Question

In the figure below, O is the centre of the circle and P, Q are points on the circle. Compute <P and <Q​

Answer 1

Answer:

Here, O is the centre of circle.

PQ and PT are tangents to the circle from a point P

R is any point on the circle. RT and RQ are joined.

∠TPQ=70∘

Now,

Join TO and QO

∠TOQ=180∘−70∘=110∘

Here, OQ and OT are perpendicular on QP and TP.

∠TOQ is on the centre and ∠TRQ is on the rest part.

∠TRQ=1/2∠TOQ=1/2(110∘)=55∘

Therefore , ∠TRQ=55 degrees

Step-by-step explanation:

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Answer 2

answer:

∠POQ=120°

step-by-step explanation:

We know, that radius is perpendicular to a tangent .

∴ ∠OPR=90°

⇒ ∠OPQ+∠QPR=90°

⇒ ∠OPQ+60°=90°

⇒ ∠OPQ=90° −60°

⇒ ∠OPQ=30°

⇒ OP=OQ [ Radii of a circle ]

⇒ ∠OPQ=∠OQP=30°

[ Base angles of equal sides are also equal ]

In △POQ,

⇒ ∠OQP+∠POQ+∠OPQ=180°

[ Sum of angles of a triangle is 180° ]

⇒ 30°+∠POQ+30° =180°

⇒ ∠POQ+60°=180°

⇒ ∠POQ=120°