Question

ABCD is a rhombus whose diagonals AC and BD intersect at O. If angleOAB:angleOBA = 3:2, find the measure of all the angles of ∆COD ​

Answer 1

Given :

• Ratio of Angles OAB and OBA = 3:2
• ABCD is a rhombus

To Find :

We have to find all the angles of Triangle COD .

Property of Rhombus to be applied:

We will apply the Property of Rhombus , relayed to its diagonal, I have also given it in my last answer in your Question . The property is :

• The Diagonals of a rhombus bisect each other at 90° .

Through this Property, we get that the value of Angle AOB = 90° .

Angle COD = 90°

Process :

We will First find the Angle AOB (Already found in the above part). Then we will use the angle sum property of triangle to find 3x and 3x . Thus we can Find the measure of all angles of Triangle COD

Solution :

As we can clearly see , AB and CD are parallel( Also a property of Rhombus), So we can say that :

Angle , OAB = Angle OCD

Because they are alternative angles

Angle OBA = Angle ODC

Because they are also Alternatives angles .

Hence, Angle OCD = 3x

Angle ODC = 2x

As from the Angle sum property of the triangle, we get that :

OCD+ODC+COD = 180°

or, 3x+2x+90° = 180°

or, 5x = 180°-90°

or, 5x = 90°

or, x = 90°/5

or, x = 18°

Hence, Angle OCD = 54°

Angle ODC = 36°

Angle COD = 90°

Hence, the above are the measures of All angles of Triangle COD .

Hope this answer Help you !

Have a great Learning!

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Answer 2

I dunno the explanation but, these are the answers :

OCD = 54°

DOC = 90°

ODC = 36°

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