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​

Answer 1

Answer:

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Answer 2

Answer:

Answers: angle cab = 35°

angle cab = 35° angle aob =110°

angle cab = 35° angle aob =110° angle acb = 55°

Step-by-step explanation:

Given figure shows a rectangle abcd.

AC is produced to E and hence it is a straight line. Therefore angle DCA + angle DCE =180° {linear pair}

It means angle DCA + 145° = 180°

Therefore, angle DCA = 35°............(1)

Now let's consider triangle AOB.

angle OAB =angle DCA because both are alternate angle.

As diagonals of a rectangle bisect each other and are equal, AO = BO.

Therefore, angle OAB = angle OBA

Hence, angle OBA = angle OAB = angle DCA= 35°

By angle sum,

angle AOB + angle OAB + angle OBA= 180°

It means angle AOB + 35° + 35° = 180°

Therefore, angle AOB = 70°............(2)

Angle BCD = 90° [every angle of a rectangle is a right angle]

Hence, angle DCA + angle ACB = 90°

35°+angle ACB =90°

Therefore angle ACB = 55°................(3)

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