Question

2) In the figure PA is the tangent to the circle with
centre O. If LAPO=35° then LAOP is
A) 55°
B) 65°
C) 75°
D) 45°​

Answer 1

Answer:

a)55°

Step-by-step explanation:

one of the angle is 90° (angle between radii and tangent)

angle AOP=35°

35°+90°+AOP=180° (angle sum property)

125°+AOP= 180°

angle AOP=180-35= 55°

Answer 2

Step-by-step explanation:

Given- O is the centre of a circle to which PA&PB are two tangents drawn from a point P at A&B respectively. ∠APO=35

o

.

To find out- ∠AOP=?

Solution- ∠OAP=90

o

=∠OBP since the radius through the point of contact of a tangent to a circle is perpendicular to the tangent. Also PA=PB since the lengths of the tangents, drawn from a point to a circle, are equal.

So, between ΔPOB & ΔPOA, we have

PA=PB,

PO common,

∠OAP=∠OBP.

thereforeΔPOB≅ΔPOA⟹∠BPO=∠APO=35

o

.(by SAS test) .

So, in ΔPOB, we have ∠POB=180°

−90°

−35°

=55°

.

Ans- Option A