Question 197170
You have a triangle made from of two sides of the rhombus AB and BC equal to 5cm and the diagonal AC equal to 8cm.


The Cosine Rule for triangle ABC states


AC^2 = AB^2 + BC^2 - 2 AB BC cos ABC


Therefore


64 = 25 + 25 - 50 cos ABC	


64 = 50 - 50 cos ABC


14 = - 50 cos ABC


-7/25 = cos ABC


Angle ABC = approx 106.3 degrees as is the opposite angle CDA of the rhombus.


The other two angles BCA and CAB of the triangle (isosceles) are equal and half the angle made in the corner by the rhombus since a triangle on one side of a diagonal of a rhombus is a mirror image of the triangle on the other side of the diagonal.


Alternatively you could say that since the opposite sides of a rhombus are parallel then adjacent angles add up to 180 degrees and opposite angles are equal.


Thus angles DAB and BCD are approximately 180 – 106.3 = 73.7 degrees.