Question

prove that the angle between the two tangents drawn from external point to a circle is supplementary to the angle subtended by the line segment joining the point of contact at the center​

Answer 1

Answer:

Let us consider a circle centered at point O. Let P be an external point from which two tangents PA and PB are drawn to the circle which are touching the circle at point A and B respectively and AB is the line segment, joining point of contacts A and B together such that it subtends ∠AOB at center O of the circle.

Step-By-Step Explanation;-

• It can be observed that
• OA (radius) ⊥ PA (tangent)
• Therefore, ∠OAP = 90°
• Similarly, OB (radius) ⊥ PB (tangent)
• ∠OBP = 90°
• In quadrilateral OAPB,
• Sum of all interior angles = 360º
• ∠OAP +∠APB+∠PBO +∠BOA = 360º
• 90º + ∠APB + 90º + ∠BOA = 360º
• ∠APB + ∠BOA = 180º
• Hence, it can be observed that the angle between the two tangents drawn from an external point to a circle is supplementary to the angle subtended by the line-segment joining the points of contact at the centre.
• plZ mark it brainlist.

Answer 2

Assume-

Let us consider a circle with centre O and M be an external point from which two tangents are drawn which touches the circle at A and B point such that AM and BM are two tangents.

AB is a line segment which joins both A and B point such that it subtends ∠AOB at the centre of the circle.

Solution-

From above assumption it is observed that ..

$\Rightarrow\:OA\:\perp\:MA$

$\Rightarrow \: \angle{OAM} \: = \: {90}^{ \circ}$

Similarly,

$\Rightarrow\:OB\:\perp\:MB$

$\Rightarrow \: \angle{OBM} \: = \: {90}^{ \circ}$

In quadrilateral OAMB

Sum of all interior angles is 360°

$\implies\:\angle{OAM}\:+\:\angle{AMB}\:+\:\angle{MBO}\:+\:\angle{BOA}\:=\:360^{\circ}$

$\implies\:90^{\circ}\:+\:\angle{AMB}\:+\:90^{\circ}\:+\:\angle{BOA}\:=\:360^{\circ}$

$\implies\:\angle{AMB}\:+\:\angle{BOA}\:+\:180^{\circ}\:=\:360^{\circ}$

$\implies\:\angle{AMB}\:+\:\angle{BOA}\:=\:360^{\circ}\:-\:180^{\circ}$

$\implies\:\angle{AMB}\:+\:\angle{BOA}\:=\:180^{\circ}$

Hence, proved.