Math, asked by liyutsazirange1674, 11 months ago

In the given figure, ABCD is a rectangle, whose diagonals intersect at O. Diagonal AC is produced to E and Angle DCE = 145°. Find : angle Cab and Angle AOB and Angle ACB

Answers

Answered by shatrughnthakur1969
4

Answer:

Step-by-step explanation:

Send the pic of figure

Answered by AditiHegde
5

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°

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