Question

In a circle with center O, AD is a diameter, ABC is a chord, BO = 5, and $\angle ABO={60}^0$. Then the length of BC is

• A. 3
• B. $3+\sqrt{3}$
• C. $5-\frac{\sqrt{3}}{2}$
• D. 5

Answer 1

The correct option is D 5


In $\Delta BOA$
Angles are in ratio of 30:60:90 then opp sides i.e OB: OA: BA to these angles will be in ratio of
$1:\sqrt{3}:2$

OB = 5 is already given,
$\Rightarrow OA=5\sqrt{3}$, BA = 10
In $\Delta BOA$ and $\Delta DCA$,
$\angle BOA=\angle DCA={90}^0$
$\angle A$is common
$\Delta BOA\simeq \Delta DCA,$
$\text{By A.A similarity Criterion}\frac{BO}{DC}=\frac{BA}{DA}=\frac{OA}{CA}$,

$CA=\frac{DA\times OA}{BA}=\frac{2\times 25\times 3}{10}=15$
$BC=CA-BA=15-10=5unit$
Hence, the correct option is d.