Question

explain woth full steps???​

Answer 1

$\large\underline{\underline{\red{\sf Given :-}}}$

• In the given figure , O is the centre of circle.
• AB and AC are tangents to circles at B and C .
• Value of angle BAC = 62° .

$\large\underline{\underline{\red{\sf To\:Find:-}}}$

• The value of angle BDC .

$\large\underline{\underline{\red{\sf Step-by-step\: Explanation :-}}}$

Given that , $\angle$ BAC = 62° . We know that radius is perpendicular to the tangent at the point of contact . Hence here , $\angle$ OBA = $\angle$ OCA = 90° .

Now, in quadrilateral ABOC ,

$=> \angle ABO + \angle BOC +\angle OCA+\angle CAB = 360^{\circ}\\=> 90^{\circ}+90^{\circ}+62^{\circ}+\angle BOC = 360^{\circ}\\=>242^{\circ}+\angle BOC = 360^{\circ}\\=> \angle BOC = 360^{\circ}-242^{\circ}\\=>\bf \angle BOC =118^{\circ}$

Now , we know that measure of a complete angle is 360° . Hence here ,

$=>\angle BOC+ext.\angle BOC = 360°\\=> ext.\angle BOC =360^{\circ}-118^{\circ}\\=> \bf ext\angle BOC = 242^{\circ}$

Theorem :-

The angle subtended by an arc of a circle at its centre is twice of the angle it subtends anywhere on the circle's circumference.

Using this we have ;

$=> \angle BDC =\dfrac{\angle BOC}{2}\\=> \angle BDC = \dfrac{242^{\circ}}{2}\\=>\bf \angle BDC = 121^{\circ}$

$\Large\boxed{\pink{\sf \red{\bigstar}\angle BDC = 121^{\circ}}}$