Question

Two Tangents are drawn from an external point P such that angle OBA=10 degree. Then angle BPA is ?
(a) 10
(6) 20
(d) 40
(c) 30
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Answer 1

Step-by-step explanation:

Given that:

Two Tangents are drawn from an external point P such that angle OBA=10 degree. Then angle BPA is ?

(a) 10

(b) 20

(d) 40

(c) 30

Solution:

To find the value of

$\angle \: BPA$

first of all one have to find the value of

$\angle \: BOA$

as in ∆ OAB,

OA= OB(Radius of circle)

∆OAB is thus an isosceles triangle,so

$\angle \: OAB= 10°$

In ∆OAB,

$\angle \: OAB + \angle \: OBA + \angle \: BOA = 180° \\ \\ \angle \: BOA= 180° - 20° \\ \\ \angle \: BOA = 160°$

As we also know that,Tangent and radius of circle meet at 90°

$\angle \: OBP= \angle \: OAP = 90°$

In quadrilateral OABP, three angles are known and sum of all interior angles are 360°

$\angle \: BPA = 360° - 160° - 90° - 90° \\ \\ \angle \: BPA= 360° - 340° \\ \\ \bold{\angle \: BPA = 20°}$

Thus angle BPA= 20°,

Option b is correct.

Hope it helps you.

Answer 2

the answer is attached in the pic

hope it helps