Question

A monument ABCD stands at a point A on a level ground. At a point P on the ground the portions AB, AC, AD subtend angles $\alpha ,\beta ,\gamma$ respectively. If $AB=a,AC=b,AD=c,AP=x$ and $\alpha +\beta +\gamma ={180}^o$ then

• A. $x^2=\frac{(a+b+c)}{abc}$
• B. $x^2=\frac{abc}{(a+b+c)}$
• C. $\tan \alpha +\tan \beta +\tan \gamma =\frac{(a+b+c{)}^{\frac{3}{2}}}{\sqrt{abc}}$
• D. $\tan \alpha \tan \beta \tan \gamma =\frac{(a+b+c{)}^{\frac{3}{2}}}{\sqrt{abc}}$

Answer 1

Answer: (B), (C), (D)

The correct options are
B $x^2=\frac{abc}{(a+b+c)}$
C $\tan \alpha +\tan \beta +\tan \gamma =\frac{(a+b+c{)}^{\frac{3}{2}}}{\sqrt{abc}}$
D $\tan \alpha \tan \beta \tan \gamma =\frac{(a+b+c{)}^{\frac{3}{2}}}{\sqrt{abc}}$

$a=AB=AP\tan \alpha =x\tan \alpha$.
$b=AC=x\tan \beta$
$c=AD=x\tan \gamma$.
Now
$a+b+c=x(\tan \alpha +\tan \gamma +\tan \beta )$
Now
$\alpha +\beta +\gamma =\pi$
Hence
$\tan \alpha .\tan \beta .\tan \gamma =\tan \alpha +\tan \gamma +\tan \beta$.
Hence
$a+b+c=x(\tan \alpha .\tan \beta .\tan \gamma )$ ...(i)
And
$abc=x^3(\tan \alpha .\tan \beta .\tan \gamma )$...(ii)
From i and ii, we get
$\frac{a+b+c}{x}=\frac{abc}{x^3}$
Or
$x^2=\frac{abc}{a+b+c}$.
Now
$\tan \alpha +\tan \beta +\tan \gamma$
$=\frac{a+b+c}{x}$

$=\frac{(a+b+c)\sqrt{a+b+c}}{\sqrt{abc}}$

$=\frac{(a+b+c{)}^{\frac{3}{2}}}{\sqrt{abc}}$.