Question

In triangle OPA, OA=8 and angle POA=30 degree and angle OAP=90 degree then find
OP+AP=​

Answer 1

Answer:

ANSWER

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- ∠POB=?

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

o

−90

o

−35

o

=55

o

.

Ans- Option B...

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