Question

In the given figure , ABCD is a rectangle, whose diagonals intersects at '0'. Diagonals AC is produced to E and angle DCE =145°
Find angle CAB ANGLE AOB ANGLE ACB

Answer 1

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Answer 2

Answer:

Step-by-step explanation:

in the given figure, ABCD is a rectangle, whose diagonals intersect at O.

Consider the attached figure while going through the following steps.

Given,

ABCD is a rectangle, whose diagonals intersect at O.

Diagonal AC is produced to E

∠  DCE = 145°

∠ DCE + ∠ DCO = 180°   (form  a straight line)

145°  + ∠ DCO = 180°  

∠ DCO = 180° - 145°  = 35°

∠ DCO = ∠ OAB   (alternate angles)

∠ OAB = 35°

∠ OAB = ∠ OBA   (angles opposite to equal sides are equal)

∠ OBA = 35°

∠ OBA = ∠ ODC  (alternate angles)

∠ ODC = 35°

∠ CAB = 35°

In Δ ODC,

∠ ODC + ∠ DCO + ∠ COD = 180°

35° + 35° + ∠ COD = 180°

70° + ∠ COD = 180°

∠ COD = 180° - 70° = 110°

∠ COD  = ∠ AOB =  110°  (vertically opposite angles)

In Δ ABC,

∠ ABC + ∠ ACB + ∠ BAC = 180°

90°  + ∠ ACB + 35° = 180°

∠ ACB = 180° - 90° - 35°

∠ ACB = 55°