Question

In the following figure find ZCOD, given LAOB = 125°
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Answer 1

Answer:

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Step-by-step explanation:

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Answer 2

Answer:

Given−

AB,BC,CD & AD are tangents drawn from the points

A,B,C  to the circle with centre O.

OA,OB,OC & OD are the lines joining O with

A,B,C & D respectively.

∠AOB = 125 °

 To find out−

∠DOC=?

Step by step explanation :

AB & AD are tangents drawn from the point

A to the circle with centre O.

OA is the line joining O with A.

∴∠OAB = ∠OAD   .....(i)

Similarly it can be shown that

∠OBA = ∠OBC    ........(ii),

∴From (i)  & (ii) ∠OAB+∠OBA=∠OAD+∠OBC  .....(iii)

Similarly  ∠ODA = ∠ODC & ∠OCB = ∠OCD

∴∠ODA+∠OCB  =  ∠ODC+∠OCD

i.e ∠ODC+∠OCD =  1/2(∠ADC+BCD)   ........(iv)

Again∠AOB=125°  ⟹∠OAB+∠OBA = 180°−125° = 55°

                                                                      (sum of the ∠s  of a  Δ=180°)

and ∠AOB = 125°⟹reflex∠AOB = −360°−125° = 235°   ...(v)

Now  AOBCD  is a pentagon.

∴reflex ∠AOB+(∠OAD+∠OBC)+(∠ADC+BCD) = 540°

                                                                  (sum of the angles of a pentagon)

⟹235°  +55°+2(∠ODC+∠OCD) = 540°     (using iii,iv,v)

⟹∠ODC +∠OCD =125°

⟹∠COD=180° −125° = 55°