# Find the smaller diagonal of a rhombus whose sides are 49 and the acute angle is 60 °.

Let ABCD be a rhombus, AC and BD diagonals (AC> BD), angle A = angle C = 60 degrees. The diagonals intersect at point O.

Consider the triangle AOB: angle BOA = 90 degrees (since the diagonals of the rhombus intersect at an angle of 90 degrees), AB is the hypotenuse (since it lies opposite an angle of 90 degrees), angle ABO = angle B / 2 (since the diagonals of the rhombus are bisectors of the angles ).

By the sum theorem, the catch of a quadrangle:

angle A + angle B + angle C + angle D = 360 degrees;

60 + angle B + 60 + angle D = 360 (angles B and D are equal, we denote them as x);

2x = 360 – 120;

2x = 240;

x = 240/2;

x = 120.

Angle B = 120 degrees.

Then:

angle ABO = 120/2 = 60 degrees.

By the theorem on the sum of the angles of a triangle:

OAB angle + ABO angle + BOA angle = 180 degrees;

angle ОАВ + 60 + 90 = 180;

angle ОАВ = 180 – 150;

angle ОАВ = 30 degrees.

BО lies opposite an angle equal to 30 degrees, therefore:

BО = AB / 2;

BO = 49/2 = 24.5.

The diagonals are halved by the intersection point, so the smaller diagonal BD is equal to:

BD = BO + OD = 2 * BO = 2 * 24.5 = 49.

Answer: BD = 49.