot \beta =\frac{x}{h}$
$x=\cot \beta \times h$
$x=h\sqrt{{2.6}^2-1}$
$x=2.4h$
$x^2+y^2=100000$
$\Rightarrow (2.4h{)}^2+(3.2h{)}^2={100}^2$
$\Rightarrow h^2\left(16\right)={100}^2$
$\Rightarrow h=25$