Question

AB is the diameter of the circle centre
at 0. A tangent PQ drawn at Pintersect
the subtended part of AB at Q. If ZAPQ
= 25°, find ZPQO​

Answer 1

Answer:

Given, PT is a tangent

∠PBA=300

Now,

∠BPA=900        (angle in semicircle)

∠OPT=900        (angle between the radius and tangent)

Also in ΔBPA

∠PAB=1800−(∠300+∠BPA)

              =1800−300−900

              =600⟶(1)

Now look into the ΔOPA,

OP=OA (radius of circle)

⇒∠OAP=∠OPA=600        (∵∠OAP=∠BAP=600)

∴   ∠APT=∠OPT−∠OPA=900−600=300

∴   ∠PAT=1800−(600)=1200          (supplementary angle)

In ΔPAT,

∠PAT+∠PTA+∠APT=1800

⇒1200+∠PTA+30