Question

In the figure above, AB and AC are the tangents to the circle at B and C respectively. IfO is the centre of the

circle, OB=8 cm and AB = 16 cm, then find the area of the quadrilateral OCAB.​

Answer 1

Answer:

$\huge\bold \blue{answer}$⤵️

OC⊥DA ∴ ∠OCD=90∘

In △OCD,∠ODC=180−∠OCD−∠DOC

∠ODC=180∘−(90∘+65∘)

∠ODC=180∘−155∘=25∘

Also OB⊥AB, ∴ ∠OBA=90∘

In △DBA,∠DAB=180∘−(∠BDA+∠DBA)

   =180∘−(25∘+90∘)

   =180∘−115∘=65∘

∴ ∠OAB=21×∠CAB(or ∠DAB)=21×65∘=32.5∘

(∴ Tangents are equally inclined to the line joining the external pt. and the centre of the circle)

In △OBA,X=180−∠OAB−∠OBA=180∘−(90∘+32.

$\huge\mathfrak \blue{hope you like }$