Question

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
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Answer 1

Answer:

angle CAB is 35

angle AOB is 110

angle acb is 55

Answer 2

Given:- ABCD is a rectangle

Angle DCE = 145°

To Find:- Angle CAB

Angle AOB

Angle ACB

Solution:-

- (i) ∠ECD + ∠OCD =180°......(AE is a straight line)

145° + ∠OCD = 180°

∠OCD = 180°- 145°

∠OCD = 35°

- Now, ∠OCD = ∠OAB or ∠CAB = 35°

( alternate angles)

- (ii) Diagnols of a rectangle are equal and bisect each other at O.

AO = BO

∠OAB = ∠OBA

- Now, in ΔAOB

∠OAB + ∠OBA + ∠AOB = 180°...(angle sum property of triangle)

35°+ 35° + ∠AOB = 180°

∠AOB =180° - 70°

∠AOB = 110°

- (iii) As we know that the reactangle all four angles arw of 90°.

So, ∠ACD + ∠ACB =90°

35° + ∠ACB = 90°

∠ACB = 90° - 35°

∠ ACB = 55°

- *So, The angles Angle CAB, AOB, ACB are 35°, 110° ,55°*