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  • ABCD is a rhombus such that ACB = 40º. Then ADB is (A) 40º (B) 45º (C) 50º (D) 60º

ABCD is a rhombus such that \angle ACB = 40^{\circ}. Then \angle ADB is
(A) 40^{\circ}          (B) 45^{\circ}         (C) 50^{\circ}           (D)60^{\circ}

Answers (1)

Answer:    [C]  50^{\circ}

Solution.


Given : ABCD is a rhombus such that \angle ACB=40^{\circ}\Rightarrow \angle OCB=40^{\circ} 
To Find : \angle ADB
Since AD \parallel BC
\Rightarrow \angle AOC=\angle ACB=40^{\circ}            {alternate interior angles}
\angle AOD=90^{\circ}                                      {diagonals of a rhombus are perpendicular to each other}
We know that sum of angles of a triangle is  180^{\circ}.
In \triangle AOD,
\angle AOD+\angle OAD+\angle ADO=180^{\circ}
90^{\circ}+40^{\circ}+\angle ADO=180^{\circ}
\angle ADO=180^{\circ}-90^{\circ}-40^{\circ}
\angle ADO=50^{\circ}=\angle ADB
Hence Option C is correct.

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