Question

At a point on a horizontal line through the base of a monument the angle of elevation of the top of the monument is found to be such that its tangent is $\frac{1}{5}$. On walking 138 metres towards the monument the secant of the angle elevation is found to be $\frac{\sqrt{193}}{12}$. The height of the monument (in metre) is:

• A. 35
• B. 49
• C. 42
• D. 56

Answer 1

The correct option is C 42

Let $AB$ be the monument$=h$m

$DC=138$m

$BD=x$m

$\tan \alpha =\frac{1}{5}$(given)

$\sec \beta =\frac{\sqrt{193}}{12}$ (given)

$\Rightarrow \tan \beta =\sqrt{{\sec }^2\beta -1}=\sqrt{{\left(\frac{\sqrt{193}}{12}\right)}^2-1}=\sqrt{\frac{193-144}{144}}=\sqrt{\frac{49}{144}}=\frac{7}{12}$

From $△ABC,\tan \alpha =\frac{AB}{BC}$

$\frac{1}{5}=\frac{h}{x+138}$

$\Rightarrow 5h=x+138$ ....$\left(1\right)$

From $△ABD,\tan \beta =\frac{h}{x}$

$\Rightarrow \frac{7}{12}=\frac{h}{x}$

$\Rightarrow x=\frac{12h}{7}$

From eqn$\left(1\right)\Rightarrow 5h=\frac{12h}{7}+138$

$\Rightarrow 35h-12h=138\times 7$

$\Rightarrow 23h=138\times 7$

$\Rightarrow h=\frac{138\times 7}{23}=42$m