Question

in the adjoing rohmbus ABCD,Angle BAC=35° find
angle ABO,
angle BCD,
angle CDA​

Answer 1

Answer:

.

Step-by-step explanation:

diagonals are bisecting the angles.so, DAB=90*. opposite angles are equal so DCB=90*. CBA and CDA are opposite angles.take them as "X".

70+70+X+X=360*

2X=220

X=110.

Answer 2

∠ABO = 55°

∠BCD = 110°

∠CDA = 110°

Given :

Rhombus ABCD

Angle BAC= 35°

To find ;

∠ABO

∠BCD

∠CDA

Solution :

We know the properties of a rhombus .

In a rhombus opposite sides are parallel , all sides of a rhombus are equal in length .

Diagonals of a rhombus bisect each other at right angles.

In the given fig rhombus ABCD

Diagonals AC and BD bisect

∠DAB and ∠DCB

As we are given

m ∠BAC = 35°

Hence m∠DAB = 2∠BAC

There fore m∠DAB = 35 + 35 = 70°.

As we know opposite angles of rhombus are congruent so m∠DAB = m∠BCD

There fore m∠BCD = 70°.

Sum of four angles of a quadrilateral is 360 °

Let m∠ADC = m∠BCD = x °

Now

x + x + 70 + 70 = 360°

2x + 140 = 360

2x = 360 - 140

2x = 220

x= 110.

therefore m∠BCD = 110°

m∠CDA = 110°

In triangle AOB , m∠AOB = 90° ( diagonals of rhombus bisect each other at right angles)

m∠ABO + ∠AOB + ∠OAB = 180° ( sum of angles of a triangle is 180° )

m∠ABO = 180 - ( 35 + 90 )

= 180 - 125

= 55°

hence m∠ABO = 55°

Hence

m∠BCD = 110°

m∠CDA = 110°

m∠ABO = 55° .

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