Question

in the given figure ABCD is a rhombus and triangle EDC is equilateral. If angle BAD is equal to 78 degree, calculate:
a. Angle CBE
b. Angle DBE​

Answer 1

∠CBE = 21° and ∠DBE  = 30°  

Step-by-step explanation:

• Given that

        ABCD is a rhombus. So

        AB =BC = CD = DA

        Opposite angle are equal in rhombus.

        ∠DAB = ∠DCB = 78°              ...1)

• ∠DAB + ∠ABC = 180°   (Sum of adjacent angle is 180°)

        78° + ∠ABC = 180°

        ∠ABC = 180° -78° =102°

• As we know that diagonal of rhombus bisect the angle of vertices B.

        So

        $\mathbf{\angle DBC=\frac{\angle ABC}{2}=\frac{102^{\circ}}{2}=51^{\circ}}$       ...2)

• Now in equilateral triangle ΔECD

        We know that all angle of equilateral triangle is 60°

         ∠EDC =∠DCE =∠CED = 60°          ...3)

• Now consider triangle ΔEBC, where EC = CB

        So it is an isosceles triangle.

        ∠CEB = ∠ CBE    (∵angle opposite to equal side are equal)

• Now using triangle angle property in  ΔEBC

        ∠CEB +∠ CBE + ∠ ECB = 180°

        ∠ CBE +∠ CBE + ∠ DCE+ ∠ DCB = 180°      ..4)

         From equation 1) and equation 3)

         2∠CBE + 60° +78° =180°

         On solving, we get

         ∠CBE = 21°      Answer              ...5)

• Now come to equation 2)

        ∠DBC = 51°

        Which can be written as

        ∠DBE + ∠CBE =51°

        ∠DBE + 21°=51°

        ∠DBE  = 51° -21° =30°      Answer

Answer 2

Answer:

∠CBE = 21° and ∠DBE = 30°

Step-by-step explanation:

Given that

ABCD is a rhombus. So

AB =BC = CD = DA

Opposite angle are equal in rhombus.

∠DAB = ∠DCB = 78° ...1)

∠DAB + ∠ABC = 180° (Sum of adjacent angle is 180°)

78° + ∠ABC = 180°

∠ABC = 180° -78° =102°

As we know that diagonal of rhombus bisect the angle of vertices B.

So

\mathbf{\angle DBC=\frac{\angle ABC}{2}=\frac{102^{\circ}}{2}=51^{\circ}}∠DBC=

2

∠ABC

=

2

102

∘

=51

∘

...2)

Now in equilateral triangle ΔECD

We know that all angle of equilateral triangle is 60°

∠EDC =∠DCE =∠CED = 60° ...3)

Now consider triangle ΔEBC, where EC = CB

So it is an isosceles triangle.

∠CEB = ∠ CBE (∵angle opposite to equal side are equal)

Now using triangle angle property in ΔEBC

∠CEB +∠ CBE + ∠ ECB = 180°

∠ CBE +∠ CBE + ∠ DCE+ ∠ DCB = 180° ..4)

From equation 1) and equation 3)

2∠CBE + 60° +78° =180°

On solving, we get

∠CBE = 21° Answer ...5)

Now come to equation 2)

∠DBC = 51°

Which can be written as

∠DBE + ∠CBE =51°

∠DBE + 21°=51°

∠DBE = 51° -21° =30° Answer