Question

In Figure 4, ABCD is a rectangle. Its diagonals AC and BD intersect each
other at O. AC is produced to E such that ECD is 140°. Find the measure of the
angles of triangle AOB.
​

Answer 1

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Answer 2

Measure of all the angles in ∆AOB;

/_AOB = 80°

/_OBA = 40°

/_OAC = 60°

GIVEN

ABCD is a rectangle.

/_ ECD = 140°

Diagonal = AC & BD

TO FIND

Measure of angles of triangle AOB

SOLUTION

We can simply solve the above problem as follows -

/_ ECD = 140°

We know,

/_ECD + /_DCO = 180°

( sum of linear pair of angles in 180° )

/_DCO = 180 - 140 = 40°

We know,

Diagonals bisects the angle in equal halves.

/_ BCO = /_DCO = 40°

NOW In ∆ BOC ,( isoceles triangle )

OB = OC.

/_ OBC = /_OCB = 40°

/_BOC + /_OCB + /_OBC = 180°

(sum of all the angles in a triangle is 180°)

/_BOC = 180°- (40° + 40° ) = 100°

NOW ,

/_ BOC + /_AOB = 180°

(sum of linear pair of angles in 180°)

/_ AOB = 180 - 100 = 80°

Since the diagonal BD bisects the /_ ABD into two equal halves .

/_ABO = /_OBC = 40°

In ∆AOB

/_ AOB+ /_OBA + /_ BAO = 180°

80 + 40 + /_OAC = 180°

/_OAC = 180- 120 = 60°

Hence, in ∆AOB;

/_AOB = 80°

/_OBA = 40°

/_OAC = 60°

# spj2

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