Question

The given figure shows a circle with centre O and BCD is tangent to it at C. Then, $\angle ACD+\angle BAC=$ __________

• A. $\angle ACD={90}^0$
• B. $\angle ACD={40}^0$
• C. $\angle ACD={70}^0$
• D. $\angle ACD={10}^0$

Answer 1

The correct option is A

$\angle ACD={90}^0$

Given - A circle with centre O and BCD is a tangent at C.

$\angle ACD+\angle BAC$

Construction - Join OC.

BCD is the tangent and OC is the radius

$\therefore OC\perp BD\Rightarrow \angle OCD={90}^0$

$\Rightarrow \angle OCA+\angle ACD={90}^0$ ...(i)

But in $\Delta$ OCA

OA=OC (radii of the same circle)

$\therefore \angle OCA=\angle OAC$ [from (i)]

$\angle OAC+\angle ACD={90}^0$

$\Rightarrow \angle BAC+\angle ACD={90}^0$

Q.E.D.