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.
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.
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