Question

16. In the figure PB is the tangent to a circle with centre Alf ABP = 40° then PAB
a. 700 b. 400
C. 500
d. 80°
​

Answer 1

Answer:

where is that figure?

Answer 2

Given :- In the figure PB is the tangent to a circle with centre A . lf ABP = 40° then PAB

a. 70°

b. 40°

C. 50°

d. 80°

Solution :-

given ,

→ centre of circle = A

→ PB = tangent .

→ AP = radius .

so, in ∆PAB we have,

→ ∠ABP = 40°

and,

→ ∠APB = 90° { Tangent to the circle is perpendicular to the radius. }

then,

→ ∠ABP + ∠APB + ∠PAB = 180° { By angle sum property. }

→ 40° + 90° + ∠PAB = 180°

→ 130° + ∠PAB = 180°

→ ∠PAB = 180° - 130°

→ ∠PAB = 50° (c) (Ans.)

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