Question

8.
In the given figure, O is the centre of the circle and
AB is the diameter of the circle. If difference of
angle COD and angle DBC is 25°, then find the measure of angle DEB
​

Answer 1

Answer:

In the figure given below, O is the center of the circle. Chord CD is parallel to diameter AB. If angle ABC = 35 degrees, how would you calculate the angle CED?

∠BCD = ∠ABC (alternate ∠s in AB//CD) . . . so ∠BCD = 35°

∠OCB = ∠OBC (base ∠s of isosceles ∆OCB) . . . so ∠OCB = 35°

∠OCD = ∠OBC + ∠BCD = 35° + 35° . . . so ∠OCD= 70°

∠ODC = ∠OCD (base ∠s of isosceles ∆OCD) . . . so ∠ODC = 70°

∠COD = 180° - ∠OCD - ∠ODC (sum of interior ∠s of ∆OCD) = 180° - 70° - 70° = 40° . . . so ∠COD = 40°

Finally, ∠CED = ∠COD2 (∠ at circumference = half ∠ at centre)

So ∠CED = 40°2=20°

Step-by-step explanation:

hope it helps

Answer 2

Given :  O is the centre of the circle and AB is the diameter of the circle.  difference of ∠COD and ∠DBC is 25°

To Find : the measure of ∠DEB.

Solution:

∠COD = 2∠DBC   angle be same arc CD at center and remaining circle  

∠COD = ∠DBC + 25°

=>  2∠DBC  = ∠DBC + 25°

=> ∠DBC  = 25°

∠ADB = 90°   as AB is diameter  

∠CDB = 180° - ∠ADB   ( linear pair)

=> ∠CDB =90°

∠DBC + ∠CDB + ∠DEB. = 180°  ( sum of angles of triangle)

=> 25° + 90° + ∠DEB. = 180°

=>  ∠DEB. = 65°

measure of ∠DEB.  = 65°

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