Question

*If angle between two radii of a circle is 130º, then the angle between the tangents at the ends of these radii is ………………..*

1️⃣ 90°
2️⃣ 50°
3️⃣ 70°
4️⃣ 40°​

Answer 1

Answer:

We know tangents are ⊥ to radius at point of contact.

∴ ∠CAO=∠CBO=90 °

Now in quadrilateral ABC,

⇒ ∠1+∠2+∠3+∠4=360 ° [ Sum of four angles of a quadrilateral is 360 °]

⇒ ∠1+90 °+90 ° +130 ° =360 °

⇒ ∠1+310 ° =360 °

⇒ ∠1=360 ° −310 °

∴ ∠1=50 °

∴ Required measure of an angle is 50°

.

Answer 2

Step-by-step explanation:

Given :

< AOB = 130°

Find :

< APB

Solution :

OA Perpendicular AP

OB Perpendicular BP

Remark < A = 90°

< B = 90°

Extra Information

Radii Of Tangent is 90°

AOBP Is Quadrilateral

Sum Of all the angle of Quadrilateral is 360°

< A + < AOB + < B + < APB = 360°

Putting Value < A and < B and < AOB

90° + 130° + 90° + < APB = 360°

< APB + 310° = 360°

< APB = 360° - 310°

< APB = 50°

So The Angle Is APB 50°

The Option Second 50° is right

Some Property Of Quadrilateral

If each pair Of Opposite side of a

Quadrilateral is equal then it is a

Parallelogram

If a Quadrilateral , each pair of opposite

angle is equal then it is a parallelogram

If a diagonal of a Quadrilateral Bisect each

other , then it is a parallelogram

A Quadrilateral is a parallelogram if a pair of

opposite side is equal and parallel