Question

A circle inscribed in a triangle ABC touches the side AB at D such that AD = 5 and BD = 3. If $\angle A={60}^{\circ },$ then the value of $\left[BC/3\right]$ (where $[\cdot ]$ represents the greatest integer function) is ________ .

• A. $4$
• B. $5$
• C. $6$
• D. None of these

Answer 1

The correct option is A $4$

Let the circle inscribed in $\Delta ABC$ touches $AB$ at $D$,$BC$ at $E$ and $CA$ at $F$ respectively.

Given $AD=5,BD=3$ and $A=\frac{\pi }{3}$

$\Rightarrow AD=AF=5,BD=BE=3$ and $CE=CF=x$

Applying cosine rule gives
$\cos A=\frac{b^2+c^2-a^2}{2bc}$

$\Rightarrow \frac{1}{2}=\frac{(5+x{)}^2+64-(3+x{)}^2}{16(5+x)}$

$\Rightarrow 8(5+x)=4x+80$

$\Rightarrow x=10$

$\therefore \left[BC/3\right]=\left[13/3\right]=4$ where [.] is a greatest integer funtion.

Hence, option A.