Question

In the figure pq is the tangent to
the circle with centre 0 and op is.
its radius and angle OQP =40° then angleOPQ=?and anglePOQ=?​

Answer 1

Answer:

60 degree

Step-by-step explanation:

beacuse ...

Answer 2

$\large{\underline{\underline{\bf{Given:-}}}}$

• PQ is a tangent
• OP is a radius
• ∠OQP = 40°

$\large{\underline{\underline{\bf{To Find:-}}}}$

• ∠OPQ = ?
• ∠POQ = ?

$\large{\underline{\underline{\bf{Solution:-}}}}$

⇢ In ∆OPQ ( Tangent is always perpendicular to the radius.)

⇢ ∠P = 90°

In ∆OPQ (by Angle Sum Property of ∆ )

⇢ ∠OPQ + ∠OQP + ∠POQ = 180°

⇢ 90° + 40° + ∠POQ = 180°

⇢ 130° + ∠POQ = 180°

⇢ ∠POQ = 180° - 130°

⇢ ∠POQ = 50°

$\fbox{∠OPQ = 90°, \: ∠POQ = 50°}$