Given: ABCD-rhombus AC and BD-diagonals Angle 1 is less than angle 2 by 30 ° Find: angles

Let us denote by x the value of the smaller angle, then the value of the larger angle is (x + 30˚). The sum of all the angles of any quadrangle, including the rhombus, is 360˚. Also, the opposite angles of the rhombus are respectively equal to each other.
Let’s compose the equation for the sum of all the angles of the rhombus:
x + (x + 30˚) + x + (x + 30˚) = 360˚;
x + x + 30˚ + x + x + 30˚ = 360˚;
4x + 60˚ = 360˚;
4x = 360˚ – 60˚;
4x = 300˚;
x = 300˚: 4;
x = 75˚.
The two smaller angles are equal to 75˚. Let’s find two big corners:
75˚ + 30˚ = 105˚.
ANSWER: The angles of the rhombus are as follows: 75˚, 75˚, 105˚, 105˚.




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