Question

5. In a rectangle ABCD, <AOB = 70°. Find <ODC and <0AB

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Answer 1

Answer is <ODC = <OAB = 55°

Step by step explanation :-

We know that diagonals of rectangle bisect each other and equal.

AB = CD

1/2AB = 1/2CD

OA = OB

Therefore ∆OAB is isocellus triangle.

In ∆OAB,

<AOB + <OAB + <OBA = 180°

70° + 2( <OAB ) =180°

$\color{Red}{angleOAB \:=\:55°}$

<ODC = <OAB

$\color{Red}{angleOBC \:=\:55°}$

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Answer 2

The measure of ∠OAB and ∠ODC is 55°

GIVEN

ABCD is a rectangle

∠AOB = 70°

TO FIND

∠ODC and ∠OAB

SOLUTION

We can simply solve the above problem as follows;

We know that

ABCD is a rectangle

Therefore,

AB = CD (Opposite sides of a rectangle are equa)

We know that,

DB and AC are diagonals of the rectangle which divides the vertices of the rectangle in half.

Therefore,

OA = OB

Now,

ΔOAB is an Isosceles triangle.

Therefore,

∠OAB = ∠OBA (Angles of equal sides are equal)

We know that,

∠OAB + ∠OBA + ∠AOB = 180° (Sum of interior angle of triangle)

Therefore,

70 + 2∠OAB = 180°

2∠OAB = 180-70 = 110°

∠OAB = 110/2 = 55°

We know that,

∠BAC = ∠ABC (Angles of rectangle are equal)

Therefore,

∠OAB = ∠ODC = 55°

Hence, The measure of ∠OAB and ∠ODC is 55°

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